# A Rectangle Has Area 50 Cm2 . Find Its Dimensions When Its Perimeter is the Least - Mathematics and Statistics

A rectangle has area 50 cm2 . Find its dimensions when its perimeter is the least

#### Solution

Let x cm and y cm be the length and breadth of the rectangle.Then its area is xy = 50

y=50/x

Perimeter of the rectangle= 2(x + y)

= 2(x+50/x)

Let f(x)=2(x+50/x)

then  f'(x)=2(1-50/x^2) and f''(x)=2(100/x^3)

if f'(x)=0,then 2(1-50/x^2)=0

x^2=50

x=+-5sqrt2

But x can not be negative and hence x =5sqrt2

and f''(5sqrt2)=200/(5sqrt2)^3>0

f has a minimum value at x = 5sqrt2

For x=5sqrt2, y=50/(5sqrt2)=5sqrt2

x=5sqrt2cm,y=5sqrt2cm

Concept: Maxima and Minima - Introduction of Extrema and Extreme Values
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