I don't know if the first part is made simpler by calculus, but as I don't know much about geometry or rules for similar triangles it's the only way I managed it. As a bonus you don't have to think about anything, you just write out an equation for the area of a triangle twice and rearrange it. Also, I think I used algebra not calculus. For simplicity I did my calculations using height times width as if there were two triangles forming a rectangle so I don't have to divide everything by 2 at each step. short side of blue rectangle = a, long side = b, so ab = area of blue rectangle 1. Multiply the length and width of the triangle to get an area double the size of the triangle. (a + 4) * (b + 3) 2. Area of the blue rectangle + area of the top triangle + area of right triangle. All of the areas are doubled to match what I did with the first equation 2ab + 4b + 3a 3. Equate the two formulas. They both calculate an area twice the size of the triangle. (a + 4) * (b + 3) = 2ab + 4b + 3a 4. expand the first equation ab + 3a + 4b +12 = 2ab + 4b + 3a 5. subtract 3a, 4b, and ab from both sides 12 = ab
For part 2, although using calculus to find the minimum of the area works for those who know calculus, you can actually visualise the minimum. In the big total triangle, cut off the little triangle to the right of the rectangle (which has sides y and 3), and rotate it 180 degrees about its upper-left vertex, and stick it back on so that it looks like a dormer window built onto the roof of the other piece (this is easier to draw than to describe in words!). Then the area of the total is equal to the bottom rectangle plus another identical rectangle above it, plus the remaining bit of the top triangle that sticks above the line. The two rectangles have total area 24, so the total area is 24 plus whatever surplus triangle sticks out. So the way to minimise the total area is to pick the configuration in which the surplus triangle vanishes - namely when the small triangle you rotated exactly matches the top triangle and leaves no surplus. (Maybe one of the 633 other commentators already said this?) Yours JOHN HARTLEY.
I would just like to add on that, the excess part is not always from the top triangle. When y > 4, the excess part is actually from the right triangle that you cut and pasted, and the excess area is actually in the shape of a trapezium if you try to visualise it. You can pause the video at around 3:08, at the part where h > 8, because when [h > 8]= [y+4 > 8] = [y > 4]. If you do your cut and paste on the triangle when h>8, the excess area is in the shape of a trapezium from the right triangle.
Sorry,I know I am a little late but I did not understand the method you stated and can you tell me is there a way to solve these kind of questions which requires maximum/minimum areas,without using calculus.
@@nareshkumar-uo7nz The only reliable and easy way to do it is using Calculus; without it, you'll have to do a bunch of Geometry grunt-work. I personally only know the Calculus way to do it, as that's the easiest way.
One more thing I like about this channel is that it leaves the viewers the other correct ways of solving the given problems and arriving at the right answer, which is a good learning opportunity.
For the first part, if you write the area of the large triangle as (x+3)(y+4)/2 and set it equal to the sum of the areas of the small triangles and the rectangle, 2x+3y/2+xy, then the equation simplifies to xy=12.
I used the equation for a slope: y = ax + b. Since the slope of the hypotenuse is the same for both triangles, so 4= ax, y= a3; a = 4/x = y/3 and then y = 12/x.
Oh... this brings back memories, I remember this coming up when I applied for my first job as a truck driver, lucky I remembered my calculus. Questions like this are probably the reason there is a shortage of truck drivers today!
I cannot imagine asking this question for someone who wants to be a truck driver. Sure, you need some basic geometry, but come on lol ask questions like "what to do during a jackknife" or "how should a load be distributed for maximum trailer stability"
What confuses me is, who are you going to ask this question? What job are they interviewing for? Maybe if you're looking for an engineer right out of college?
You can use the inequality of means to solve the second part without calculus: We know that x.y =12 and we want to minimize (4x+3y)/2 +12. We can substitute a=4x and b=3y, giving us a.b=144 We now use that (a+b)/2 >= sqrt(a.b) and we find that (a+b)/2>=12, therefore, the minimum will happen at the equality. Inputing the value in the area equation gives us 24 as the minimum area.
We broke math here is why you just proved that the minimum area is 24 but when you realize that with4/x=y/3 than you can get 12x=y and since xy=12 and since the only pairs for x and y that fit that equation are x=1 y=12 than the area of the triangle must be 32
(Assuming the answer would be an integer) You can also take ordered pairs on ways of writing x.y=12, they are (12,1) (6,2) (4,3) (3,4) (2,6) (1,12), now if you take the area of triangle using these values, you'll find (4,3) to givr you the lowest and that is exactly 24 sq untis.
But whats with x=y=0? Then you've got all the conditions, no rectangle and only one triangle with an area of 6 (i think the formula is only right assuming there is a rectangle with a size bigger then x=0, cause x is in the denominator)
First part was easy. For the second part I tried taking the min of the function 2y+(3/2)x+12 directly. Took a while to notice I needed to use xy=12 and solve it by minimising 2y+18/y+12. In the end finally solved it 😅
Rather than use a derivative, I guessed that the minimum would occur when the two opposing terms 2y and 18/y are equal. It could have been wrong, but it wasn't.
Yes I got first part in 5 seconds as I assumed it was a 3, 4, 5 triangle and ignored the graphical lengths as a distraction. Second part was not so easy and guessed at 12 again. I was only 100% wrong lol
@@w1swh1 You calculated the area for just one of the infinite number of triangles without checking any of the other possibilities. So you got the first part right by luck rather than by proof.
Basically the first derivative is for finding the slope. And we set that slope equal to zero to find the max peak or min crate values of that graph/function
I used to be able to get max and min without using calculus but have forgotten how to do it.I'd have to look up my notes on this which are around 55 to 60 years old.
I think this sum can be solved in easy way without applying calculus I am currently in 11th std and don't know about calculus but have heard about it I cracked the logic by using Basic proportionality theorem and theorem on ratios of similar triangles
@@ivornworrell thats a triangle that satisfies the condition with the 3 and the 4. In other words, any blue area thats 12 will satisfy. in 6x9, blue would be 6x2= 12
You can solve this by pure geometry. Imagine the hypotenuse of the biggest triangle balancing on the corner of the rectangle. If the two smaller triangles are of different sizes we can rotate the hypotenuse to make the larger triangle smaller and the smaller triangle larger. Thinking in terms of limits it's easy to see that a very small change in angle will make the larger triangle decrease by more than the smaller triangle increases. So the two smaller triangles must be congruent at the minimum. Pick one of those triangles up, flip it around and put it against the other small triangle and we get a bigger rectangle of total area 2xy. So the area of the big triangle is 2x12 = 24. Job done. No algebra or calculus needed.
2nd part has an easy solution . just think about for which triangle it is going to be minimum . yes , it is isosceles triangle . now in that case the white sub triangles(small triangle) are going to be same to each other and each of their area will be the half of the rectangle(in this case the rectangle is going to be a square).you can think of folding the white part over the blue area. thus you can understand, "sum of the 2 small(white ) triangle = area of blue square " so the area of the triangle will be 2*(area of square)=2*12=24. simple, you don't need calculus
area of triangle = 1/2 * (4 + y) * (3 + x) = 12 + 1/2(4x + 3y) use arithmetic geometric mean inequality on 4x + 3y you get 4x+3y >= 2 * sqrt(4x* 3y) where equality holds when 4x = 3y rest is easy
Funny, I find mathematics, especially geometry very fascinating and solved the first part of the problem in a minute, then got halfway through the second half and got stuck, so I decided to watch the video. From the word "derivative" onward, I understand exactly zero percent of what's being said and done.
It's just that if you differentiate a function and set it equal to 0 then you will get a value of x such that f(x) is either maximum or minimum. To decide it's maximum or minimum you see the double differentiation of f(x). Learning Derivatives is fairly easy you can learn it in a week easily.
Yeah, his way of solving the problem here is _way_ overcomplicated. Would be much easier to use the arithmetic-geometric-mean inequality to minimize f.
@@ribozyme2899 but once you know calculus, there's little reason not to use it everywhere. It's just so damn powerful, and derivatives like this are so simple. Why wouldn't you use calculus whenever the question is "maximize/minimize the value of this function"
We can also minimise the area using AM ≥ GM Area(∆) = 4x/2 + 3y/2 + 12, Let (4x/2 + 3y/2) = k So area(∆) = k + 12 ------- (1) Looking at (4x/2 + 3y/2) , and by AM ≥ GM, we can say, k/2 ≥ [(4x/2)(3y/2)]^½ on putting xy=12 and solving we get, k ≥ 12 Putting this in (1), we get Area(∆) ≥ 24 Therefore, min area(∆) = 24
I saw another similar problem on the UK IMC a few years ago, another nice solution! Solving questions like these inspire me to share my own maths tricks, like you do, as well!
@@arshtewari2833 I checked again - it was a similar diagram, one of the 20-25 questions. In terms of other ways to prepare for the paper, I'd recommend checking out the 'Intermediate Problems' book on the UKMT website, it's very good practice. I achieved a Gold in the challenge and qualified for the Hamilton Olympiad last year (achieved a Merit), so I'm gonna be making videos covering questions, how to prepare, the ideal thought process etc. Stay tuned - I'm sure it'll benefit you!
@@AliKhanMaths Wow this is so surreal because I also received a merit in the Hamilton last year! Thank you for recommending the book and I'll be sure to check it out. Good luck for next year and I hope we both qualify for the Maclaurin :)!
