You used a and b as parametters of the quadratic function and after that you used a and b again as the lenght and wide of the rectangle. So you create confusion .Better to use 2.xfor the lenght and y for the wide of the rectangle.
You could save a little time and effort by taking advantage of properties of parabolas. The given conditions imply that the parabola is symmetric about the y-axis, so the quadratic’s other zero must be at x=4. Thus, its formula is of the form y=a(x-4)(x+4). Now adjust the scale factor a so that y=4 when x=0. Interestingly, the maximum-area rectangle has the same width regardless of the value of a, i.e., we could set the parabola’s y-intercept anywhere on the positive y-axis without changing the width of the maximal inscribed rectangle.
That was amazing! I get the area under the curve usage case, and I love that we can use the Second Derivative proof to find if that area of the function at the specific point is actually the maximum. Thank you! Will buy a hoodies or something one day, you really made this accessible and understandable for a student that needs this maths understanding fairly soon.
My attempt. Graph 0, (x-4)(x+4) Make it negative -(x-4)(x+4) When foiled out it will equal -x^2+16 You need to turn that 16 into a 4 so it crosses the y axis at the right point, so divide the whole function by 4 -(1/4)(x-4)(x+4) Now the area of the rectangle is 2x(as the base of the rectangle is subscribed to the x axis) times y. So the area of the rectangle is (2x)(-(1/4)(x-4)(x+4) So foil it out ((-2x^3)/4)+8x Find the derivative to find the slope ((-6x^2)/4)+8=m When m=0, we’ll find the x coordinate where the slope is maximized or minimized. Sqrt(32/6)=2.309… = x Input 2.309 in the area of the rectangle. And you get the maximum area, which is 12.31… It makes sense as a maximum cause if you put 4 as the height, you’d have no base for the rectangle so the area would be 0, and if you put 8 at the base, the height would be 0, so the area would be 0. But somewhere in the middle is a real area The bottom of the rectangle is 2.309*2 The height is 2.667 if we put the x value in the equation we made. Edit: Good point to mention the substitution, that side length a is determined by the function of y or f(b)
Firstly, I found this quadratic function: f(x) = -1/4 * (x-4) * (x+4). Then l - length, w - width; A = l * w. I saw, that w = f(l/2) = -1/16 * l^2 + 4 and A(l) = -1/16 * l^3 + 4 * l. Found the derivative: A'(l) = -3/16 * l^2 + 4 and set it equal to 0. I found maximum at l = 8/sqrt(3). Then w = 8/3. Now I will watch the video to see if I am correct.
Jens, this is such a good little problem! Only need enough calculus to be dangerous, and a little bit of a different problem from the usual optimizations! Should working out the optimal area of a rectangle inscribed between a parabola and the x-axis IN GENERAL be left as an exercise to the viewer? Or do you have that in the queue.
I got 8sqrt(3)/3 for the first side and 8/3 for the second side, this was such a fun question please bring us more questions like this, hope I got it right :)
Possible extension problem. If we keep the x-intercepts at (4,0) and (-4,0) then what would the y-intercept need to be such that the largest inscribed rectangle is in fact a square?
Obviously it is a maximum. If b were 0 or 4, then the area would be 0. By construction, we know the area we've found is positive. Ergo, it's a maximum, not a minimum.
Assuming one of the sides of the rectangle is at the x-axis, this is true. But if you were to try bending the rules, that is not required in the assignment. It just needs to be parallel to the x-axis. So one even larger rectangle would be the one found in the video, but for example with the lower side shifted by 1 in the -y direction would have sidelengths 8/sqrt(3) and 11/3. And even if one side needs to be at the x-axis itself, we can still shift the upper side to the negative half of the coordinate system. So the largest rectangle I can think of matching the assignment would have horizontal sidelength 8 (from the left to the right intersection of the parabola with the x-axis) and vertical sidelength infinity...
@mistersilvereagleEuclidean plane geometry (as implied by the definitions of our objects here) entails that distances have measurable quantity, which is not a thing that infinities do. A rectangle is definitionally an object for which geometric and algebraic relationships hold (variations on Pythagoras, etc.) that infinity renders meaningless.
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You used a and b as parametters of the quadratic function and after that you used a and b again as the lenght and wide of the rectangle. So you create confusion .Better to use 2.xfor the lenght and y for the wide of the rectangle.
You could save a little time and effort by taking advantage of properties of parabolas. The given conditions imply that the parabola is symmetric about the y-axis, so the quadratic’s other zero must be at x=4. Thus, its formula is of the form y=a(x-4)(x+4). Now adjust the scale factor a so that y=4 when x=0. Interestingly, the maximum-area rectangle has the same width regardless of the value of a, i.e., we could set the parabola’s y-intercept anywhere on the positive y-axis without changing the width of the maximal inscribed rectangle.
