Question

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

Answer 1

Answer:

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