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A piece of wire 40 units long is to be cut into two pieces. One piece will be bent to form a circle; the other will be bent to form a square. Find the minimum and maximum values of the area.

I found that $$Area(x)=\left(\dfrac x4\right)^2+\left(\frac{40-x}{2\pi}\right)^2\pi$$ and after usual process of optimization that $$Area(0)=127.324…$$ the critical value is 15.559 so $$Area(15.559)=62$$ and $$Area(40)=100$$

But in my textbook it says that the critical value is 17.596 and that $Area(17.596)=56$. I checked all my steps again and again but found no error, so who's wrong?

user181609
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  • why is the area of the square $\left(\frac{x}{4}\right)^2$? – Dr. Sonnhard Graubner Oct 07 '14 at 18:32
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    the perimeter of the square is $x$ so one side of the square is $x/4$ so area is side*side=$(x/4)^2$ – user181609 Oct 07 '14 at 18:33
  • Can you post your derivative? Because I get a critical $x$ as $x=\frac{160}{4+\pi}$ based on your correct area equation. (Which is $40-17.596$ which means they used $x$ to represent the amount bent into a circle, not a square). – BeaumontTaz Oct 07 '14 at 18:34
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    Could you insert your usual process of optimization here so we can figure out what you did wrong? – Alice Ryhl Oct 07 '14 at 18:35
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    the side length of the square is $x$ and we have $4x+2\pi r=40$ – Dr. Sonnhard Graubner Oct 07 '14 at 18:36
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    Derivative $(x/4)+(x-40)/(2\pi)$ critical value $80/(2+\pi)$ – user181609 Oct 07 '14 at 18:38
  • from here you got $r$ and the sum of the area is $x^2+\pi\left(\frac{40-4x}{2\pi}\right)^2$ – Dr. Sonnhard Graubner Oct 07 '14 at 18:38
  • @user181609, recheck your derivative for the first term. Note that if you square before taking the derivative, you can avoid using the chain rule to complicate your calculus: $(x/4)^2=x^2/16$ – BeaumontTaz Oct 07 '14 at 18:38
  • @Dr.SonnhardGraubner, his approach is identical to yours except he defined $x$ to be the perimeter so that it is the length of the wire that is made into a square. Substitute $x/4$ for $x$ in your equation (the relationship between your $x$ and his $x$), you have the exact function that he has. His error is in taking the derivative of that function. – BeaumontTaz Oct 07 '14 at 18:47
  • After checking I have critical value is 22.404 and Area(22.404)=56 – user181609 Oct 07 '14 at 18:47
  • @user181609, that is correct. It turns out that $40-22.404=17.596$. So basically, their "$x$" was the perimeter of the circle, not the perimeter of the square. With that corrected derivative, your work is correct. :) – BeaumontTaz Oct 07 '14 at 18:48

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