Optizimation Question

Question

Four feet of wire is to be used to form a square and a circle. How much of the wire should

be used for the square and how much should be used for the circle to enclose the

(a)maximum total area?

(b) minimum total area?

Answer 1

For a square, P = 4s, and area = s². For a circle, P = 2 π r and area = π r²

Then, if x of the wire is used for the circle, 4 - x is used for the square. Of course, 0 <= x <= 4.

If x is used for the circle, P = 2 π r , so x = 2 π r, and r = x/(2 π)

4 - x is used for the square, so P = 4s, and 4s = 4 - x, and s = 1 - x/4

Then, the area of the circle plus the area of the square is

π r² + s² = π (x/(2 π))² + (1 - x/4)² = x²/(4π) + 1 - x/2 + x²/16 =

(1/4π + 1/16)x^2 - x/2 + 1

As a check of this formula, when x = 0, all 4 ft of the wire are used for the square, each side of the square is 1, and 1 * 1 = 1.

When x = 4, (1/(4π) + 1/16)x^2 - x/2 + 1 = (1/(4π) + 1/16)(16) - 4/2 + 1 = 4/π + 1 - 2 + 1 = 4/π

All 4 ft of the wire are on the circle, so the radius is 4/(2π) = 2/π, and the area is π r² = π (2/π)² = 4/π

Then, now we find the max and min of (1/4π + 1/16)x^2 - x/2 + 1.

From our check, we know the value at the two endpoints of the interval.

This is a parabola with a positive leading coefficient, so we know that it will possibly have a global minimum in the interval, but at least one of the endpoints will be the maximum.

As this is a parabola, we may either use calculus or just complete the square.

f(x) = (1/(4π) + 1/16)x^2 - x/2 + 1 = (1/(4π) + 1/16)(x -1/4/(1/(4π) + 1/16))² - (1/(4π) + 1/16)(1/4/(1/(4π) + 1/16))² + 1 = (1/(4π) + 1/16)(x - 1/(1/π + 1/4))² - 1/(4/π + 1) + 1 =

(1/(4π) + 1/16)(x - 1/(1/π + 1/4))² + 4/(4 + π )

Then, we check that 1/(1/π + 1/4) is between 0 and 4. We know that it will be less than π, so this will be the case

x = 1.75960338595377

Then, the squared term will be zero there.

The minimum area will be 4/(4 + π ) = 0.560099153511557

The maximum area, as 4/π > 1, is 4/π