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