To minimize expression (4x+3y) , where xy=12, you can use AM-GM inequality - (a+ b)>= 2sqrt(ab), no need for derivative. So (4x+3y)>=2sqrt(4x*3y)=2*12=24, equality holds true when 4x=3y, i.e. 4x =3*12/x ; => x =3.
1) by similarity 2) we have xy=12 We get factors (1,12) (12,1) (4,3) (3,4) (6,2) (2,6) Here we get minimum value when x=y or the difference between x and y is minimum so in this case (3,4) or (4,3) are if we put (3,4) we get 24 and if we put (4,3) we get 24.5 For max we take (1,12) and (12,1) (1,12)=32 (12,1)=37.5 Minimum=(3,4)=24 Maximum=(12,1)=37.5 :) ✌️
For all my peeps who thought "Why not just set both x and y to zero?" Remember that there is the requirement that there be an inscribed rectangle touching the hypotenuse. For that to happen, x becomes infinitely large as y goes to zero and vice versa. If both drop to zero, you no longer meet the requirements of the problem.
I got the answer to the second part instantly with geometric intuition, which works basically like this: First, note that the inner triangles are similar, and have the same area if and only if they are congruent. 2a. Now assume the triangles have different area. Rotate the smaller triangle 180 degrees about the top right corner of the rectangle. The two triangles form a rectangle congruent to the blue rectangle, plus some extra. This total area is 24 plus some positive area. 2b. Now assume the triangles are congruent. The same rotation now forms only the rectangle, with a total area of 24.
This reminds me of the electrical circuit where you want to maximize the power to a load, given a source impedance. The answer is that the load impedance must equal to the source impedance. This case reminds because the area of the triangles is equal to the area of the rectangle. The rectangle area is given & to minimize the total area, the area of the triangles must equal the rectangle's. A stretch, yeah, I'll admit.
I have to admit I would walk out of any interview for a job that required me to solve this inappropriate and useless skill that has no bearing on any problem in real life! With geometry algebra trigonometry calculus - all of it can go in the bin I'd rather work out the best way to dress a Pizza and then cook and eat it - satisfaction rating ***** mathematics minus *****
@@theprior46 What are you talking about? If this stuff is of no interest to you, then you are not applying for the types of jobs where such questions are appropriate.
@@theprior46 It probably would have been better if you just said it had no bearing on the job you were applying for. Assuming you could even know that. Saying that it has no bearing on any problem in real life is just ridiculous. Human beings wouldn't have come up with this stuff if there wasn't a reason for it out there somewhere.
I quickly saw the AM-GM inequality way to solve this just by labeling x,y for the rectangle, we have reduced the problem into finding the min value of (x+3)(y+4) with xy = 12 which can easily be done by using am-gm for 2 numbers.
But whats with x=y=0? Then you've got all the conditions, no rectangle and only one triangle with an area of 6 (i think the formula is only right assuming there is a rectangle with a size bigger then x=0, cause x is in the denominator)
@@neutronenstern. x=y=0 does not work. A key requirement is that the upper right corner of the rectangle lies along the hypotenuse. This means if you have y=0, i.e. the height of the rectangle is 0, then x, the length of the rectangle, must be the entire width of the triangle. But we know the width of the triangle is x+3. This means we need x+3=x, which is impossible, so y cannot be 0. The same applies for x=0. This means x>0 and y>0.
The AM GM inequality is a great alternative to the derivative technique for quadratic functions, but its proof depends on theory of quadratics; so I prefer to keep in practice by completing the square.
When you did your calculation you assumed x and y not zero. But x=y=0 also works with the triangle a 3 4 5 right triangle. I think that makes a better "tricky" solution.
@@hemph9772 Why can't they? We have no condition on possible solutions. If you make one equal to 0 and the other equal to the length of its side, for example x=0, y=4, you get that the rectangle becomes a segment but it's still a degenerate rectangle. Not sure about both variables being 0 as it wouldn't fit the drawing, the degenerate rectangle and the third side would have no point of contact. Edit: Nope, just a brainfart
@@martinovilla218 please draw this triangle where the degenerate rectangle (with vertical height = 0). The base of the rectangle and the base of the triangle would be the same line segment. This contradicts the definition of the problem whereby the base of the triangle must be 3 units longer than the base of the rectangle.
@@TPinesGold Yeah, you're right. I was thinking too hard about how a degenerate triangle could fit that I forgot how the problem was actually formulated
@@TPinesGold in the video, the problem is defined as the distance from the top of the rectangle to the top vertex of the triangle is 4 units, and the distance from the right of the rectangle to the right vertex is 3 units. A degenerate rectangle would fit this criteria, thus the area is 0 and the minimum triangle area is 6.
Great video! I especially enjoyed the solution to the first problem, although it would have been nice if you explained how you got the solution to the second problem instead of just handing the method down from above. Logic puzzles are your forte in this channel, so keep doing those!
For the one, I did it in two ways. One way was the similar triangles. The other way is to make the triangle into a square by adding another equal and opposite triangle, and then the question becomes quite easy as there are two triangles that are exactly the same as the other triangles, and then a rectangle with area 12 (4*3) which is the answer.
It's easy though. Solved it by assuming rectangle length and breadth as x and y. Used similarity property and then used differentiation to find minimum area.
There's easy one in second part, Just take two sides as, 'x+3' and '4+y or 4+12/x(since xy = 12)', and that's it you have a function for area, f(x) = (1/2)(x+3)(4+12/x),. and solve for it's derivative equal to zero🙂
For part 1, I dove in with looking at the two different ways of calculating the area of the large triangle. One way is (x+3)(y+4)/2, the other is 4x/2+3y/2+xy. Set those two equal to one another, and everything cancels out quickly leaving xy=12.
Another exciting question of the given problem would be to find the areas of two sub-triangles. I tried to get them, and I enjoyed the problem. Happy solving!
Yeah. This looked like a trick question, but I've been drinking. I saw the 3-4-5 relationship and just thought make the box zero. I think the problem needs more rules, or is it okay to ... drink ... what was I saying?
For the first part you can mirror the triangle along the hypotenuse. The part mirrored along the hypotenuse of 4*x will be equal and the part mirrored along the hypotenuse of 3*y will be equal, so the leftover part will be equal to the rectangle, which we now know the side lengths of, which will be 4 and 3 and that will give us an area of 12
4/x = y/3 gives xy=12. Then you can find the answer to the second question without calculus by using substitution values for the function f(x)=2x+18/x+12. f(3) gives the lowest product for the function.
I easily found the area of the right triangle by using the similar triangles method you used, but I used calculus to find the minimum. Let x be the length and y be the height of the triangle. We need to find the minimum area of (x+3)(y+4)/2 which works out to be (xy + 3y + 4x + 12)/2. Substitute xy for 12 and y for 12/x, we get (36/x + 4x + 24)/2 = 18/x + 2x + 12. Finding the derivative, it will be -18/x^2 + 2. Setting it equal to zero to find the minimum, we have a quadratic. -18/x^2+2=0, so -18+2x^2=0, so x = 3. y=12/3. Our original right triangle has a length of 6 and a height of 8, so it will have an area of 24.
I've been wanting to buy a new toaster but couldn't come to a decision as to what price I was willing to pay. But, thanks to this video, I can now figure out how much money I can spend on a new toaster. Thanks, this really helps.
The very top angle and the top angle of the smaller triangle are the same(angle "A"). So we can use trigonometry to find the width W of the rectangle as the opposite of the right angled triangle, as tanA=w/4>>w=4tanA. Since we know the smaller triangle has angle A too, we can find the length L of the rectangle since it is the base of the triangle, which is tanA=3/L>>L=3/tan A. The area of the rectangle is LW so 4tanA*3/tanA. Leaving the answer as 12
@schrodingers cat Since B is a right angle, and the center of the semicircle is on BC, it can only be tangent to AB if the tangency point is B. Then the triangle can be described with three parameters: y = BC - 20, z = AB, and x = AC - z. x is the length from C to the tangency point on AC, since the tangents from A are equal, and one tangent is AB = z. Now, using the pythagorean theorem twice gives x^2 = y(y+20) and (y+20)^2 = zx. Then z = 10sqrt(1+20/y). The area of ABC is z(y+20)/2, which can then be fully parametrized by y. Using calculus, the minimum is found at y = 10, and the value of the area is then 150sqrt(3). It should be noted that the resulting triangle is a 30-60-90 triangle, which indicates there is probably a more elegant solution :)
I have a simpler solution that doesn't require derivative or AM-GM So the area of the triangle is (x+3)(y+4)/2=(xy+4x+3y+12)/2 xy we already know, 12 and 2 is a constant so we can ignore those, so the area of the triangle is minimal when 4x+3y is minimal. because 4x and 3y is positive real number then (4x-3y)^2>=0 =>16x^2 - 24xy + 9y^2 >= 0 => 16x^2 + 24xy + 9y^2 >= 48xy => (4x + 3y)^2 >= 576 => 4x +3y >= 24 then apply to the function above then we have min(S)=(12 +24+12)/2=24
I think in 2nd question you can use AM - GM inequality to finding the minimum value of f(x) because all numbers are greater than 0. Then you can condition to "=" is occurred. If I don't true in some point, hope everyone can help me.
