A piece of wire 16 m long is cut into two pieces. One piece is bent into a square and the other is bent into a circle

Question

a) How much wire should be used for the square in order to maximize the total area?

b) How much wire should be used for the square in order to minimize the total area?

Answer 1

Setup the equations that define this problem

x + y = 16

y = 2πr

s = x/4

A = s² + πr²

1. A, total area
2. x, length of wire used to make square
3. y, length of wire used to make circle
4. r, radius of circle made with wire y
5. s, length of side of square made with wire x

Now, let's setup the equation A(x) by combining and substituting other equations.

A = (x/4)² + π[(16 - x)/(2π)]²

Now, let's take the derivative of this equation with respect to x.

dA/dx = x/8 - (16 - x)/(2π)

Set the equation equal to zero and solve for x.

0 = x/8 - 8/π + x/(2π)

8/π = x/8 + x/(2π)

16 = πx/4 + x

64 = πx + 4x

32 = x(π + 4)

64/(π + 4) = x

x = 8.96

Now, let's find the area at our local extrema and the boundary conditions

x = 0, x = 8.96, x = 16

A(0) = 64/π = 20.37

A(8.96) = 8.96

A(16) = 16

a) x = 0 m

b) x = 8.96 m