The perimeter of a rectangle is 100 cm. Find the dimensions
of the rectangle when its area is maximum.
Solution : Let the length of the rectangle be 'x' cm
and breadth be 'y' cm.
Let its area be A sq. cm.
Now given that perimeter = 100 cm
\ 2 ( x + y) = 100
\ x + y = 50
\ y = 50 - x
Then A= xy = x (50 - x)
By Differentiating w.r.to x, we get
= 50 - 2x
and = 0
gives 50 - 2x = 0
i.e. x = 25 (critical point)
also, f "(x) = -2
Þ f "(25) = -2
By the second derivative test, A is maximum at x=25 Þ
\ when the area is maximum,
the rectangle is a square of side 25 cm.
Divide 15 into two parts such that the product of the
square of one and the cube of other is maximum.
By the second derivative test ‘f’ has a relative maximum when x = 9
Also, when x = 9
then y = 15 - 9
\ The two required parts of
15 are 9 and 6.