But whats with x=y=0? Then you've got all the conditions, no rectangle and only one triangle with an area of 6 (i think the formula is only right assuming there is a rectangle with a size bigger then x=0, cause x is in the denominator)
First part can be solved easily by taking tan(θ) of the upper triangle and lower triangle , from there you get the values of x = 3tan(θ) and y = 4/tan(θ), after multiplying x and y area will be 12.
First part can also be solved by using Pythagoras theorem on triangle on the bottom right since we know that if 1 side of triangle is 3 and 1 of the angles is 90° then length of hypotenuse has to be 5. From this we know that side y=4. Then we can easily figure out area of Rectangle.
I did part 1 completely different. Just draw a triangle against the given triangle, so that both triangles form a rectangle. Three of the points of the original triangle form the corner points of the new rectangle. Then continue lines x and y. Now it's easy to see that there are two equal triangles in the upper left section of the new rectangle, and two equal triangles in the lower right section of the new rectangle. This means that the two smaller rectangles must be equal in size too. The upper right rectangle is 4 x 3, so the lower left rectanble must also be 12.
Well, that's interesting. My immediate thought was that when I set x = 0 and y = 0 I would get the minimum solution with a right triangle with sides 3,4,5.
I thought there is no minimum size for the rectangle, so it could be reduced to zero, so surely the minimum area of the triangle is 1/2 x 4 x 3 = 6 units squared.
@@CrankyTim my very simple point to the was to the OP (not you) that x=0 & y=0 doesn't meet the basic criterion that the area of the rectangle = 12 (although you do get a triangle).
@@CrankyTim It's entirely unnecessary to look at angles to see x=0 & y=0 doesn't work, which is what the OP first thought of. Read what he wrote "My immediate thought was that when I set x = 0 and y = 0". You are overcomplicating things.
@@CrankyTim if you set x & y to 0, then you most certainly do get a 3:4:5 triangle, just not with one with a 0 x 0 rectangle which doesn't, of course, touch the hypotenuse.
Did it exact the same way. Nice Problem. Maybe there is an elegant way to solve the second part: As we know (do we?) the largest rectangle in a triangle has half the area of the triangle. So the minimum area of the triangle must be twice the rectangle.
Got this. But for part one I used a slightly different way without noticing the similar triangles (equating the area of the big triangle with that of its three component shapes): Let A be the area of the large triangle. Then 2A = (4+y)(3+x) = 2xy + 4x + 3y. The x and y disappear and leave xy = 12.
I think there are other answers that were neglected. The equation derived from the similar triangles is only valid if x or y are different from 0. But if you take x = 0 and y = 0, the rectangle is only one point with an area of 0 and therefore the minimum area is 6 (ie 4*3/2) Would you say that these answers are also acceptable ?
I also saw this solution, and would respectfully suggest that 6 is, indeed, the only correct answer to part two. Surely we should not assume that the illustration of the problem is drawn to scale, unless we are told so. That was my initial reaction, but now I'm not so sure. If, as is suggested in some of the comments further down, you cannot theoretically have a rectangle with side length 0, then I would have to agree that this solution is not valid. Can anyone give a definitive answer about the validity of a rectangle with side length 0? I cannot find an answer to this on the web, although I seem to remember being told (in a computer graphics class) that a rectangle which has one side of length 0 is a line, and a line of length of zero is a point. I may be misremembering, of course, but would appreciate any help.
No. Because we know that ONE of the angles of the rectangle is on the hypotenuse we cannot have all of the angles coincide with the right angle of the triangle.
The rectangle with side lengths 0, 0 cannot be inscribed in the triangle in the way shown in the video, since the triangle has positive side lengths (>3 and >4, respectively). Therefore, (0, 0) is not a solution at all. This geometric condition ensures that all side lengths are positive and the math presented in the video makes sense.
The two triangles are similar. Using x as the width of the rectangle and y as the height, the proportion for the triangles is 3/y=x/4. By cross multiplying, we get the area of the rectangle xy = 12.
Hello Presh, I've followed you since I was in school. Now I'm training to become a math teacher. I've always loved the way you present the problem and the animations you create. please could you let me know what software you use for your videos so I could make my own to help my students understand math better?
I, without reading the question properly, found the minimum area of the two small triangles pretty easily. In terms of angle B it is area = 4.5 * tan(B) + 8 / tan (B). Differentiate and you get the minimum at tan(B) = 4/3, easy after that to solve everything.
I found this one fairly easy, save that I just took the area of the triangle as (4+y)(3+x)/2 rather than as three separate areas (why do that). Then substitute in for y to get (4+12/x)(3+x)/2 and simplify to get area equals 12+2x+18/x. We might as well throw away the common factor of 2 and the constant as they play no part in finding a minimum, so take the first differential of x+9/x and get 1-9/x^2. Solve that for the value of 0 to get a min/max and the only positive root is 3, so an 8 x 6 right angle triangle. It's a general rule when looking for minimum/maximums you can strip out those common factors and fixed constants when differentiating the function. It's a short-cut, but then that's what interview questions are surely about. However, it might be wiser to do it in full for an examination answer. This is about the level of question that was in the calculus part of the "O" level examination in England when I took it back in the summer of 1966 aged 16. It's a bit basic for university entrance in my view, albeit in most of the UK university courses are specialist subjects so if you are going for a science, then the standard required for mathematics is a lot higher than, say, English Literature. The mathematics requirements for the latter will normally be fairly low.
Mark Levi's book describes a super useful way of solving problems like the second one almost instantly. The idea is to think about the triangle as filled with a vacuum and being squeezed by air pressure. Then the area of the triangle is proportional to its energy, so the minimal area configuration is always an equilibrium state. But if the corner of the rectangle doesn't touch the hypotenuse at its midpoint, then the hypotenuse will clearly feel a torque, and hence the state cannot be in equilibrium. From there the solution is immediate.
After getting f(X)=2x+18/x+12 we can use AM-GM inequality on 2x+18/x to get that (2x+18/x)/2 >=root(18*2)=6....Thus we can directly find the min of f(x) as 12+12=24
@@benvanrensburg4261 Just like checking at both sides of a value of a formula by +/- than the x value, check with tiny extension to the right and big upward and tiny upward and big to the right, fold over the rectangle, and "see" that there is a lot of leftovers... No leftovers when foldovers tips fall on right angle corner. * I don't know how to say this concisely in Mathian... :)
First part was easy, second part was interesting but can also be solved without differentiating, by using AM-GM: 2x+18/x>=2sqrt(2x*18/x) 2x+18/x>=2sqrt(36) 2x+18/x>=12 2x+18/x+12>=24 so the minimum is 24 very interesting problem though. edit: differentiating not integrating
@@shubhammishra8286 It uses the AM-GM inequality, that states that for any two positive real numbers A and B the inequality (A + B)/2 ≥ √(AB) holds, and the equality appears if and only if A equals B.
FOR the part two you can just apply arithmetic-geometric mean inequality it is easy to do！ surface = 1/2*(3x+4y)+12 and 1/2*(3x+4y) >= √12xy = 12 → surface >= 12+12 = 24~~
I do not know how to explain exactly, but I had an intuition that this would be the solution! Only when the two triangles overlap will there be a minimum area for the large triangle. This is the midpoint in their relationship. Any change here or there, (it will be symmetrical) will only increase the area of the large triangle.
With 3 and 4, obviously x=y=0! The restriction that x is not null comes way too late during the proof, it is not part of the initial question, so it should not be taken into account. 3, 4, 5 is the easy right triangle...
Never thought that calculus could be used for these types of geometry problems. A great tool, I remember something, one man's tool is another man's weapon
Let w and h be the width and height of the blue rectangle respectively, since the two triangles are similar, w/4 = 3/h => w*h = area of the blue rectangle = 12 To derive the minimum triangle that satisfies the condition, we use derivative on the minimum area of the sum of the two small triangles. The area of the triangles = 3*h/2 + 4*w/2 = 3*h/2 + 4(12/h)/2 = 3(h+16/h)/2 d(3/2[h + 16/h])/dh = 0 => 3/2[1 - 16/h^2] = 0 => h^2 = 16 => h = 4 => w = 12/h = 3 So the minimum large triangle that satisfies the condition has width = 3+3 = 6 and height = 4 + 4 = 8
the question is very easy, you can write x=3tanB, y=4cotB. then area of rectangle is x.y= 3tanB.4cotB=12. and are of triangle= (1/2)(4+3tanB).(3+4cotB). diffentiating and equating to zero will give us tanB=4/3,-4/3. rejecting -4/3. and substituting value of tanB=4/3 in area will give us write answer.
So we know the total area is the sum of the 3 areas. With x and y being found initially from part 1. The area of the square is 12, with question is, if we vary x and or y, such that they still equal 12, what is the smallest total area possible? The total area is the summation of the 2 triangles and the area of the square. We substitute 12=xy or y=12/x into the total area so now the total area is only a function of x. Which is fine since y is dependent on x because of 12=yx, so now we look into minimizing the function based on x alone. Taking the first derivative gets the equation shown, by setting it equal to zero, it's where the tangent line is zero. Assuming you have no calculus experience this is helpful for finding the minimum or the maximum. Imagine a circle, a tangent line will have a slope of zero, at the top of the circle or the bottom of the circle. We solved for this and get x = 3, x could be -3 because of the square root. But areas aren't negative, so only 3. Taking the second derivative is how we determine if this tangent is either a maximum or a minimum. Is the lineon the top or bottom of the circle? Imagine you have a circle cut in half left to right. When the circle is open facing down, like an upside down U. Then the tangent line will represent the maximum. The opposite is true. If the tangent is on the semi circle facing up, it's the minimum. The second derivative tells us if it's the top circle or the bottom circle. If by plugging x = 3 into the second derivative the value is negative, < 0, then we have the maximum value. But since the value was positive, > 0, it's the bottom of the half circle and therefore the minimum. So now we know that x=3 is the minimum x value. Plugging x back into 12 = xy, we get y = 4. And plugging 3 and 4 into x and y for the total area gets us 24. The minimum. Hope that helps.
