Question

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.

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

M = 0

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

M= ?

I've tried 16.8 but its wrong.

Answer 1

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

b=30-a

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:

l=a/4

Then, the area of the square is:

$A_s=l^2=(a/4)^2=a^2/16$

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

$2\pi r=b=30-a$

$r=\frac{30-a}{2\pi}$

The area of the circle will be:

$A_c=\pi r^2=\pi (\frac{30-a}{2\pi})^2=\frac{900-60a+a^2}{4\pi}$

The total area enclosed in this two figures is:

$A=A_s+A_c=\frac{a^2}{16}+\frac{900-60a+a^2}{4\pi}$

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

$dA/da=\frac{2a}{16}+\frac{-60+2a}{4\pi}=\frac{a}{8}+\frac{a-30}{2\pi}=\frac{a\pi +4a-120}{8\pi}=0$

$a\pi+4a-120=0$

7.14a=120

a=16.8

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):

$d^2A/da^2=\frac{\pi +4}{8\pi}>0$

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:

$A=\pi r^2=\pi (\frac{30}{2\pi})^2=71.62$

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

$A=a^2/16=900/16=56.25$

The maximum value happens for a=0.