Question

You have a wire that is 71 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

Answer:

The circumference of the circle when A is a minimum is 31.21 cm.

Step-by-step explanation:

If we construct a square of side "a" and a circle of diameter "D", we divide the length of the wire in the sum of both perimeters.

The perimeter of the square is

$P_s=4\cdot a$

The perimeter of the circle is

$P_c=\pi D$

The length of the wire is

$L=4a+\pi D=71$

Then we can express a in function of D:

$4a+\pi D=71\\\\a=\frac{71}{4}-\frac{\pi}{4}D$

The area A is the sum of the area of the circle and the square. We can then replace a by D as the independent variable:

$A=A_c+A_s=\frac{\pi D^2}{4}+a^2= \frac{\pi D^2}{4}+(\frac{71-\pi D}{4})^2\\\\ A= \frac{\pi D^2}{4} +\frac{71^2-2*71*\pi D+(\pi D)^2}{16}$

To optimize the area we derive and equal to 0:

$\frac{dA}{dD} =\frac{2\pi}{4} D+\frac{-142\pi+2\pi^2 D}{16}=0\\\\\frac{8\pi D-142\pi+2\pi^2 D}{16} =0\\\\\frac{\pi}{16} (8D-142+2\pi D)=0\\\\(8+2\pi)D=142\\\\D=142/(8+2\pi)=142/14.28=9.94$

The area is mimimum when the D=9.94.

Then, the circunference for this diameter is:

$P_c=\pi D=3.14*9.94=31.21$