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.

Answer 1

Answer:

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:

$a+b=30\\\\b=30-a$

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:

$2\pi r=b=30-a\\\\r=\dfrac{30-a}{2\pi}$

The area of the circle will be:

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

The total area enclosed in this two figures is:

$A=A_s+A_c=\dfrac{a^2}{16}+\dfrac{900-60a+a^2}{4\pi}=\left(\dfrac{1}{16}+\dfrac{1}{4\pi}\right)a^2-\dfrac{60a}{4\pi}+\dfrac{900}{4\pi}$

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

$\left(\dfrac{1}{16}+\dfrac{1}{4\pi}\right)a^2-\dfrac{60a}{4\pi}+\dfrac{900}{4\pi}\\\\\\\dfrac{dA}{da}=\left(\dfrac{1}{16}+\dfrac{1}{4\pi}\right)(2a)-\dfrac{60}{4\pi}+0=0\\\\\\\left(\dfrac{1}{8}+\dfrac{1}{2\pi}\right)a=\dfrac{15}{\pi}\\\\\\\dfrac{\pi+4}{8\pi}\cdot a=\dfrac{15}{\pi}\\\\\\\dfrac{\pi+4}{8}\cdot a=15\\\\\\a=15\cdot \dfrac{8}{\pi+4}\approx 16.8$

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):

$\dfrac{d^2A}{da^2}=\left(\dfrac{1}{16}+\dfrac{1}{4\pi}\right)(2)-0=\dfrac{1}{8}+\dfrac{1}{2\pi}>0$

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:

$A=\pi r^2=\pi\left( \dfrac{30}{2\pi}\right)^2=\dfrac{900}{4\pi}\approx71.62$

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

$A=a^2/16=30^2/16=900/16=56.25$

The maximum value happens for a=0.