A piece of wire 40 cm long is to be cut into two pieces. One piece will be bent to form a circle; the other will be bent to form a square.

(a) Find the lengths of the two pieces that cause the sum of the area of the circle and the area of the square to be a minimum.

(b) How could you make the total area of the circle and the square a maximum

a) 17.6cm and 22.4cm

b) the total area of the circle and the square can be maximum by differentiating and equating it to zero

Step-by-step explanation:

The length of the wire = 40

Let x be the circumference of the circle.

x = 2πr

r = x/2π (r = radius)

Let the perimeter of the square = (40-x)

4L = 40-x

Where L = lenght

L = (40-x) / 4

Area of a circle = πr^2

A = π (x/2π) ^2

A = π (x^2 / (4π) ^2)

A = x^2 / 4π

Area of a square = L^2

= [ (40-x) / 40]^2

= (x^2 - 80x + 1600) / 16

Total area = area of circle + area of square

= x^2/4π + (x^2 - 80x + 1600) / 16

= 0.0796x^2 + 0.0625x^2 - 5x + 100

A = 0.1421x^2 - 5x + 100

Differentiate A with respect to x

dA/dx = 0.2842x - 5

Total area is minimum when dA/dx = 0

0.2842x - 5 = 0

0.2842x = 5

x = 5/0.2842

x = 17.6cm

The circumference of the circle is 17.6cm

40-x = 40 - 17.6 = 22.4cm

The perimeter of the square is 22.4cm.

b) the total area of the circle and the square can be maximum by differentiating and equating it to zero