i have a calc exam and i need at least a 90% to get the grade i want and a seventy something to pass the class. i struggle so hard with these optimization problems but i think i got it after this and i might just get a 100% on this exam. if i do im not forgetting all the help you gave me . thankyou, Your the bestest person in the universe
It's interesting because that's an incredible amount of math for an answer that's pretty intuitive. Of course it's important to actually prove that it's the case, but it seems like the best guess is that the largest rectangle would use a 45 degree angle, and since x and y are equal, you can see (from Pythagoras) that x^2 = r^2 - x^2, or that 2(x^2) = r^2. Since A = 2xy and x and y are equal, A must equal r^2. It's the kind of problem most people would be able to guess but not solve.
i guess this is just a hard example of demonstrating the power of calculus. of course by symmetry one would think of relating a square inscribed in a full circle that has the largest area. or to reduce the amount of calculus, polar coordinates is a better choice. after getting the formula for the rectangle's area, either calculus or trigonometry will do, and it leads directly to the 45 degree angle you mentioned.
it could be the min or the max. you wouldn't know for sure unless you tested it. i didn't show it in the video because i tested it on my own and wanted to save time in the video, but the critical point in this video is the max. :)
r is a constant value and usually it is given like 10 cm or 12 cm(etc) to not confuse you. The "r" is a constant value and yes we are taking the derivative in terms of x. If you wish you can take the derivative in terms of y but that would be the same thing with only a naming difference (you say y rather than x)
use parametric form, base = 2rcosx height= rsinx where r is radius, therefore area is 2rrsinxcosx = rrsin(2x) which has max at 2x=pi/4 ie area = rrsin(2pi/4) = rr
This can also be seen geometrically: if you look specifically at the first quadrant, viewing r as a vector with components x and y, r will maximize at 45 degrees when x=y, this is because cos(45)=sin(45). with equations r^2=x^2+y^2, x=y and A=2xy you can then solve for A=r^2.
@integralCALC- Excellent presentation particularly the references and description of optimization and constraint equations. I've always use just a subscript of '(max)' or '(min)' for the equations that I would apply d/dx to. I was helping my daughter with college Calculus 1 on a similar type problem. Three times, double checked distribution and that a negative didn't get dropped, which are common type mistakes, still got same answer all three times. I guess the online site where she input her answers didn't like the format. I must be a bit old fashioned but when did instructors get so lazy that they rely on programs to auto-correct homework problems? I was lucky; my college professors (or their assistants) actually looked at my work and pointed out where I made my mistakes in logic or arithmetic so I would LEARN TO NOT REPEAT MY ERRORS. Students learn nothing when a program gives to a green check or red 'X' as a response. Good luck on that final with this type of course interaction. My advise to students is to beware of these type of courses; and, if this is your only option, make sure your instructor has plenty of office hours to assist you with your questions.
I have watched a ton of videos on different maths, and you have done an excellent job on the calculus videos you have uploaded for us. Good job, and thanks so much. For the young ladies out there, you are a great role model, as you do not see many videos of professors for that matter that are female, so keep it up! Anyone can learn maths with determination and practice, it's not just for us dudes hahahaha! Subbed fo sho!
What if you were looking for the smallest area of a rectangle inscribed in a semicircle? I'm confused wouldn't the process be the exact same and result in the same answer?
Very good illustration, especially wit all of the algebra involved. However, that said, this type problem becomes much clearer to the CALC I student if actual numbers are assigned to the figures---in other words, a DOMAIN of (0,4)--and then working out the final answer for length and width to have actual numeric values. NONETHELESS, a very clear presentation of the multi-steps involved.
Why wouldn't you go on & divide the initial equation you wrote for the circle (x^2+y^2=r^2) by 2 to account for the fact that we're working with a semicircle on this problem instead of a whole circle?
Because if we did that, then we'd still just cancel out the 1/2 for each term by multiplying both sides by 2 when solving for y. So in the end it doesn't matter; it's just an unnecessary extra step.
it is too lengthy but if we convert it by parametric form like the coordinate (rcosx,rsinx) instead the x and y distance . it will come by length 2rcosx and breadth rsinx .multiply both to get the area of the rectangle .i.e r^2sin2x .in order to get maximum area sin2x should be 1 i.e x=45 degree. therefore max area is r^2
just one question, you know how you find the values for y and x, can you show how you would get the domain of x, because lets say the largest values were at the endpoints of the function? plz and thank you
thx for doing part of my homework. But for real, I was having a ton of trouble with this single step stuff. I'm somehow better with the more complicated problems, but I couldn't for the life of me get this. Thanks for the help.
