Question

A wire of length 120 cm is bent in the form of a rectangle. Find its dimensions if the area of the rectangle is maximum

Answer 1

Let the length and breadth of a rectangle be x cm and y cm

∴ Perimeter of rectangle = 2(x + y) = 120 cm

∴ x + y = 60       .......(i)

Let A be the area of the rectangle.

∴ A = xy

= x(60 − x)   .......[From (i)]

= 60x − x² 

Differentiating w. r. t. x, we get

`("dA")/("d"x)` = 60 − 2x

∴ `("d"^2"A")/("d"x^2)` = −2

For maximum area, `"dA"/("d"x)` = 0

∴ 60 − 2x = 0

∴ x = 30

For x = 30,

`(("d"^2"A")/("d"x^2))_(x = 30)` = − 2 < 0

When x = 30, area of the rectangle is maximum.

and y = 60 − 30 = 30   .......[From (i)]

∴ Area of the rectangle is maximum if length = breadth = 30 cm.