That was amazing! I get the area under the curve usage case, and I love that we can use the Second Derivative proof to find if that area of the function at the specific point is actually the maximum. Thank you! Will buy a hoodies or something one day, you really made this accessible and understandable for a student that needs this maths understanding fairly soon.
My attempt.
Graph 0,
(x-4)(x+4)
Make it negative
-(x-4)(x+4)
When foiled out it will equal
-x^2+16
You need to turn that 16 into a 4 so it crosses the y axis at the right point, so divide the whole function by 4
-(1/4)(x-4)(x+4)
Now the area of the rectangle is 2x(as the base of the rectangle is subscribed to the x axis) times y.
So the area of the rectangle is (2x)(-(1/4)(x-4)(x+4)
So foil it out
((-2x^3)/4)+8x
Find the derivative to find the slope
((-6x^2)/4)+8=m
When m=0, we’ll find the x coordinate where the slope is maximized or minimized.
Sqrt(32/6)=2.309… = x
Input 2.309 in the area of the rectangle.
And you get the maximum area, which is 12.31…
It makes sense as a maximum cause if you put 4 as the height, you’d have no base for the rectangle so the area would be 0, and if you put 8 at the base, the height would be 0, so the area would be 0. But somewhere in the middle is a real area
The bottom of the rectangle is 2.309*2
The height is 2.667 if we put the x value in the equation we made.
Edit:
Good point to mention the substitution, that side length a is determined by the function of y or f(b)
Good old optimization problems. Gotta love them
Thanks for the nice and easy problem to start my day!
Firstly, I found this quadratic function: f(x) = -1/4 * (x-4) * (x+4). Then l - length, w - width; A = l * w. I saw, that w = f(l/2) = -1/16 * l^2 + 4 and A(l) = -1/16 * l^3 + 4 * l. Found the derivative: A'(l) = -3/16 * l^2 + 4 and set it equal to 0. I found maximum at l = 8/sqrt(3). Then w = 8/3.
Now I will watch the video to see if I am correct.
Happy New Year Papa Flammy! Great to see you're back to uploading :)
Jens, this is such a good little problem! Only need enough calculus to be dangerous, and a little bit of a different problem from the usual optimizations!
Should working out the optimal area of a rectangle inscribed between a parabola and the x-axis IN GENERAL be left as an exercise to the viewer? Or do you have that in the queue.
I got 8sqrt(3)/3 for the first side and 8/3 for the second side, this was such a fun question please bring us more questions like this, hope I got it right :)
gj! =)
I solved such question just yesterday
Only the function was sine
It is quite interesting to get known to calc 2
Nice question! And I'm you aren't completely gone from YT.
I can't explain how excited I am after seeing papa flammy again I am jumping with joy ❤❤❤ he kept my promise.Love from India 🇮🇳🇮🇳❤️❤️
Possible extension problem. If we keep the x-intercepts at (4,0) and (-4,0) then what would the y-intercept need to be such that the largest inscribed rectangle is in fact a square?
Only one inscribed rectangle would be a square; it would trivialize the "Multivariable" issue-area would equal 4b or, in other words, -x^2+ 16.
64/sqrt(27)
I had a question like this on my math exam in december
Obviously it is a maximum. If b were 0 or 4, then the area would be 0. By construction, we know the area we've found is positive. Ergo, it's a maximum, not a minimum.
thank you sir
Assuming one of the sides of the rectangle is at the x-axis, this is true. But if you were to try bending the rules, that is not required in the assignment. It just needs to be parallel to the x-axis. So one even larger rectangle would be the one found in the video, but for example with the lower side shifted by 1 in the -y direction would have sidelengths 8/sqrt(3) and 11/3.
And even if one side needs to be at the x-axis itself, we can still shift the upper side to the negative half of the coordinate system. So the largest rectangle I can think of matching the assignment would have horizontal sidelength 8 (from the left to the right intersection of the parabola with the x-axis) and vertical sidelength infinity...
And therefore not...a rectangle
@@HighKingTurgon Why not?
@mistersilvereagleEuclidean plane geometry (as implied by the definitions of our objects here) entails that distances have measurable quantity, which is not a thing that infinities do. A rectangle is definitionally an object for which geometric and algebraic relationships hold (variations on Pythagoras, etc.) that infinity renders meaningless.
@ Of course. With "infinity" I meant take any positive real number as large as you wany
@mistersilvereagle got it. I can dig it. But surely without constraint "optimization" is a meaningless exercise.
How about for a general parabola?
There are more trivial ways to solve this in seconds
bruh I got 8sqrt(3)/3 and 8/3, must have gotten some algebra wrong along the way... don't care enough to check
First
The solution ist trivial if u solve is this way it onlshows how bad ur brain is.
bro i solved it in 3 mins
its 10th grade math
@@Pepegalordexactlyy
Nice job bro. But he is explaining the reasoning behind every step. That takes time.
@@elshons1576 facts! being a teacher is hard
@@Pepegalordrather 11th