Question

A rectangular plot is enclosed by 200 m of fencing and has an area of A square metres. Show that :

a) A= 100x-x^2 where x m is the length of one of its sides
b) the area is maximised if the rectangle is a square

Answer 1

Let x be one side and l be the other side of the plot.
Perimeter = 2(x + l)
200 = 2(x + l)
l = 100 - x

Area of this rectangle:
A = (x)(l)
A = x(100 - x)
A = 100x - x²

To find the maximum value of the area, we differentiate with respect to x and equate to 0
dA/dx = 100 - 2x
0 = 100 -2x
x = 50
l = 100 - 50 = 50
This means the area will be maximized when the sides of the rectangle are equal; that is, when it is a square.

Answer 2

The perimeter of plot is 200 m. Denote sides of plot with x and y.


`P=2(x+y) \\200=2(x+y) \\100=x+y \\y=100-x`

The area of plot is
`A=xy`

a)
`A=xy=x(100-x)=100x-x^2`

b) The area is maximal for A'=0

`A'=(100x-x^2)'=100-2x`

`100-2x=0 \\100=2x \\x=50 \\ \\y=100-x \\y=100-50 \\y=50`

The sides of rectangle are x=50 and y=50. Therefore, the area is maximal if the rectangle is a square.