Minimize Combined Area Cut Wire

Question

A piece of wire of length 62 is cut into two pieces. One piece is bent into a square and the other is bent into a circle. If the sum of the areas enclosed by each part is a​ minimum, what is the length of each​ part?

To minimize the combined​ area, the wire should be cut so that a length of ( balnk) is used for the circle and a length of (blank) is used for the square.

Answer 1

To solve this problem, let's think of the total area as a function of the length of the wire used to make the square, x.

Perimeter of square = x

Area of square = (x/4)^2

Circumference of circle = 2*pi*r = 62 - x, so r = (62 - x)/(2*pi)

Area of circle = pi*r^2 = pi*((62 - x)/(2*pi))^2

Total Area = (x/4)^2 + pi*((62 - x)/(2*pi))^2 = x^2/16 + (62 - x)^2/(4*pi)

To minimize this area function, we can take the derivative and set the derivative equal to 0 (this tells us where the function as a minimum or maximum point). We can use power rule and chain rule to take this derivative.

d/dx Total Area = 2x/16 - 2*(62 - x)/(4*pi)

d/dx Total Area = 0

0 = 2x/16 - 2*(62 - x)/(4*pi)

2x/16 = 2*(62 - x)/(4*pi)

8*pi*x = 32*(62 - x)

pi*x = 4*(62 - x)

pi*x = 248 - 4x

4x + pi*x = 248

x = 248/(4 + pi)

x is approximately equal to 34.73

Does this value of x minimize the area, or maximize the area? One way that we can check is by taking the second derivative of the area function:

d^2/dx^2 Total Area = 2/16 + 2/(4*pi) = (4 + pi)/(8*pi)

(4 + pi)/(8*pi) > 0

Because the second derivative is always greater than 0, we know that the derivative is always increasing, and thus the critical point that we identified, x = 248/(4 + pi), must be a minimum value of the area function.

This tells us that the wire should be cut at a length of 248/(4 + pi), or approximately 34.73, so that the perimeter of the square is 34.73 and the circumference of the circle is 27.27.

Hope this is helpful! Please let me know if you are interested in setting up a session to discuss further questions.

Answer 2

Since you make two shapes with the wire you need to sum of the perimeters to equal the total length of wire.

Perimeter of square = 4 x and perimeter (circumference) of circle = 2πr so the constraint is

4x +2πr =62 so x = (62-2πr)/4 = (31-πr)/2

Since we need to minimize area. We can express the combined area in terms of one of the variables

(In this case I used r so it is less messy with the fractions)

A = x² + πr² = (31-πr)²/4 +πr² = (961-62πr+π²r²+4πr²)/4 so A(r) = [961-62πr+ (π²+4π)r²]/4

which is minimized when A'(r) = [-62π+2(π²+4π)r]/4 =0

You should get with some algebra that r = 31/(π²+4π) so you can use the value to get the two pieces of wire equal to 2πr and 4x = 62-2πr