Calculus I Optimization

Question

 A wire of length 15 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

Since the length of 15 is divided between the circumference of the circle and the perimeter of the square,

 

2πr + 4s = 15 ⇒ s = (15 - 2πr ) / 4

 

A = πr² + s² = πr² + [ (15 - 2πr ) / 4]² = πr² + (225 - 60πr + 4π²r²) / 16

 

A = π(1 + π/4)r² - (15/4)πr + 225/16

 

dA/dr = 2π(1 + π/4)r - (15/4)π = 0 ⇒ r = (15/4)π / [2π(1 + π/4)] = 15 / (8 + 2π)

 

Since A(r) is concave up, it has a minimum at

 

r = 15 / (8 + 2π)

 

Since  s = (15 - 2πr ) / 4 and s > 0, the maximum r is 15/(2π).  This same r will also maximize the area.