A piece of wire 30 m long is cut into two pieces. One piece is bent into a square and the other is bent into a circle. (a) How much wire should be used for the square in order to maximize the total area?m
(b) How much wire should be used for the square in order to minimize the total area? m

a) 0 m

b) 16.8 m

Step-by-step explanation:

A piece of wire, 30 m long, is cut in two sections: a and b. Then, the relation between a and b is:

The section "a" is used to make a square and the section "b" is used to make a circle.

The section "a" will be the perimeter of the square, so the square side will be:

Then, the area of the square is:

The section "b" will be the perimeter of the circle. Then, the radius of the circle will be:

The area of the circle will be:

The total area enclosed in this two figures is:

To calculate the extreme values of the total area, we derive and equal to 0:

We obtain one value for the extreme value, that is a=16.8.

We can derive again and calculate the value of the second derivative at a=16.8 in order to know if the extreme value is a minimum (the second derivative has a positive value) or is a maximum (the second derivative has a negative value):

As the second derivative is positive at a=16.8, this value is a minimum.

In order to find the maximum area, we analyze the function. It is a parabola, which decreases until a=16.8, and then increases.

Then, the maximum value has to be at a=0 or a=30, that are the extremes of the range of valid solutions.

When a=0 (and therefore, b=30), all the wire is used for the circle, so the total area is a circle, which surface is:

When a=30, all the wire is used for the square, so the total area is:

The maximum value happens for a=0.