What if you reflect the same rectangle over the x-axis and try to maximize the area of this new rectangle (the 2 rectangles added) in a full circle, will you get the maximum area to be 2r^2? Or is it wrong to just add r^2 + r^2 ?
I hope you might be able to shed some light for me, Im praying actually lol. Im in pre-cal right now and we have had some of these optimization problems show up in our homework. The issue Im having is that we have not learned about derivatives and their rules or what they even are. I have been watching videos on this and only find derivatives being used. Is their a method that would be more appropriate for myself to find the maximum and minimum values without needing to use the derivative? Any insight on this would be so helpful.
I'd have to know more about what your problems look like, but if you have a graph of the equation you're dealing with, you can always get extrema information from the graph. Or if you don't have the graph, but the equation you're given is something you easily graph (like a quadratic, which turns into a parabola when you graph it), then you can find the extrema. Those are just some ideas off the top of my head! :)
Thank you so much for replying back, I was thinking that there might have been a different way of going about showing the work than with derivatives, but graphing the extreme is definitely possible at this point, I appreciate your help!
In order for it to be a semicircle, the radius has to be the same everywhere, and that wouldn't be the case for any parabolic function. So while it doesn't kind of look like a semicircle, technically it isn't. :)
How come at 9:57 when you set the equation to zero and simplify from there, it doesn't work? I tried it but ended up getting x=0. Can someone explain? Or was the x=0 probably due to algebraic errors?
I'm not solving optimization or anything, but I just want to know why length is = 2x and not just x. I know the area of a rectangle is Length x Width so I don't conceptually understand why length is 2x and my math teacher has been pretty useless just telling me to refer to the book.
When you've been asked to find the largest area, that means you need to maximize area. In order to find maximum area (or maximum or minimum anything), the procedure is to take the derivative of what you're trying to optimize, and then set that derivative equation equal to 0. That allows you to find critical points, which are the only points where the value can potentially be a maximum or minimum.
I think about it as a square inside a whole circle, which is the biggest area because is a square, then use basic unit circle stuff to the the exact dimensions.
In my calc book it asks me to find the dimensions of the rectangle that can be inscribed in a semi circle with a radius r what would be the difference for finding the dimensions.
since we are looking for the critical points from our derivative equation we set it equal to zero. and since in a fraction it only takes for the numerator to be zero in order to have a zero value we can ignore the denominator bc if we have 0/(r-x) it'll be 0. also if you wanted to you could go and find what makes the denominator 0 which ends up being 3. meaning there are vertical asymptotes at -3 & 3. but we already know this bc in the problem the radius is 3 which is the limit of our semi circle.
She multiplied by -2x which made the whole second term negative. She could have written +(-2x^2.....) but it was faster and easier to just write -(2x^2....).
Hi Orlando! Thanks for subscribing. I know I haven't published as many videos lately. I have a couple of projects I'm working on at the moment, but I do plan to upload many more!
Hands down best video on UA-cam that explains optimizing a rectangle inscribed in a semi-circle.
You have absolutely saved me!! My professor loves to assign problems like these without any clear explanation as to how to solve. Thanks
+2407Summer You're welcome, so glad the videos are helping!
I agree
i have a calc exam and i need at least a 90% to get the grade i want and a seventy something to pass the class. i struggle so hard with these optimization problems but i think i got it after this and i might just get a 100% on this exam. if i do im not forgetting all the help you gave me . thankyou, Your the bestest person in the universe
Good luck on the exam!
He failed, he got 37%
It's interesting because that's an incredible amount of math for an answer that's pretty intuitive. Of course it's important to actually prove that it's the case, but it seems like the best guess is that the largest rectangle would use a 45 degree angle, and since x and y are equal, you can see (from Pythagoras) that x^2 = r^2 - x^2, or that 2(x^2) = r^2. Since A = 2xy and x and y are equal, A must equal r^2.