I got this right away by remembering that 3-4-5 makes a right triangle, and that the minimum total area is the one that makes the rectangle closest to a square, so with the given lengths, a 3 × 4 rectangle minimizes the total area, making the case of 2 congruent 3-4-5 right sub-triangles the minimum ... done.
"What is the minimum area for a triangle that satisfies these conditions." This part of the question is ambiguous. There are three triangles, the large one and the two smaller ones. Either the question isn't precise enough, or the solution is incomplete. To answer this question as it's presented would require three solutions, one for the overall triangle (which is presented), one for the upper triangle and one for the lower right triangle. An interesting question though. :)
Yeah it should have been worded "minimum area for the triangle that contains the rectangle" which I assumed was the question, but it could have been made more completely clear.
I just used Pythagorean triple, base = 6,side length = 8 and hypotenuse = 10,so 4x3=12 for the rectangle and 8x6/2= 24 for minimum area of the whole triangle
one should note that this works for arbitrary values a,b instead of 3 and 4. we always get xy=ab and its minimized for x=a and y=b (or vice versa, depending how you labeled it)
After creating the equations you could have used ARITHMETIC MEAN >= GEOMETRIC MEAN. => (2x + 18/x)÷2 >= √[(2x)*(18/x)] which gives => ( 2x + 18/x) >= 12 For min. we will take "12" and hence min. area will be 12+12 = 24.
I understand the second part up to the derivitive part (im 12 btw). I know how to take derivitives, but can somebody explain why he took the derivitive and stuff like that.
We took the derivative of the equation of the area of the triangle in order to find its rate of change. So as x changes the area changes by the derivative. So when the derivative equation is equal to zero we know that the original equation for the area is at a maximum or minimum. Since the area equation is positive(the graph of it looks like a bowl) we know we have the minimum at x=3.
Cannot really teach calculus in a single comment. But you can look up that the derivative gives you the slope of the curve, and when the slope is 0 then you have either a maximum or a minimum (or a saddle point).
@@ayushrudra8600 take any function Use a graphing calculator and take single derivative and then a double derivative. U will understand Or u can use the same function as used in the problem and u will see how double derivative shows whether the number was a minima or maxima
@@ayushrudra8600 The poor man's way (the way I did it) was just try a solution slightly off of the one with derivative of 0 and saw that it gave a larger area.
Here´s a problem you may find interesting and somewhat challenging: Three circles of radii 1,2 and 3 are externally tangent. What is the radius of a fourth circle inscribed between the other three? (answer: 6/23)
I was pretty proud I was able to solve this. I don't remember much calculus but I did remember what the derivative of x^n is. And similar triangles show up so frequently on these problems by Mind Your Decisions and Premath that anyone who tries to solve very many of these will definitely know that. Great problem in that you only need to remember a couple of basic concepts, yet it's not trivial to solve.
For the second part you can use a weighted AM-GM inequality: the area is equal to 2x + 3y/2 + 12 (since xy = 12 is given). So we want to maximise this. Using the AM-GM inequality splitting 2x into x + x and 3y/2 into 3y/4 + 3y/4 you can simplify the inequality to 2x + 3y/2 is greater than or equal to 12 with equality when the variables we input namely x and 3y/2 are equal. This solves to give us x=3 and y=4 and the total area is 12+12=24
I did the first part using the property of similarity of triangles. Then solved for x and y by substituting possible factors and got x=3 and y=4 and came to the answer i.e., 24. But the values could be any number and not integers only. Lol. But since the question demanded to know the minimum area, I understood that my answer is right but one of many right answers. Using differentiation to find minimum value never crossed my mind. Thanks. A nice question indeed.
I simply used the Inequality of arithmetic and geometric means in 4x and 3y and I solved it Knowing that the area of the larger triangle is (4x+12+xy+3y)/2.
Angles of a triangle equal 180 degrees, 90 for one and 45 degrees for the other two. 3 a bit less than 4, I am going with the rectangle is 2 length and 4 width. You can measure the three triangle if you want. Length and width would be the number not added to the 4 and 3.
the sides of rectangle can assumed 4cot@ and 3tan@ where @ is the base angle so the and of first part is directly 4cot@*3tan@=12 and the area traingle is 1/2 (4cot@+3)(3tan@+4) open the brackets and minimise it vy using arithmetic mean greater than geometric mean inequality and get the and for second part
here's how i did it. by proportion upper triangle area is 4*4z/2 lower triangle area is 3*(3/2z). the product of these is a constant (36). if two numbers have a constant product and we wish to minimise the sum, then the numbers are equal. so the areas are equal., hence z=3/4, each small triangle has area 6, total area 24
I'm glad I have my own business, cause I answered 'tomato' for the second part...
Correct answer
Is that tomayto or tomahto?
LoL
ROFL!!
@@marioluigi9599 tomahto is the correct answer don’t even try to fight me
I'm constantly fascinated by the amount of mental gymnastics that could be avoided simply by using calculus.
How? In this example?
I don't know if the first part is made simpler by calculus, but as I don't know much about geometry or rules for similar triangles it's the only way I managed it. As a bonus you don't have to think about anything, you just write out an equation for the area of a triangle twice and rearrange it. Also, I think I used algebra not calculus.
For simplicity I did my calculations using height times width as if there were two triangles forming a rectangle so I don't have to divide everything by 2 at each step.
short side of blue rectangle = a, long side = b, so ab = area of blue rectangle
1.
Multiply the length and width of the triangle to get an area double the size of the triangle.
(a + 4) * (b + 3)
2.
Area of the blue rectangle + area of the top triangle + area of right triangle. All of the areas are doubled to match what I did with the first equation
2ab + 4b + 3a
3.
Equate the two formulas. They both calculate an area twice the size of the triangle.
(a + 4) * (b + 3) = 2ab + 4b + 3a
4.
expand the first equation
ab + 3a + 4b +12 = 2ab + 4b + 3a
5.
subtract 3a, 4b, and ab from both sides
12 = ab
@@bartholomewlyons I think you could do part 2 with a limit. Also calculus is used in this video when he takes the first derivative of the expression.
_simply using calculus_
Check E J Rupp above. It is a really simple one-step solution.
For part 2, although using calculus to find the minimum of the area works for those who know calculus, you can actually visualise the minimum. In the big total triangle, cut off the little triangle to the right of the rectangle (which has sides y and 3), and rotate it 180 degrees about its upper-left vertex, and stick it back on so that it looks like a dormer window built onto the roof of the other piece (this is easier to draw than to describe in words!). Then the area of the total is equal to the bottom rectangle plus another identical rectangle above it, plus the remaining bit of the top triangle that sticks above the line. The two rectangles have total area 24, so the total area is 24 plus whatever surplus triangle sticks out. So the way to minimise the total area is to pick the configuration in which the surplus triangle vanishes - namely when the small triangle you rotated exactly matches the top triangle and leaves no surplus. (Maybe one of the 633 other commentators already said this?) Yours JOHN HARTLEY.
I would just like to add on that, the excess part is not always from the top triangle. When y > 4, the excess part is actually from the right triangle that you cut and pasted, and the excess area is actually in the shape of a trapezium if you try to visualise it.
You can pause the video at around 3:08, at the part where h > 8, because when [h > 8]= [y+4 > 8] = [y > 4]. If you do your cut and paste on the triangle when h>8, the excess area is in the shape of a trapezium from the right triangle.
@Pooshan HalderChess Hell yeah, bro
Sorry,I know I am a little late but I did not understand the method you stated and can you tell me is there a way to solve these kind of questions which requires maximum/minimum areas,without using calculus.
@@nareshkumar-uo7nz The only reliable and easy way to do it is using Calculus; without it, you'll have to do a bunch of Geometry grunt-work. I personally only know the Calculus way to do it, as that's the easiest way.
@@daddy_myers Thank You for replying but can you explain me this method I didn't get it. It will be appreciated.
One more thing I like about this channel is that it leaves the viewers the other correct ways of solving the given problems and arriving at the right answer, which is a good learning opportunity.
For the first part, if you write the area of the large triangle as (x+3)(y+4)/2 and set it equal to the sum of the areas of the small triangles and the rectangle, 2x+3y/2+xy, then the equation simplifies to xy=12.
I used the equation for a slope: y = ax + b.
Since the slope of the hypotenuse is the same for both triangles, so 4= ax, y= a3; a = 4/x = y/3 and then y = 12/x.
I used similar triangles.
Exactly what I did, no trig rules needed!
@@robbymoonshot4975 just try the method then you find that RHS =LHS =0 each and every value will be cut down
I used simple ratios 4/x = y/3
Oh... this brings back memories, I remember this coming up when I applied for my first job as a truck
driver, lucky I remembered my calculus.
