A rectangle has a perimeter of 20m. Let x m be the length of one side.

Question

A rectangle has a perimeter of 20m. Let x m be the length of one side. Find a formula for the area A of the rectangle in terms of x. Hence find the maximum area of A

Answer 1

Hi Melanie,

 

Let the other side of the rectangle be x. Then,

2(x+y) = 20

x+y = 10

y = 10 - x  (we express the other side in terms of x)

A = x(10-x) = -x^2 +10x

A will be maximized when the rectangle has equal sides, hence a square. Remember that every square is also a rectangle.

maximum area is 5*5 = 25 m^2

Answer 2

If the length of one side of the rectangle is x, then the side opposite and parallel to it also has a length of x. Since the perimeter of the rectangle is 20, then the other two sides have a measure of (20 - 2x), and each one has a measure of (20 - 2x)/2 = 10 - x.

 

The area of a rectangle = length x width = x(10 - x) = 10x - x²

 

The maximum area can be found by computing the derivative of the area formula

 

f(x) = 10x - x²

f'(x) = 10 - 2x

 

Find the critical points. Since f'(x) is defined for all real x, then the critical points occur where f'(x) = 0

 

10 - 2x = 0

 

10 = 2x

 

x = 5

 

So the other side = 10 - x = 10 - 5 = 5

 

The maximum area is 5 x 5 = 25 m²