It's the kind of problem most people would be able to guess but not solve.
this has been published 10 years ago. I needed it for college in 2022....lol
Thanks by the way . I will pray for you
Thanks for letting me know! I'm so glad I'm able help! :D
Thank you so much I am doing the same problem in my text and I could't understand it. Clear, concise, and perfect! Thank you for your help!!!
Every view should've been a like on the video! Super underrated! Thank you so much! Instant sub!
Glad you like the videos! I use a tablet and the chalkboard is just an image I write on. :)
Thank you for explaining everything step-by-step.
You're welcome, Nicholas, I'm so glad it helped! :)
i guess this is just a hard example of demonstrating the power of calculus. of course by symmetry one would think of relating a square inscribed in a full circle that has the largest area. or to reduce the amount of calculus, polar coordinates is a better choice. after getting the formula for the rectangle's area, either calculus or trigonometry will do, and it leads directly to the 45 degree angle you mentioned.
That is a beautiful solution. Love it.
This helped me on my homework! Thank you
This was a question that showed up on my Calculus homework this weekend. Thank you for helping me get through it!
You're welcome, I'm so glad it helped! These optimization problems can be tricky. :)
it could be the min or the max. you wouldn't know for sure unless you tested it. i didn't show it in the video because i tested it on my own and wanted to save time in the video, but the critical point in this video is the max. :)
I use a bunch of different books that I've collected over the last couple of years, plus online resources and making up my own problems. :)
Thanks!
great explanation, thank you!
r is a constant value and usually it is given like 10 cm or 12 cm(etc) to not confuse you. The "r" is a constant value and yes we are taking the derivative in terms of x. If you wish you can take the derivative in terms of y but that would be the same thing with only a naming difference (you say y rather than x)
use parametric form, base = 2rcosx height= rsinx where r is radius, therefore area is 2rrsinxcosx = rrsin(2x) which has max at 2x=pi/4 ie area = rrsin(2pi/4) = rr
Awesome! So glad I could help! :)
This can also be seen geometrically: if you look specifically at the first quadrant, viewing r as a vector with components x and y, r will maximize at 45 degrees when x=y, this is because cos(45)=sin(45). with equations r^2=x^2+y^2, x=y and A=2xy you can then solve for A=r^2.
awesome! thank you so much, i'm so glad you like my videos!! :D
This video honestly helped me so much.
TBH, I think that you should have more followers than Taylor Swift... You have earned a subscriber!!!
Milky Custard Thank you very much!
Thanks a lot , this was really helpful and helped me to understand deeply about this section :)
You're welcome, Station! I'm so glad it helped! :D
gr8 video , your videos are also helpful for engg students like me.
@integralCALC- Excellent presentation particularly the references and description of optimization and constraint equations. I've always use just a subscript of '(max)' or '(min)' for the equations that I would apply d/dx to.
I was helping my daughter with college Calculus 1 on a similar type problem. Three times, double checked distribution and that a negative didn't get dropped, which are common type mistakes, still got same answer all three times. I guess the online site where she input her answers didn't like the format.
I must be a bit old fashioned but when did instructors get so lazy that they rely on programs to auto-correct homework problems? I was lucky; my college professors (or their assistants) actually looked at my work and pointed out where I made my mistakes in logic or arithmetic so I would LEARN TO NOT REPEAT MY ERRORS. Students learn nothing when a program gives to a green check or red 'X' as a response. Good luck on that final with this type of course interaction.
My advise to students is to beware of these type of courses; and, if this is your only option, make sure your instructor has plenty of office hours to assist you with your questions.
I have watched a ton of videos on different maths, and you have done an excellent job on the calculus videos you have uploaded for us. Good job, and thanks so much. For the young ladies out there, you are a great role model, as you do not see many videos of professors for that matter that are female, so keep it up! Anyone can learn maths with determination and practice, it's not just for us dudes hahahaha! Subbed fo sho!
What if you were looking for the smallest area of a rectangle inscribed in a semicircle? I'm confused wouldn't the process be the exact same and result in the same answer?
😱 . If i had the money, I have hired you already from the beggining of my academic life. So much thanks for this.