Questions like this are probably the reason there is a shortage of truck drivers today!
I cannot imagine asking this question for someone who wants to be a truck driver. Sure, you need some basic geometry, but come on lol ask questions like "what to do during a jackknife" or "how should a load be distributed for maximum trailer stability"
@@NeoNoggie guess you can’t understand when someone is making a joke
r/woooooosh lmao
What confuses me is, who are you going to ask this question? What job are they interviewing for? Maybe if you're looking for an engineer right out of college?
@@soaringvulture Not really...this is important for cleaning staff...when working out the surface area they have to sweep.....
You can use the inequality of means to solve the second part without calculus:
We know that x.y =12 and we want to minimize (4x+3y)/2 +12. We can substitute a=4x and b=3y, giving us a.b=144
We now use that (a+b)/2 >= sqrt(a.b) and we find that (a+b)/2>=12, therefore, the minimum will happen at the equality.
Inputing the value in the area equation gives us 24 as the minimum area.
We broke math here is why you just proved that the minimum area is 24 but when you realize that with4/x=y/3 than you can get 12x=y and since xy=12 and since the only pairs for x and y that fit that equation are x=1 y=12 than the area of the triangle must be 32
(Assuming the answer would be an integer) You can also take ordered pairs on ways of writing x.y=12, they are (12,1) (6,2) (4,3) (3,4) (2,6) (1,12), now if you take the area of triangle using these values, you'll find (4,3) to givr you the lowest and that is exactly 24 sq untis.
But whats with x=y=0?
Then you've got all the conditions, no rectangle and only one triangle with an area of 6
(i think the formula is only right assuming there is a rectangle with a size bigger then x=0, cause x is in the denominator)
I also thought of this inequality, wells spotted!
I used the same method, it becomes very easy to solve this problem using this method
First part was easy. For the second part I tried taking the min of the function 2y+(3/2)x+12 directly. Took a while to notice I needed to use xy=12 and solve it by minimising 2y+18/y+12. In the end finally solved it 😅
Rather than use a derivative, I guessed that the minimum would occur when the two opposing terms 2y and 18/y are equal. It could have been wrong, but it wasn't.
I f**k up on the derivative of 18/x :-( otherwise I was good
Yes I got first part in 5 seconds as I assumed it was a 3, 4, 5 triangle and ignored the graphical lengths as a distraction. Second part was not so easy and guessed at 12 again. I was only 100% wrong lol
@Aadishree Chaurasiya I'm studying as an Electrical engineer (2nd semester). Not any class 😅😂
@@w1swh1 You calculated the area for just one of the infinite number of triangles without checking any of the other possibilities. So you got the first part right by luck rather than by proof.
When I heard "minimum" I knew I had to do derivatives. I guess it's an instinct as a math student. It takes me back to my optimization lesson.
Basically the first derivative is for finding the slope. And we set that slope equal to zero to find the max peak or min crate values of that graph/function
Me too bro
I used to be able to get max and min without using calculus but have forgotten how to do it.I'd have to look up my notes on this which are around 55 to 60 years old.
@@shadrana1 The way to do it is to use AM-GM inequality in this case
I think this sum can be solved in easy way without applying calculus
I am currently in 11th std and don't know about calculus but have heard about it
I cracked the logic by using Basic proportionality theorem and theorem on ratios of similar triangles
I nailed the first part quickly, then equally quickly stalled..... this was a nice problem.....
how come the base of the triangle is 9 & the height 6?
@@ivornworrell thats a triangle that satisfies the condition with the 3 and the 4. In other words, any blue area thats 12 will satisfy. in 6x9, blue would be 6x2= 12
how is comment 5 days ago?
@@aashsyed1277 the comp. was reset
is it weird that I can't solve part 1 but easily solve part 2 😅
You can solve this by pure geometry. Imagine the hypotenuse of the biggest triangle balancing on the corner of the rectangle. If the two smaller triangles are of different sizes we can rotate the hypotenuse to make the larger triangle smaller and the smaller triangle larger. Thinking in terms of limits it's easy to see that a very small change in angle will make the larger triangle decrease by more than the smaller triangle increases. So the two smaller triangles must be congruent at the minimum. Pick one of those triangles up, flip it around and put it against the other small triangle and we get a bigger rectangle of total area 2xy. So the area of the big triangle is 2x12 = 24. Job done. No algebra or calculus needed.
Finally an answer I can comprehend and use in the future
Ik this is easy but without diagrams, this doesn't seem that easy to understand..
Calculus is simply not taught in the standard curriculum in many countries, so the derivative at 4:31 is a huge leap for so many viewers
2nd part has an easy solution . just think about for which triangle it is going to be minimum . yes , it is isosceles triangle . now in that case the white sub triangles(small triangle) are going to be same to each other and each of their area will be the half of the rectangle(in this case the rectangle is going to be a square).you can think of folding the white part over the blue area. thus you can understand,
"sum of the 2 small(white ) triangle = area of blue square "
so the area of the triangle will be 2*(area of square)=2*12=24.
simple, you don't need calculus
area of triangle = 1/2 * (4 + y) * (3 + x) = 12 + 1/2(4x + 3y)
use arithmetic geometric mean inequality on 4x + 3y
you get 4x+3y >= 2 * sqrt(4x* 3y) where equality holds when 4x = 3y
rest is easy
Yeah like why do they do differential
Funny, I find mathematics, especially geometry very fascinating and solved the first part of the problem in a minute, then got halfway through the second half and got stuck, so I decided to watch the video. From the word "derivative" onward, I understand exactly zero percent of what's being said and done.
yeah if you havent learnt calculus its very difficult. once you've learnt it this is actually a fairly simple problem
It's just that if you differentiate a function and set it equal to 0 then you will get a value of x such that f(x) is either maximum or minimum. To decide it's maximum or minimum you see the double differentiation of f(x). Learning Derivatives is fairly easy you can learn it in a week easily.
Yeah, his way of solving the problem here is _way_ overcomplicated. Would be much easier to use the arithmetic-geometric-mean inequality to minimize f.
@@ribozyme2899 but once you know calculus, there's little reason not to use it everywhere. It's just so damn powerful, and derivatives like this are so simple.
Why wouldn't you use calculus whenever the question is "maximize/minimize the value of this function"
I am sure there is a more intuitive solution to the second part
We can also minimise the area using
AM ≥ GM
Area(∆) = 4x/2 + 3y/2 + 12,
Let (4x/2 + 3y/2) = k
So area(∆) = k + 12 ------- (1)
Looking at (4x/2 + 3y/2) , and by AM ≥ GM, we can say,
k/2 ≥ [(4x/2)(3y/2)]^½
on putting xy=12 and solving we get, k ≥ 12
Putting this in (1), we get
Area(∆) ≥ 24
Therefore, min area(∆) = 24
I saw another similar problem on the UK IMC a few years ago, another nice solution! Solving questions like these inspire me to share my own maths tricks, like you do, as well!
ua-cam.com/video/XV2Dg-9pGFQ/v-deo.html ho
@schrodingers cat 150*sqrt(3) cm
do you know which question, as in 1-15/ 16-20/ 21-25. Still got one last chance at the paper this year and want to re-establish what i should expect.
@@arshtewari2833 I checked again - it was a similar diagram, one of the 20-25 questions. In terms of other ways to prepare for the paper, I'd recommend checking out the 'Intermediate Problems' book on the UKMT website, it's very good practice.
I achieved a Gold in the challenge and qualified for the Hamilton Olympiad last year (achieved a Merit), so I'm gonna be making videos covering questions, how to prepare, the ideal thought process etc. Stay tuned - I'm sure it'll benefit you!
@@AliKhanMaths Wow this is so surreal because I also received a merit in the Hamilton last year! Thank you for recommending the book and I'll be sure to check it out. Good luck for next year and I hope we both qualify for the Maclaurin :)!
To minimize expression (4x+3y) , where xy=12, you can use AM-GM inequality - (a+ b)>= 2sqrt(ab), no need for derivative. So (4x+3y)>=2sqrt(4x*3y)=2*12=24, equality holds true when 4x=3y, i.e. 4x =3*12/x ; => x =3.
For the part 2 question, we can solve by using inequality of arithmetic and geometric means
1) by similarity
2) we have xy=12
We get factors (1,12) (12,1) (4,3) (3,4) (6,2) (2,6)
Here we get minimum value when x=y or the difference between x and y is minimum so in this case (3,4) or (4,3) are if we put (3,4) we get 24 and if we put (4,3) we get 24.5
For max we take (1,12) and (12,1)
(1,12)=32
(12,1)=37.5
Minimum=(3,4)=24
Maximum=(12,1)=37.5
:) ✌️
For all my peeps who thought "Why not just set both x and y to zero?" Remember that there is the requirement that there be an inscribed rectangle touching the hypotenuse. For that to happen, x becomes infinitely large as y goes to zero and vice versa. If both drop to zero, you no longer meet the requirements of the problem.
I got the answer to the second part instantly with geometric intuition, which works basically like this:
First, note that the inner triangles are similar, and have the same area if and only if they are congruent.
2a. Now assume the triangles have different area. Rotate the smaller triangle 180 degrees about the top right corner of the rectangle. The two triangles form a rectangle congruent to the blue rectangle, plus some extra. This total area is 24 plus some positive area.
