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A rectangle has area 50 cm^{2} . Find its dimensions when its perimeter is the least

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#### 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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