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Application Of Derivatives

Question

Wired Faculty App

A figure consists of a semi-circle with a rectangle on its diameter. Given perimeter of the figure, find the dimensions in order that the area may be maximum.

Solution

Let x, y be ength, breadth of the rectangle ABCD. Let

straight x over 2

be the radius of semi-circle with centre at O.

Let P be the perimeter of figure.

therefore space space space space space space straight x plus 2 straight y plus straight pi straight x over 2 space equals space straight p rightwards double arrow space space space space space 2 straight x plus 4 straight y plus πx space equals space 2 space straight p rightwards double arrow space space space space 4 straight y space equals 2 straight p minus left parenthesis straight pi plus 2 right parenthesis space straight x rightwards double arrow space space space space space straight y space equals fraction numerator 2 straight p minus left parenthesis straight pi plus 2 right parenthesis space straight x over denominator 4 end fraction 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... left parenthesis 1 right parenthesis

Let A be area of the figure.

therefore space space space space straight A space equals space xy plus 1 half straight pi open parentheses straight x over 2 close parentheses squared space equals space straight x open square brackets fraction numerator 2 straight p minus left parenthesis straight pi plus 2 right parenthesis space straight x over denominator 4 end fraction close square brackets space plus space πx squared over 8 therefore space space space space space straight A space equals space 1 fourth left square bracket 2 space straight p space straight x space minus space left parenthesis straight pi plus 2 right parenthesis space straight x squared right square bracket space plus space straight pi over 8 straight x squared therefore space space space space space dA over dx space equals space 0 space space space gives space us space 1 fourth left square bracket 2 straight p space minus space 2 left parenthesis straight pi plus 2 right parenthesis space straight x right square bracket space plus space straight pi over 4 straight x space equals space 0 rightwards double arrow space space space space space space space 2 straight p minus 2 left parenthesis straight pi plus 2 right parenthesis space straight x space plus space πx space equals space 0 space space space space space rightwards double arrow space space space space 2 straight p minus left parenthesis straight pi plus 4 right parenthesis space straight x space equals space 0 rightwards double arrow space space space space space space space space straight x space equals space fraction numerator 2 straight p over denominator straight pi plus 4 end fraction

fraction numerator straight d squared straight A over denominator dx squared end fraction space equals 1 fourth left square bracket 0 minus 2 left parenthesis straight pi plus 2 right parenthesis plus straight pi over 4 space equals fraction numerator negative 2 straight pi minus 4 over denominator 4 end fraction plus straight pi over 4 space equals space fraction numerator negative straight pi minus 4 over denominator 4 end fraction space equals space minus fraction numerator straight pi plus 4 over denominator 4 end fraction

When

straight x space equals fraction numerator 2 space straight p over denominator straight pi plus 4 end fraction comma space space fraction numerator straight d squared straight A over denominator dx squared end fraction space equals space minus fraction numerator straight pi plus 4 over denominator 4 end fraction less than 0

therefore space space space space space straight A space is space maximum space when space straight x space equals space fraction numerator 2 straight p over denominator straight pi plus 4 end fraction and space straight y space equals space fraction numerator 2 straight p minus left parenthesis straight pi plus 2 right parenthesis. space begin display style fraction numerator 2 straight p over denominator straight pi plus 4 end fraction end style over denominator 4 end fraction space equals fraction numerator 2 pπ plus space 8 straight p minus 2 pπ space minus 4 straight p over denominator 4 left parenthesis straight pi plus 4 right parenthesis end fraction space equals fraction numerator 4 straight p over denominator 4 left parenthesis straight pi plus 4 right parenthesis end fraction equals fraction numerator straight p over denominator straight pi plus 4 end fraction

therefore space space space space space area space of space figure space is space maximum space when space space space space space space space space space length space of space rectangle space space equals space straight x space equals space fraction numerator 2 straight p over denominator straight pi plus 4 end fraction comma space space space space space space space space space breadth space of space rectangle space space equals space straight y space equals space fraction numerator straight p over denominator straight pi plus 4 end fraction comma space space space space space space space space space space radius space of space semi minus circle space equals space straight x over 2 space equals fraction numerator straight p over denominator straight pi plus 4 end fraction

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