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