Question

A 30-in. piece of string is cut into two pieces. One piece is used to form a circle and the other to form a square. How shouls the string be cut so that the sum of the areas is a minimum? Round to the nearest tenth, if necessary.

Answer 1

Let x be the amount used to form the circle, and 30-x is used to form the square. The total enclosed area is
A = pi x^2 + (30-x)^2
= (pi+1)x^2 -60 x + 900
Differentiate that with respect t x and set the derivative equalo to zero.
2 (pi+1)x -60 = 0
x = 30/(1 + pi) inches

Check my work.

Answer 2

Thanks drwls. I came to the same conclusion after racking my brain over it for some time. I made the problem more difficult than it really was. Thanks again for responding.

Answer 3

I made two rather bad mistakes. The circular area that can be made piece of length x is
pi * [x/(2 pi)]^2 and the square area that can be enclosed with length (30-x) is
[1/4)(30-x)]^2 The total is
A = x^2/(4 pi) + [7.5 - (x/4)]^2
= x^2[1/(4 pi) + 1/16] -3.75x +7.5^2

x [(1/(2 pi) + 1/8] = 3.75
x[1/pi + 1/4] = 7.5
x(4/pi + 1) = 30
x = 30/[(4/pi) + 1]

Tha maximum area is formed when all of string is used to form a circle. This is a minimum area that can be formed.
Set dA/dx = 0

Answer 4

This used to be one of my favourite max/min type of questions when I taught this stuff a long time ago.

Here is a slightly different approach from the solution done by drwls.

let the radius of the circle be r,
then the side of the square is (30-2(pi)r)/4 = (15-(pi)r)/2

A = [(15-(pi)r)/2]^2 + (pi)r^2

finding A' and setting this equal to zero gives r = 15/(pi+4)

then 2(pi)r, the length needed for the circle is 2(pi)(15/(pi+4)) = 30pi/(pi+4) = 13.2 , the same result as drwls