Perimeter = 40 m
2(l + b) = 40 ⇒ l + b = 20
Let l = x m
b = (20 – x)m
Area = l × b = x(20 – x) = 20x – x2
To find the maximum area
A(x) = 20x – x2
A'(x) = 20 – 2x
A”(x) = -2
A'(x) = 0 ⇒ 20 – 2x = 0
⇒ x = 10
x = 10 is a maximum point
Therefore, maximum area = x(20 – x)
= 10(20 – 10)
= 10 × 10 = 100 sq.m.