A rectangle is constructed with its base on the diameter of a semicircle with radius 5 cm and with two vertices on the semicircle. What are the dimensions of the rectangle with maximum area?
I'm guessing the picture is just a regular rectangle with one side being attached to a semicircle.
I'm having problems with figuring out what/how is my constraint going to look like? First I wrote it in terms of the perimeter on the semicircle, then the rectangle, and now I just have no idea. I wrote the area function as A(x) = xy.
Has anyone done this one before? The answer in the book says 10/root2 cm by 5/root2 cm.


10312012, 04:30 PM #1
Math geniuses I need your help again with this optimization problem gtfih (1.7k reps)
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10312012, 04:37 PM #2"I'm bloated, holding water, and can look a little puffy. I can have slight gyno, some acne, and at times rage and mood swings. But If you cant handle me when im bulking, you sure as hell don't deserve me when I'm shredded." Marilyn Munbro
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10312012, 04:38 PM #3

10312012, 04:39 PM #4
I didn't think of that but I'll try it. I suck at optimization problems that don't clearly give you the constraint. Like this other one says that the perimeter of 3 fences is 200 meters, that's easy, but this one just gives you the radius of a semicircle attached to a rectangle, wtf cuz...
Hello! I am very sorry that you are reading this page right now instead of posting hilarious comedy on the internet. I'm sure this issue will resolve itself soon.


10312012, 04:40 PM #5
currently doing this problem in calculus
saving spot and typing my answer
please wait.....
actually idk but the hint i can give u is the area of the semi circle is [5 pie] idk about the rectangle tho x should be 10 but the question doesn't seem clear enough to discover the Y. All of my problems specifically say the x and y boundaries are bound by blah blah blahLast edited by 1thPost; 10312012 at 04:50 PM.

10312012, 04:42 PM #6"I'm bloated, holding water, and can look a little puffy. I can have slight gyno, some acne, and at times rage and mood swings. But If you cant handle me when im bulking, you sure as hell don't deserve me when I'm shredded." Marilyn Munbro
*Alpha height CREW*
*Alberta Oil & Gas CREW*
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10312012, 04:44 PM #7

10312012, 04:44 PM #8
also brah, a quick google search gave me the following link
http://ca.answers.yahoo.com/question...3102326AACStfk
the question 19 seems to be identical to your problem, take a look at it, might help
hope I could help"I'm bloated, holding water, and can look a little puffy. I can have slight gyno, some acne, and at times rage and mood swings. But If you cant handle me when im bulking, you sure as hell don't deserve me when I'm shredded." Marilyn Munbro
*Alpha height CREW*
*Alberta Oil & Gas CREW*
*Blink 182 CREW*
*Taylor Swift CREW*
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10312012, 04:45 PM #9


10312012, 04:48 PM #10
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let x and y be the sides of the rectangle
its easy to see that (r)^2 = x^2 + y^2 eq(1) where r is the radius of the semicircle
the area of the retacngle is A=xy eq(2)
we know that y=sqrt(r^2)x^2) from(1)
(1)>(2)
A=x*sqrt(r^2)x^2)
A=sqrt(x^2(r^2)x^4)
sqrt(x^2(r^2)x^4)
A is now a A(x) , A is max when sqrt(x^2(r)^2)x^4) is max and sqrt(x^2(r^2)x^4) is max when x^2(r^2)x^4 is max
f(x)=x^2(r^2)x^4
f'(x)=2(r^2)x 4x^3 =0 (we want max value)
4x^2=2(r^2)
x^2=2(r^2)/3=2(r^2)/3
x=sqrt(2/4)*r=sqrt(2)*r/2
wich is the max value of side x...now u get max area.
if you are a calculus student justification is a bit more complex , you'd have to analize the graph of the function A(x) and justify that the value x=sqrt(2)*r/2 is not just a local maximum (it has to be a global maximum , but remember the problems restrictions.)
edit erros...Last edited by mathwizz; 10312012 at 04:57 PM.

10312012, 04:50 PM #11

10312012, 04:53 PM #12
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Whenever they are asking for the max/min of something, it's all about setting the derivative to 0.
What you need to do is find the derivative of A(x) = 2*x*y, set it equal to 0, and solve for x. Perhaps, you're confused because there are two variables?
No worries. Write y in terms of x. Equation of a circle is x^2+y^2 = 25 > y = sqrt(25x^2). Great! Now, we can take the derivative of A(x) = 2x(25x^2)^(1/2).
When you set d[A(x)]/dx = 0, and solve for x, you get x = 5/sqrt(2). Substitute that back into y = sqrt(25x^2), and you also get y = 5/sqrt(2). To bet the base, you multiply x by 2 since the rectangle will span two quadrants. Now you get 10/sqrt(2) by 5/sqrt(2) with an area of 25.

10312012, 04:55 PM #13

10312012, 05:08 PM #14
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