anyone know how to calculate the max area of a rectangle..........?

Question

the rectangle has a semicircle on each end, to make a 200 meter running track, what are the dimensions that will make area the area of the rectangular region as large as possible?

Answer 1

equations a and b are the length and width.
200 =2a+pi*b
A(max)=a*b
Amax=(200a-2a^2)/pi
differentiate set to zero to find max

a=50, b=31.82

Answer 2

Draw out the shape of the rectangle with the semi-circular ends.

Let the length of the rectangle be y and the width of the rectangle be x

The perimeter of the shape = 2y + pi* x

200 = 2y + pi*x

y = 100 - (pi/2)*x

The area of the rectangle (A) is given by y *x

Substituting for y gives A = [100 -(pi/2)*x] x

A = 100x - pi/2*x^2

A is max when dA/dx = 0 (Hope you have done calculus)

dA/dx = 100 - pi*x

100 - pi*x = 0
x = 100/pi = 31.830 = 31.8 (3 sf)

Since y =100 - (pi/2)*x

y = 100 - (pi/2) * 31.830

y = 50

Answer: The rectangle will have a max area when x = 31.8 m and y = 50 m

Answer 3

The dimensions of the rectangle will be 2r and (100-pi*r)
A = 200r - 2pi r^2
So dA/dr = 200-4*pi*r
Set above =0 and solve for r
4*pi *r = 200
r =
50/pi
The rectangle will have maximum area when the radius of semicirle is equal to 50/pi