You're welcome, so glad it helped! :)
Very good illustration, especially wit all of the algebra involved. However, that said, this type problem becomes much clearer to the CALC I student if actual numbers are assigned to the figures---in other words, a DOMAIN of (0,4)--and then working out the final answer for length and width to have actual numeric values. NONETHELESS, a very clear presentation of the multi-steps involved.
You are amazing! And you should know that you are my Calculus lecturer :)...Keep up the good service! I like your smile too; its magical!
Does this mean that any problem asking for the area of a rectangle in a semicircle is generally expressed A=r^2?
Splendid and many thanks indeed
But i would like to know what if the question was to find the smallest area of the in a circle inscribed rectangle
I love math geniuses and I love u ... U saved me
Why wouldn't you go on & divide the initial equation you wrote for the circle (x^2+y^2=r^2) by 2 to account for the fact that we're working with a semicircle on this problem instead of a whole circle?
+Hayley V.W.
It already is a semicircle when you solved for Y.
You took the positive root
Because if we did that, then we'd still just cancel out the 1/2 for each term by multiplying both sides by 2 when solving for y. So in the end it doesn't matter; it's just an unnecessary extra step.
Enjoyed watching this video. As I said recently in another calculus video of yours, these are excellent refreshers!
it is too lengthy but if we convert it by parametric form like the coordinate (rcosx,rsinx) instead the x and y distance . it will come by length 2rcosx and breadth rsinx .multiply both to get the area of the rectangle .i.e r^2sin2x .in order to get maximum area sin2x should be 1 i.e x=45 degree. therefore max area is r^2
THANK YOU FOR SAVING ME.
you are helping my cramming for my final tomorrow so much
you're welcome! good luck on your final, i hope it goes great!!
you're very sweet! i'm glad the videos are helping!! :D
Excellent explanation!!
This is the just greatest explanation. Thanks =)
You're welcome, I'm glad you liked it!
you're welcome!! :D
just one question, you know how you find the values for y and x, can you show how you would get the domain of x, because lets say the largest values were at the endpoints of the function? plz and thank you
Thank you, you are the ISH!!!
Hi! Are we taking the derivative of the area in terms of "x"? I mean A'(X). Also why is "r" treated as a constant?
Thanks!
You are an angel. Thank you so much!
thx for doing part of my homework. But for real, I was having a ton of trouble with this single step stuff. I'm somehow better with the more complicated problems, but I couldn't for the life of me get this. Thanks for the help.
You're welcome, I'm glad it helped! :)
Just wondering, which would have more area, a square inscribed in a semi-circle, or 2 adjacent squares inscribed in a semicircle?
What if you reflect the same rectangle over the x-axis and try to maximize the area of this new rectangle (the 2 rectangles added) in a full circle, will you get the maximum area to be 2r^2? Or is it wrong to just add r^2 + r^2 ?
You just helped me finish my math homework! Suuuubbbed & thank you :D
+blublast3 Awesome! Thank you so much!!
You saved me! Many thanks!
You're welcome!! I'm glad the videos were helpful!! :D
I had this exact question on my final exam for calc one, except I was given the length of the radius, and asked for the dimensions, rather than area.
Yeah, I'm literally doing that right now. My radius is 2 lmfao. I got l = 2sqrt(2) & w = sqrt(2), l and w are interchangeable of course.
I hope you might be able to shed some light for me, Im praying actually lol. Im in pre-cal right now and we have had some of these optimization problems show up in our homework. The issue Im having is that we have not learned about derivatives and their rules or what they even are. I have been watching videos on this and only find derivatives being used. Is their a method that would be more appropriate for myself to find the maximum and minimum values without needing to use the derivative? Any insight on this would be so helpful.
I'd have to know more about what your problems look like, but if you have a graph of the equation you're dealing with, you can always get extrema information from the graph. Or if you don't have the graph, but the equation you're given is something you easily graph (like a quadratic, which turns into a parabola when you graph it), then you can find the extrema. Those are just some ideas off the top of my head! :)
Thank you so much for replying back, I was thinking that there might have been a different way of going about showing the work than with derivatives, but graphing the extreme is definitely possible at this point, I appreciate your help!
Thank you so much for doing this video!
You're welcome! :D
Very helpful! Thank you for posting:)
Glad I could help! :)
Yes yes yes it worked! I graphed where radius was 8, and I got the thing you had! Eyy Thansk so much!