2b. Now assume the triangles are congruent. The same rotation now forms only the rectangle, with a total area of 24.
This reminds me of the electrical circuit where you want to maximize the power to a load, given a source impedance. The answer is that the load impedance must equal to the source impedance. This case reminds because the area of the triangles is equal to the area of the rectangle. The rectangle area is given & to minimize the total area, the area of the triangles must equal the rectangle's.
A stretch, yeah, I'll admit.
Oh geez....that's just made me even more terrified of job interviews than I already am...
I have to admit I would walk out of any interview for a job that required me to solve this inappropriate and useless skill that has no bearing on any problem in real life! With geometry algebra trigonometry calculus - all of it can go in the bin I'd rather work out the best way to dress a Pizza and then cook and eat it - satisfaction rating ***** mathematics minus *****
@@theprior46 What are you talking about? If this stuff is of no interest to you, then you are not applying for the types of jobs where such questions are appropriate.
@@theprior46 how ignorant are you if you think this is useless?
@@theprior46 It probably would have been better if you just said it had no bearing on the job you were applying for. Assuming you could even know that. Saying that it has no bearing on any problem in real life is just ridiculous. Human beings wouldn't have come up with this stuff if there wasn't a reason for it out there somewhere.
@@theprior46 If you are applying for the position of a school janitor or something, then sure.
I quickly saw the AM-GM inequality way to solve this just by labeling x,y for the rectangle, we have reduced the problem into finding the min value of (x+3)(y+4) with xy = 12 which can easily be done by using am-gm for 2 numbers.
But whats with x=y=0?
Then you've got all the conditions, no rectangle and only one triangle with an area of 6
(i think the formula is only right assuming there is a rectangle with a size bigger then x=0, cause x is in the denominator)
@@neutronenstern. x=y=0 does not work.
A key requirement is that the upper right corner of the rectangle lies along the hypotenuse.
This means if you have y=0, i.e. the height of the rectangle is 0, then x, the length of the rectangle, must be the entire width of the triangle. But we know the width of the triangle is x+3.
This means we need x+3=x, which is impossible, so y cannot be 0. The same applies for x=0.
This means x>0 and y>0.
Points
(0,y+4) (x,y) (x+3,0)
Gradient
4/x = y/3
xy=12 (rectangle area)
(y+4)*(x+3)= xy + 3y + 4x + 12 = 3y + 24 + 48/y
3 - 48/(y^2) = 0
y = sqrt (16) = 4 (positive value of y)
Min Area = (3(4) + 24 + 48/(4))/2 = 24
The AM GM inequality is a great alternative to the derivative technique for quadratic functions, but its proof depends on theory of quadratics; so I prefer to keep in practice by completing the square.
When you did your calculation you assumed x and y not zero. But x=y=0 also works with the triangle a 3 4 5 right triangle. I think that makes a better "tricky" solution.
x and y cannot equal to 0.
@@hemph9772 Why can't they? We have no condition on possible solutions. If you make one equal to 0 and the other equal to the length of its side, for example x=0, y=4, you get that the rectangle becomes a segment but it's still a degenerate rectangle. Not sure about both variables being 0 as it wouldn't fit the drawing, the degenerate rectangle and the third side would have no point of contact.
Edit: Nope, just a brainfart
@@martinovilla218 please draw this triangle where the degenerate rectangle (with vertical height = 0). The base of the rectangle and the base of the triangle would be the same line segment. This contradicts the definition of the problem whereby the base of the triangle must be 3 units longer than the base of the rectangle.
@@TPinesGold Yeah, you're right. I was thinking too hard about how a degenerate triangle could fit that I forgot how the problem was actually formulated
@@TPinesGold in the video, the problem is defined as the distance from the top of the rectangle to the top vertex of the triangle is 4 units, and the distance from the right of the rectangle to the right vertex is 3 units.
A degenerate rectangle would fit this criteria, thus the area is 0 and the minimum triangle area is 6.
Great video! I especially enjoyed the solution to the first problem, although it would have been nice if you explained how you got the solution to the second problem instead of just handing the method down from above. Logic puzzles are your forte in this channel, so keep doing those!
For the one, I did it in two ways. One way was the similar triangles. The other way is to make the triangle into a square by adding another equal and opposite triangle, and then the question becomes quite easy as there are two triangles that are exactly the same as the other triangles, and then a rectangle with area 12 (4*3) which is the answer.
It's easy though. Solved it by assuming rectangle length and breadth as x and y. Used similarity property and then used differentiation to find minimum area.
There's easy one in second part,
Just take two sides as, 'x+3' and '4+y or 4+12/x(since xy = 12)', and that's it you have a function for area,
f(x) = (1/2)(x+3)(4+12/x),. and solve for it's derivative equal to zero🙂
For part 1, I dove in with looking at the two different ways of calculating the area of the large triangle. One way is (x+3)(y+4)/2, the other is 4x/2+3y/2+xy. Set those two equal to one another, and everything cancels out quickly leaving xy=12.
I did this, and then found the derivative of the f(x)=(x+3)(y+4)/2 equation (where y=12/x) to find the critical point where x=3
@@mervstar guys you dont need to use derivatives . just use the property AM>=GM
Another exciting question of the given problem would be to find the areas of two sub-triangles. I tried to get them, and I enjoyed the problem. Happy solving!
The minimum area triangle ends up being a multiple of the 3-4-5 triangle, that's awesome 😁
Yeah. This looked like a trick question, but I've been drinking. I saw the 3-4-5 relationship and just thought make the box zero.
I think the problem needs more rules, or is it okay to ... drink ... what was I saying?
ua-cam.com/video/XV2Dg-9pGFQ/v-deo.html ho
I WAS CONSTANTLY THINKING THAT
@Math427 Just a suggestion: putting the word SPOILER at the beginning of your comment (with maybe a couple of carriage returns after) would be polite.
I notice that when i used algebra for it, lol
For the first part you can mirror the triangle along the hypotenuse. The part mirrored along the hypotenuse of 4*x will be equal and the part mirrored along the hypotenuse of 3*y will be equal, so the leftover part will be equal to the rectangle, which we now know the side lengths of, which will be 4 and 3 and that will give us an area of 12
This is just secondary maths in Asian countries I believe. Wish interview questions were this easy.
4/x = y/3 gives xy=12. Then you can find the answer to the second question without calculus by using substitution values for the function f(x)=2x+18/x+12. f(3) gives the lowest product for the function.
I easily found the area of the right triangle by using the similar triangles method you used, but I used calculus to find the minimum. Let x be the length and y be the height of the triangle. We need to find the minimum area of (x+3)(y+4)/2 which works out to be (xy + 3y + 4x + 12)/2. Substitute xy for 12 and y for 12/x, we get (36/x + 4x + 24)/2 = 18/x + 2x + 12. Finding the derivative, it will be -18/x^2 + 2. Setting it equal to zero to find the minimum, we have a quadratic. -18/x^2+2=0, so -18+2x^2=0, so x = 3. y=12/3. Our original right triangle has a length of 6 and a height of 8, so it will have an area of 24.
Finally, the first problem that I've solved flawless on this channel 💯
I've been wanting to buy a new toaster but couldn't come to a decision as to what price I was willing to pay. But, thanks to this video, I can now figure out how much money I can spend on a new toaster. Thanks, this really helps.
The very top angle and the top angle of the smaller triangle are the same(angle "A"). So we can use trigonometry to find the width W of the rectangle as the opposite of the right angled triangle, as tanA=w/4>>w=4tanA. Since we know the smaller triangle has angle A too, we can find the length L of the rectangle since it is the base of the triangle, which is tanA=3/L>>L=3/tan A. The area of the rectangle is LW so 4tanA*3/tanA.
Leaving the answer as 12
The second part can also be done by AM - Gm inequality as length of sides need to be greater than 0
@schrodingers cat is it 150√3 ???
@schrodingers cat bunch of similar triangles, plus tangent-chord theorem,
@schrodingers cat 🙂
@schrodingers cat Since B is a right angle, and the center of the semicircle is on BC, it can only be tangent to AB if the tangency point is B. Then the triangle can be described with three parameters: y = BC - 20, z = AB, and x = AC - z. x is the length from C to the tangency point on AC, since the tangents from A are equal, and one tangent is AB = z. Now, using the pythagorean theorem twice gives x^2 = y(y+20) and (y+20)^2 = zx. Then z = 10sqrt(1+20/y). The area of ABC is z(y+20)/2, which can then be fully parametrized by y. Using calculus, the minimum is found at y = 10, and the value of the area is then 150sqrt(3). It should be noted that the resulting triangle is a 30-60-90 triangle, which indicates there is probably a more elegant solution :)
@schrodingers cat 150√3 , for that sides are 10√3 and 30
I have a simpler solution that doesn't require derivative or AM-GM
So the area of the triangle is (x+3)(y+4)/2=(xy+4x+3y+12)/2
xy we already know, 12 and 2 is a constant so we can ignore those, so the area of the triangle is minimal when 4x+3y is minimal.
because 4x and 3y is positive real number then (4x-3y)^2>=0
=>16x^2 - 24xy + 9y^2 >= 0
=> 16x^2 + 24xy + 9y^2 >= 48xy
=> (4x + 3y)^2 >= 576
=> 4x +3y >= 24
then apply to the function above then we have min(S)=(12 +24+12)/2=24
I think in 2nd question you can use AM - GM inequality to finding the minimum value of f(x) because all numbers are greater than 0. Then you can condition to "=" is occurred. If I don't true in some point, hope everyone can help me.
