Application - Optimization

Question

A wire of length 12 is cut into two pieces which are then bent into the shape of a circle of radius r and a square of side s. Then the total area enclosed by the circle and square is the following function of s and r

If we solve for s in terms of r, we can reexpress this area as the following function of r alone:

Thus we find that to obtain maximal area we should let r=

To obtain minimal area we should let r=

Answer 1

The first sentence translates into a constraint equation

12 = L = 2πr + 4s.

The possible ranges are thus 0 < r = (L – 4s)/2π < L/2π

and 0 < s = L/4 – πr/2 < L/4.

The area function should be

A(r,s) = πr² + s².

Given the constraint, we can write

a(r) = A(r,s(r))

= πr² + (L/4 – πr/2)²

= (π + π²/4)r² – πLr/4 + L²/16

that has a minimum as its derivative

da/dr = 2(π + π²/4)r – πL/4 → 0

vanishes that is, as r → L/(8 + 2π)).

Its local maxima are a(0) = L²/16 and a(L/2π) = L²/4π, between which a(L/2π) is the greater.