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Application Of Derivatives

Question

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Show that among rectangles of given area, the square has the least perimeter.

Solution

Let x, y be the sides of rectangle and k² be the area where k > 0.

therefore space space straight x space straight y space equals space straight k squared space space space space rightwards double arrow space space space space straight y space equals space straight k squared over straight x

Let

straight P space equals space 2 straight x plus 2 straight y space equals space 2 straight x plus 2 straight k squared over straight x

open square brackets because space of space left parenthesis 1 right parenthesis close square brackets

dP over dx equals 0 space space space give space us space 2 minus fraction numerator 2 straight k squared over denominator straight x squared end fraction space equals space 0 comma space space space space space space rightwards double arrow space space space straight x squared space equals space straight k squared space space space space rightwards double arrow space space straight x space space equals straight k comma space space minus straight k

Rejecting x = -k, we get x = k

fraction numerator straight d squared straight P over denominator dx squared end fraction space equals space fraction numerator 4 straight k squared over denominator straight x cubed end fraction

When

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therefore space space space space straight P space is space minimum space when space straight x space equals space straight k comma space space space space straight y space equals space straight k squared over straight k space equals space straight k space space space space straight i. straight e. space space when space rectangle space is space straight a space square.

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