@Vinicius Machado Pena ua-cam.com/video/X9X0UKtf7Ek/v-deo.html
But whats with x=y=0?
Then you've got all the conditions, no rectangle and only one triangle with an area of 6
(i think the formula is only right assuming there is a rectangle with a size bigger then x=0, cause x is in the denominator)
@@neutronenstern.ua-cam.com/video/nkCwCQCjALo/v-deo.html
First part can be solved easily by taking tan(θ) of the upper triangle and lower triangle , from there you get the values of x = 3tan(θ) and y = 4/tan(θ), after multiplying x and y area will be 12.
Great question. I'm going to remember the 2nd part for my Calculus class!
you don't need calculus for this question , there is a simple solution.
@@ishtiakhasan8397 True. Many optimization problems do offer more than one approach that does not involve Calculus.
@@PackerBronco you could just use the pythagorian triplet which is way easier but deriving the function is way more fun imo
First part can also be solved by using Pythagoras theorem on triangle on the bottom right since we know that if 1 side of triangle is 3 and 1 of the angles is 90° then length of hypotenuse has to be 5. From this we know that side y=4. Then we can easily figure out area of Rectangle.
What about the solution x = y = 0 ? e.g., the blue area is either 12 or... 0, and the minimum white area is... 6 when x = y = 0.
@IAmBot Nobody said that the area of the rectangle cannot be zero! A zero area rectangle is still a rectangle.
I did part 1 completely different. Just draw a triangle against the given triangle, so that both triangles form a rectangle. Three of the points of the original triangle form the corner points of the new rectangle. Then continue lines x and y. Now it's easy to see that there are two equal triangles in the upper left section of the new rectangle, and two equal triangles in the lower right section of the new rectangle. This means that the two smaller rectangles must be equal in size too. The upper right rectangle is 4 x 3, so the lower left rectanble must also be 12.
Exactly how I did it
Well, that's interesting. My immediate thought was that when I set x = 0 and y = 0 I would get the minimum solution with a right triangle with sides 3,4,5.
I thought there is no minimum size for the rectangle, so it could be reduced to zero, so surely the minimum area of the triangle is 1/2 x 4 x 3 = 6 units squared.
Please let me know how x=0 and y=0 meets the requirement that x * y = 12...
@@CrankyTim my very simple point to the was to the OP (not you) that x=0 & y=0 doesn't meet the basic criterion that the area of the rectangle = 12 (although you do get a triangle).
@@CrankyTim
It's entirely unnecessary to look at angles to see x=0 & y=0 doesn't work, which is what the OP first thought of.
Read what he wrote "My immediate thought was that when I set x = 0 and y = 0". You are overcomplicating things.
@@CrankyTim if you set x & y to 0, then you most certainly do get a 3:4:5 triangle, just not with one with a 0 x 0 rectangle which doesn't, of course, touch the hypotenuse.
Did it exact the same way. Nice Problem.
Maybe there is an elegant way to solve the second part:
As we know (do we?) the largest rectangle in a triangle has half the area of the triangle.
So the minimum area of the triangle must be twice the rectangle.
Yay I did it :D
Really didn’t expect the second part to involve calculus
@@ashishchotani2458 ua-cam.com/video/X9X0UKtf7Ek/v-deo.html
ua-cam.com/video/XV2Dg-9pGFQ/v-deo.html ho
@@omletyt8055 ua-cam.com/video/q_ePTlGkfBo/v-deo.html
Got this. But for part one I used a slightly different way without noticing the similar triangles (equating the area of the big triangle with that of its three component shapes): Let A be the area of the large triangle. Then 2A = (4+y)(3+x) = 2xy + 4x + 3y. The x and y disappear and leave xy = 12.
I think there are other answers that were neglected.
The equation derived from the similar triangles is only valid if x or y are different from 0.
But if you take x = 0 and y = 0, the rectangle is only one point with an area of 0 and therefore the minimum area is 6 (ie 4*3/2)
Would you say that these answers are also acceptable ?
I also saw this solution, and would respectfully suggest that 6 is, indeed, the only correct answer to part two. Surely we should not assume that the illustration of the problem is drawn to scale, unless we are told so.
That was my initial reaction, but now I'm not so sure.
If, as is suggested in some of the comments further down, you cannot theoretically have a rectangle with side length 0, then I would have to agree that this solution is not valid. Can anyone give a definitive answer about the validity of a rectangle with side length 0? I cannot find an answer to this on the web, although I seem to remember being told (in a computer graphics class) that a rectangle which has one side of length 0 is a line, and a line of length of zero is a point. I may be misremembering, of course, but would appreciate any help.
No. Because we know that ONE of the angles of the rectangle is on the hypotenuse we cannot have all of the angles coincide with the right angle of the triangle.
The rectangle with side lengths 0, 0 cannot be inscribed in the triangle in the way shown in the video, since the triangle has positive side lengths (>3 and >4, respectively). Therefore, (0, 0) is not a solution at all. This geometric condition ensures that all side lengths are positive and the math presented in the video makes sense.
yes
A similar task was used in the national exams in Georgia (2016). It was rated 4 points.
I learnt how to find maximum and minimum value of a function yesterday and found it very interesting and here we are now
The two triangles are similar. Using x as the width of the rectangle and y as the height, the proportion for the triangles is 3/y=x/4. By cross multiplying, we get the area of the rectangle xy = 12.
Hello Presh, I've followed you since I was in school. Now I'm training to become a math teacher. I've always loved the way you present the problem and the animations you create. please could you let me know what software you use for your videos so I could make my own to help my students understand math better?
He uses PowerPoint
I, without reading the question properly, found the minimum area of the two small triangles pretty easily. In terms of angle B it is area = 4.5 * tan(B) + 8 / tan (B). Differentiate and you get the minimum at tan(B) = 4/3, easy after that to solve everything.
Well according to the pythagorean triplet it has 2m m^2-1 and m^2+1 its way easier by that
I found this one fairly easy, save that I just took the area of the triangle as (4+y)(3+x)/2 rather than as three separate areas (why do that). Then substitute in for y to get (4+12/x)(3+x)/2 and simplify to get area equals 12+2x+18/x. We might as well throw away the common factor of 2 and the constant as they play no part in finding a minimum, so take the first differential of x+9/x and get 1-9/x^2. Solve that for the value of 0 to get a min/max and the only positive root is 3, so an 8 x 6 right angle triangle.
It's a general rule when looking for minimum/maximums you can strip out those common factors and fixed constants when differentiating the function. It's a short-cut, but then that's what interview questions are surely about. However, it might be wiser to do it in full for an examination answer.
This is about the level of question that was in the calculus part of the "O" level examination in England when I took it back in the summer of 1966 aged 16. It's a bit basic for university entrance in my view, albeit in most of the UK university courses are specialist subjects so if you are going for a science, then the standard required for mathematics is a lot higher than, say, English Literature. The mathematics requirements for the latter will normally be fairly low.
Mark Levi's book describes a super useful way of solving problems like the second one almost instantly. The idea is to think about the triangle as filled with a vacuum and being squeezed by air pressure. Then the area of the triangle is proportional to its energy, so the minimal area configuration is always an equilibrium state. But if the corner of the rectangle doesn't touch the hypotenuse at its midpoint, then the hypotenuse will clearly feel a torque, and hence the state cannot be in equilibrium. From there the solution is immediate.
After getting f(X)=2x+18/x+12 we can use AM-GM inequality on 2x+18/x to get that (2x+18/x)/2 >=root(18*2)=6....Thus we can directly find the min of f(x) as 12+12=24
You can also "flip" the triangle onto the rectangle and "see" that they cover it only when x and y are equal to 3 and 4 (and
angles are mirrored).
this was exactly my method as well. no calculus needed.
But you still need to explain why that would give you the minimum.
@@benvanrensburg4261
Just like checking at both sides of a value of a formula by +/- than the x value,
check with tiny extension to the right and big upward and tiny upward and big to the right, fold over the rectangle, and "see" that there is a lot of leftovers...
No leftovers when foldovers tips fall on right angle corner.
* I don't know how to say this concisely in Mathian... :)
@@OrenLikes I get it.
Sir
I simply used am is greater than or equal to gm
4x+3y/2>root 12xy
Now substitute xy = 12
And area of small triangles comes 12
big one =12+12=24
First part was easy, second part was interesting but can also be solved without differentiating, by using AM-GM:
2x+18/x>=2sqrt(2x*18/x)
2x+18/x>=2sqrt(36)
2x+18/x>=12
2x+18/x+12>=24
so the minimum is 24
very interesting problem though.
edit: differentiating not integrating
Can you explain the first expression , it would be very helpful 🙏.
@@shubhammishra8286 It uses the AM-GM inequality, that states that for any two positive real numbers A and B the inequality
(A + B)/2 ≥ √(AB)
holds, and the equality appears if and only if A equals B.
Did anybody else just go, "Oh, they can both be 3, 4, 5 right triangles. That'll answer both questions."?
