Calculus Applied Optimization sec 4.7

Question

Consider a wire of length 11ft. The wire is to be cut into two pieces of length x and 11-x. Suppose the length x is used to form a circle of radius r and the length 11-x is used to form a square with side length s. What value of x will minimize the sum of their areas?

x=____ ft

The minimum sum of the areas is then: _____ ft^2

Answer 1

Circumference of the circle = 2πr = x

So, r = x/(2π) and A_c = πr² = πx²/(2π)² = x²/(4π)

Each side of the square, S, is (11-x)/4

So, A_s = S² = ((11-x)/4)² = (11-x)²/16

This makes the sum of the areas

A(x) = A_c + A_s = x²/(4π) + (11-x)²/16

Taking the derivative

A '(x) = x/(2π) + 2(11-x)(-1)/16

Setting this to zero and solving for x:

x/(2π) = (11-x)/8 ===> 8x = 22π - 2πx ==> 4x = 11π - πx

x = 11π/(4+π) ft

Minimum sum of Areas, A = (11π/(4+π))²/(4π) + (11 - 11π/(4+π))²/16 ft²

Which can be simplified

Answer 2

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