Question

A piece of wire of length 5050 is​ cut, and the resulting two pieces are formed to make a circle and a square. Where should the wire be cut to ​(a) minimize and ​(b) maximize the combined area of the circle and the​ square?

Answer 1

Answer:

x should be cut at 2221.5 to minimize the total combined area, and at 5050 to maximize it.

Step-by-step explanation:

Let x be the length of wire that is cut to form a circle within the 5050 wire, so 5050 - x would be the length to form a square.

A circle with perimeter of x would have a radius of x/(2π), and its area would be

$A_c = \pir^2 = \pi (\frac{x}{2pi})^2 = \frac{x^2}{4\pi}$

A square with perimeter of 5050 - x would have side length of (5050 - x)/4, and its area would be

$A_s = (\frac{5050 - x}{4})^2 = \frac{(5050 - x)^2}{16}$

The total combined area of the square and circles is

$\sum A = A_c + A_s = \frac{x^2}{4\pi} + \frac{(5050 - x)^2}{16}$

To find the maximum and minimum of this, we just take the 1st derivative, and set it to 0

$A' = \frac{2x}{4\pi} + \frac{-2(5050-x)}{16} = 0$

$\frac{x}{2\pi} - \frac{5050 - x}{8} = 0$

Multiple both sides by 8π and we have

$4x - 5050\pi + x\pi = 0$

$x(4 + \pi) = 5050\pi$

$x = \frac{5050\pi}{4 + \pi} = 2221.5$

At x = 2221.5:

$A = \frac{x^2}{4\pi} + \frac{(5050 - x)^2}{16}$ = 392720 + 500026 = 892746 [/tex]

At x = 0, $A = 5050^2/16 = 1593906$

At x = 5050, $A = \frac{5050^2}{4\pi} = 2029424$

As 892746 < 1593906 < 2029424, x should be cut at 2221.5 to minimize the total combined area, and at 5050 to maximize it.