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Application Of Derivatives

Question

Wired Faculty App

Show that the rectangle of maximum perimeter which can be inscribed in a circle of radius is a square of side $straight a square root of 2.$

Solution

Let O be the centre of circle of radius a. Let ABCD be the rectangle inscribed in the circle such that AB = x, AD = y
Now,

AB squared plus BC squared space equals space AC squared

therefore space space space space space space straight x squared plus straight y squared space equals space 4 straight a squared therefore space space space space space space space space space space space space space space space straight y squared space equals space 4 straight a squared minus straight x squared therefore space space space space space space space space space space space straight y space equals space square root of 4 straight a squared minus straight x squared end root space space space space space space space space space space space space space space space space... left parenthesis 1 right parenthesis

Let P be the perimeter of rectangle

therefore space space space straight P space equals 2 straight x plus 2 straight y

therefore space space straight P space equals 2 straight x plus 2 square root of 4 straight a squared minus straight x squared end root

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

therefore space space space space dP over dx equals 2 plus 2 open parentheses fraction numerator negative 2 straight x over denominator 2 square root of 4 straight a squared minus straight x squared end root end fraction close parentheses therefore space space space space space space dP over dx space equals space 2 minus fraction numerator 2 straight x over denominator square root of 4 straight a squared minus straight x squared end root end fraction space space space space space space space space dP over dx space equals space 0 space space space space gives space us space space space 2 minus fraction numerator 2 straight x over denominator square root of 4 straight a squared minus straight x squared end root end fraction space equals 0 rightwards double arrow space space space straight x space equals space square root of 4 straight a squared minus straight x squared end root space space space space space space space rightwards double arrow space space space space straight x squared space equals space 4 straight a squared minus straight x squared space space space space rightwards double arrow space space space space 2 straight x squared space equals space 4 straight a squared space space space space space space rightwards double arrow space space space space straight x squared space equals space 2 straight a squared space space space space space

rightwards double arrow space space space straight x space equals space straight a square root of 2 space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space

open square brackets because space space straight x space cannot space be space negative close square brackets.

fraction numerator straight d squared straight P over denominator dx squared end fraction space space equals space 0 minus 2 space open square brackets fraction numerator square root of 4 straight a squared minus straight x squared end root. space 1 minus straight x. space open parentheses begin display style fraction numerator negative 2 straight x over denominator 2 square root of 4 straight a squared minus straight x squared end root end fraction end style close parentheses over denominator left parenthesis square root of 4 straight a squared minus straight x squared end root right parenthesis squared end fraction close square brackets

equals negative 2 open square brackets fraction numerator left parenthesis 4 straight a squared minus straight x squared right parenthesis space plus straight x squared over denominator left parenthesis 4 straight a squared minus straight x squared right parenthesis to the power of begin display style 3 over 2 end style end exponent end fraction close square brackets space equals space minus fraction numerator 8 straight a squared over denominator left parenthesis 4 straight a squared minus straight x squared right parenthesis to the power of begin display style 3 over 2 end style end exponent end fraction

When space straight x space equals space straight a square root of 2. space space fraction numerator straight d squared straight P over denominator dx squared end fraction space equals space minus fraction numerator 8 straight a squared over denominator left parenthesis 4 straight a squared minus 2 straight a squared right parenthesis to the power of begin display style 3 over 2 end style end exponent end fraction space equals space minus fraction numerator 8 straight a squared over denominator 2 square root of 2 straight a cubed end fraction space equals space minus fraction numerator 2 square root of 2 over denominator straight a end fraction less than 0

therefore space space space straight x space equals space straight y therefore space space space straight P space is space maximum space when space rectangle space becomes space straight a space square. Hence space the space result.

Some More Questions From Application of Derivatives Chapter

The radius of a circle is increasing uniformly at the rate of 4 cm per second Find the rate at which the area of the circle is increasing when the radius is 8 cm.

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