Question

A metal wire of 36 cm long is bent to form a rectangle. By completing the following activity, find it’s dimensions when it’s area is maximum.

Solution: Let the dimensions of the rectangle be x cm and y cm.

∴ 2x + 2y = 36

Let f(x) be the area of rectangle in terms of x, then

f(x) = `square`

∴ f'(x) = `square`

∴ f''(x) = `square`

For extreme value, f'(x) = 0, we get

x = `square`

∴ f''`(square)` = – 2 < 0

∴ Area is maximum when x = `square`, y = `square`

∴ Dimensions of rectangle are `square`

Answer 1

Let the dimensions of the rectangle be x cm and y cm.

∴ 2x + 2y = 36

∴ x + y = 18

∴ y = 18 – x

Let f(x) be the area of rectangle in terms of x, then

f(x) = = xy = x(18 – x) = 18x – x²

∴ f'(x) = 18 – 2x

∴ f''(x) = – 2

For extreme value, f'(x) = 0, we get

18 – 2x = 0

∴ 18 = 2x

∴ x = 9

∴ f''(9) = – 2 < 0

∴ Area is maximum when x = 9, y = 18 – 9 = 9

∴ Dimensions of rectangle are 9 cm × 9 cm.