A piece of wire 26 m long is cut into two pieces. One piece is bent into a square and the other is bent into a circle.

Question

(a)

How much wire (in m) should be used for the square in order to maximize the total area?

(b)

How much wire (in m) should be used for the square in order to minimize the total area?

Answer 1

Mark did a fantastic job figuring out the procedure to determine radius (but area should be (13-2s)²/π)

what is going to happen is the second derivative will be positive meaning critical point found from first derivative indicated local minimum. Will need to check end points to find maximum.

dA/ds=2s (1+4/π)-52/π using A=s²+(1/π)(13-2s)²

dAds=0 when s=52/(2π+8)=3.64approx.,,,,,A=23.664approx

when s=26/4, A=(26/4)²=42.5

when s=0, A=13²/π=53.794

max area when s=0

min area when s=52/(2π+8)

Answer 2

Draw and label a diagram!!

s repreents the side of the square

s² represents the area of the square

26 - 4s represents the circumference of the circle

(13 - 2s) / π = radius

(13 - 2s)² repressents the area of the circle

A(s) = s² + (13 - 2s)²

Take first deerivative to determine maximum.