Cut wire two shapes, circumference of circle when A is a minimum

Question

You have a wire that is 83 cm long. You wish to cut it into two pieces. One piece will be bent into the shape of a square. The other piece will be bent into the shape of a circle. Let A represent the total area of the square and the circle. What is the circumference of the circle when A is a minimum?

Answer 1

If the wire is 83 cm long then the perimeter of the square and the circumference of the circle must add to 83.

Let s = the side of the square so 4s is the perimeter. Let r = the radius of the circle so the circumference is 2πr so:

4s + 2πr = 83

or s = (83 - 2πr)/4

Shifting to area, the area of the square is s² and the area of the circle is πr² so the total area is given by:

A = s² + πr²

Substituting in "(83 - 2πr)/4" in place of "s" we get:

A(r) = [(83 - 2πr)/4]² + πr²

A(r) = (6889 - 166πr + 4π²r²)/16 + πr²

substituting in π = 3.1416:

A(r) = 430.5625 - 32.594r + 2.4674r² + 3.1416r²

A(r) = 5.609r² - 32.594r + 430.5625

This is a standard quadratic so the graph is an upright parabola meaning the vertex is a minimum. To find the vertex use -b/2a = 32.594/(2•5.609) = 2,906

So the minimum occurs when r = 2.906

The circumference of the circle is 2πr = 2π(2.906) = 18.256