Yet another instance where the fact that x+1/x has a minimum at x = 1 comes in handy
Lmao that's so easy to do. Takes 5 seconds to do it mentally. Ig only dumbasses have to memorise the result :(
FOR the part two you can just apply arithmetic-geometric mean inequality it is easy to do！
surface = 1/2*(3x+4y)+12 and 1/2*(3x+4y) >= √12xy = 12 → surface >= 12+12 = 24~~
for the last part, you can just use am gm
I do not know how to explain exactly, but I had an intuition that this would be the solution!
Only when the two triangles overlap will there be a minimum area for the large triangle. This is the midpoint in their relationship.
Any change here or there, (it will be symmetrical) will only increase the area of the large triangle.
With 3 and 4, obviously x=y=0! The restriction that x is not null comes way too late during the proof, it is not part of the initial question, so it should not be taken into account. 3, 4, 5 is the easy right triangle...
Never thought that calculus could be used for these types of geometry problems. A great tool,
I remember something, one man's tool is another man's weapon
How did you do the animation for the second part? Like which tools do you use?
I have asked many time but he didn't replied .
Probably a Monitor, keyboard and mouse and a PC
Let w and h be the width and height of the blue rectangle respectively,
since the two triangles are similar,
w/4 = 3/h => w*h = area of the blue rectangle = 12
To derive the minimum triangle that satisfies the condition, we use derivative on the minimum area of the sum of the two small triangles.
The area of the triangles = 3*h/2 + 4*w/2 = 3*h/2 + 4(12/h)/2 = 3(h+16/h)/2
d(3/2[h + 16/h])/dh = 0
=> 3/2[1 - 16/h^2] = 0
=> h^2 = 16
=> h = 4
=> w = 12/h = 3
So the minimum large triangle that satisfies the condition has width = 3+3 = 6 and height = 4 + 4 = 8
This makes me feel like I know nothing about maths
Don't worry. You aren't alone....
it's fine, I've always been interested in math, just keep watching videos like these and soon enough you will get them :)
Thanks for teaching me the rectangle in a triangle problem.
Why complicate it so much?
Use Pythagoras.
a = 6
b = 8
c = 10
Area of triangle = 6 x 8 divided by 2 = 24
Blue Area = (6-4) x (8-3) = 10
the question is very easy, you can write x=3tanB, y=4cotB. then area of rectangle is x.y= 3tanB.4cotB=12. and are of triangle= (1/2)(4+3tanB).(3+4cotB). diffentiating and equating to zero will give us tanB=4/3,-4/3. rejecting -4/3. and substituting value of tanB=4/3 in area will give us write answer.
you lost me on calculating the area of the triangle
So we know the total area is the sum of the 3 areas.
With x and y being found initially from part 1. The area of the square is 12, with question is, if we vary x and or y, such that they still equal 12, what is the smallest total area possible?
The total area is the summation of the 2 triangles and the area of the square. We substitute 12=xy or y=12/x into the total area so now the total area is only a function of x. Which is fine since y is dependent on x because of 12=yx, so now we look into minimizing the function based on x alone.
Taking the first derivative gets the equation shown, by setting it equal to zero, it's where the tangent line is zero. Assuming you have no calculus experience this is helpful for finding the minimum or the maximum.
Imagine a circle, a tangent line will have a slope of zero, at the top of the circle or the bottom of the circle. We solved for this and get x = 3, x could be -3 because of the square root. But areas aren't negative, so only 3.
Taking the second derivative is how we determine if this tangent is either a maximum or a minimum. Is the lineon the top or bottom of the circle?
Imagine you have a circle cut in half left to right. When the circle is open facing down, like an upside down U. Then the tangent line will represent the maximum. The opposite is true. If the tangent is on the semi circle facing up, it's the minimum.
The second derivative tells us if it's the top circle or the bottom circle. If by plugging x = 3 into the second derivative the value is negative, < 0, then we have the maximum value. But since the value was positive, > 0, it's the bottom of the half circle and therefore the minimum.
So now we know that x=3 is the minimum x value. Plugging x back into 12 = xy, we get y = 4. And plugging 3 and 4 into x and y for the total area gets us 24. The minimum.
Hope that helps.
I got this right away by remembering that 3-4-5 makes a right triangle, and that the minimum total area is the one that makes the rectangle closest to a square, so with the given lengths, a 3 × 4 rectangle minimizes the total area, making the case of 2 congruent 3-4-5 right sub-triangles the minimum ... done.
"What is the minimum area for a triangle that satisfies these conditions." This part of the question is ambiguous. There are three triangles, the large one and the two smaller ones. Either the question isn't precise enough, or the solution is incomplete. To answer this question as it's presented would require three solutions, one for the overall triangle (which is presented), one for the upper triangle and one for the lower right triangle. An interesting question though. :)
Yeah it should have been worded "minimum area for the triangle that contains the rectangle" which I assumed was the question, but it could have been made more completely clear.
I just used Pythagorean triple, base = 6,side length = 8 and hypotenuse = 10,so 4x3=12 for the rectangle and 8x6/2= 24 for minimum area of the whole triangle
3_4_5 is always a right triangle! 😊
one should note that this works for arbitrary values a,b instead of 3 and 4. we always get xy=ab and its minimized for x=a and y=b (or vice versa, depending how you labeled it)
Sometimes, I forget about congruency and similarity
3:56 using inequality of arithmetic and geometric means we get (3x+4y)/2 >=√(3x*4y) = 12, equality when 3x=4y
Oh well, back to the unemployment office for me...
After creating the equations you could have used ARITHMETIC MEAN >= GEOMETRIC MEAN.
=> (2x + 18/x)÷2 >= √[(2x)*(18/x)]
which gives => ( 2x + 18/x) >= 12
For min. we will take "12" and hence min. area will be 12+12 = 24.
It's really halwa for jee aspirants 💪💪
Power of preparing for IIT 😎💪
Wow, I have been following you for years, and this is only the second problem that I got right before I finished the video. 😁
I understand the second part up to the derivitive part (im 12 btw). I know how to take derivitives, but can somebody explain why he took the derivitive and stuff like that.
We took the derivative of the equation of the area of the triangle in order to find its rate of change. So as x changes the area changes by the derivative.
So when the derivative equation is equal to zero we know that the original equation for the area is at a maximum or minimum. Since the area equation is positive(the graph of it looks like a bowl) we know we have the minimum at x=3.
Cannot really teach calculus in a single comment. But you can look up that the derivative gives you the slope of the curve, and when the slope is 0 then you have either a maximum or a minimum (or a saddle point).
@@XJWill1 how do you know if its the maximum or minimum? is it based on if the curve curves upwards or downwards?
@@ayushrudra8600 take any function
Use a graphing calculator and take single derivative and then a double derivative. U will understand
Or u can use the same function as used in the problem and u will see how double derivative shows whether the number was a minima or maxima
@@ayushrudra8600 The poor man's way (the way I did it) was just try a solution slightly off of the one with derivative of 0 and saw that it gave a larger area.
Here´s a problem you may find interesting and somewhat challenging: Three circles of radii 1,2 and 3 are externally tangent. What is the radius of a fourth circle inscribed between the other three? (answer: 6/23)
I was pretty proud I was able to solve this. I don't remember much calculus but I did remember what the derivative of x^n is. And similar triangles show up so frequently on these problems by Mind Your Decisions and Premath that anyone who tries to solve very many of these will definitely know that. Great problem in that you only need to remember a couple of basic concepts, yet it's not trivial to solve.
For the second part you can use a weighted AM-GM inequality: the area is equal to 2x + 3y/2 + 12 (since xy = 12 is given). So we want to maximise this. Using the AM-GM inequality splitting 2x into x + x and 3y/2 into 3y/4 + 3y/4 you can simplify the inequality to 2x + 3y/2 is greater than or equal to 12 with equality when the variables we input namely x and 3y/2 are equal. This solves to give us x=3 and y=4 and the total area is 12+12=24
Check E J Rupp above. It is a really simple one-step solution.
I did the first part using the property of similarity of triangles.
Then solved for x and y by substituting possible factors and got x=3 and y=4 and came to the answer i.e., 24. But the values could be any number and not integers only. Lol. But since the question demanded to know the minimum area, I understood that my answer is right but one of many right answers.
Using differentiation to find minimum value never crossed my mind. Thanks. A nice question indeed.
Similar brother the min integral for ratio 3 4 5 right triangle
I simply used the Inequality of arithmetic and geometric means in 4x and 3y and I solved it Knowing that the area of the larger triangle is (4x+12+xy+3y)/2.
Angles of a triangle equal 180 degrees, 90 for one and 45 degrees for the other two. 3 a bit less than 4, I am going with the rectangle is 2 length and 4 width.
You can measure the three triangle if you want. Length and width would be the number not added to the 4 and 3.
the sides of rectangle can assumed 4cot@ and 3tan@ where @ is the base angle so the and of first part is directly 4cot@*3tan@=12 and the area traingle is 1/2 (4cot@+3)(3tan@+4) open the brackets and minimise it vy using arithmetic mean greater than geometric mean inequality and get the and for second part
here's how i did it. by proportion upper triangle area is 4*4z/2 lower triangle area is 3*(3/2z). the product of these is a constant (36). if two numbers have a constant product and we wish to minimise the sum, then the numbers are equal. so the areas are equal., hence z=3/4, each small triangle has area 6, total area 24