You're welcome, I'm so glad it worked!! :D
Krista King :)
Awesome lesson!
Thank you so much!
Thank you for saving my math grade
LOL, you're welcome, but I'm pretty sure YOU saved it!!
Could we consider the part of a parabola above the x axis as a semi circle( the parabola is concave down) ?
In order for it to be a semicircle, the radius has to be the same everywhere, and that wouldn't be the case for any parabolic function. So while it doesn't kind of look like a semicircle, technically it isn't. :)
ayyeee! thank you so much! this helped a lot on my calc homework!
Thanks for the videos. Love it
So helpful!! Thank-you so much :D
Glad it could help!
Awesome !! Thank you so much 😊
How come at 9:57 when you set the equation to zero and simplify from there, it doesn't work? I tried it but ended up getting x=0. Can someone explain? Or was the x=0 probably due to algebraic errors?
how would you approach the same type of problem when you're given a value for the radius?
dude, you do the same thing but just plug the value of radius for r.
I love your videos :). What program do you use for the chalkboard and do you use a drawing pad?
You are very nice.
I'm not solving optimization or anything, but I just want to know why length is = 2x and not just x. I know the area of a rectangle is Length x Width so I don't conceptually understand why length is 2x and my math teacher has been pretty useless just telling me to refer to the book.
great vid, thanks for the effort
Oh, the chalkboard has a very nice effect! Very unique XD.
Hi, could you please explain why we are taking the derivative of A and why we later set the equation to zero?
Thank you!
When you've been asked to find the largest area, that means you need to maximize area. In order to find maximum area (or maximum or minimum anything), the procedure is to take the derivative of what you're trying to optimize, and then set that derivative equation equal to 0. That allows you to find critical points, which are the only points where the value can potentially be a maximum or minimum.
Krista King oh. That is why. thank you!
Yuhsin C its an application for minimum and maximum of calculus.its a process of solving.
I think about it as a square inside a whole circle, which is the biggest area because is a square, then use basic unit circle stuff to the the exact dimensions.
@TheIntegralCALC if we could duplicate you and make you a teacher all around this world, a lot of students would understand higher mathematics
In my calc book it asks me to find the dimensions of the rectangle that can be inscribed in a semi circle with a radius r what would be the difference for finding the dimensions.
8 years later and still saving calc grades :)
can you please post some vedio about functions? please please please, and thanks! nice and helpful video.
this might help! Calculus - Functions
integralCALC oops,thank you s much!
I am still confused about where the denominator went when we set it equal to 0, we can't just get rid of it, can we?
since we are looking for the critical points from our derivative equation we set it equal to zero. and since in a fraction it only takes for the numerator to be zero in order to have a zero value we can ignore the denominator bc if we have 0/(r-x) it'll be 0. also if you wanted to you could go and find what makes the denominator 0 which ends up being 3. meaning there are vertical asymptotes at -3 & 3. but we already know this bc in the problem the radius is 3 which is the limit of our semi circle.
thanks
Wait so I could basically finish this in 10 seconds if I just know the radius and square it to get the max area
Thank you!:D
You're welcome, Bula! :)
I have to say that whoever came up with calculus must've been damn geniuses ._.
at 9:58 why is there a minus sign instead of a plus sign?
She multiplied by -2x which made the whole second term negative. She could have written +(-2x^2.....) but it was faster and easier to just write -(2x^2....).
I didn`t catch that one. Thanks!
Please explain this diagram
excellent!
wow super helpful
What if they give me the radium?
Hi. Your video is helpful and it is worth watching. Can you please upload more videos? I have already subscribed to your channel.
Hi Orlando! Thanks for subscribing. I know I haven't published as many videos lately. I have a couple of projects I'm working on at the moment, but I do plan to upload many more!
Krista King You are always welcome
Awww thanks!! :D
128√e0
i can be on youtube!! :D
awesome video
Thx alot
:)
You got some brain there, girl.
If it wasn't for integralCALC I wouldn't pass xD Love you and thank you Krista!!!!! :p
Test in 1.5 days! Gotta pass! ^_^
Study hard! And good luck. :)
Thank you! :)
Exceptional :-)
Thanks! :)
You are a total babe. and i love watching you videos. You got me through calculus! thanks math goddess
thanks babe
love it