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Application Of Derivatives

Question

Wired Faculty App

Show that the volume of the greatest cylinder which can be inscribed in a cone of height h and semi-vertical angle 30° is $4 over 81 πh cubed$

Solution

Let PR = h be the height of circular one and PQ = y be the height of right circular cylinder of radius x.
In rt.

angle straight d space increment space RQB comma space space space RQ over QB space equals space cot space 30 degree

therefore space space space space space space RQ over straight x space equals space cot space 30 degree comma space space where space QB space equals space straight x rightwards double arrow space space space space space space space RQ space space equals straight x space cot space 30 degree therefore space space space space space space space space RQ space equals space square root of 3 space straight x therefore space space space space space space space space space space space space space space straight y space equals space PR space minus space QR space space equals space straight h minus square root of 3 space straight x 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 space space space space space space space space space space space space... left parenthesis 1 right parenthesis

Let v be volume of right circular cylinder

therefore space space space space straight v space equals space πx squared left parenthesis straight h minus square root of 3 space straight x right parenthesis

open square brackets because space space volume space space equals space πr squared straight h close square brackets

dv over dx space equals straight pi open square brackets straight x squared left parenthesis negative square root of 3 right parenthesis space plus space left parenthesis straight h minus square root of 3 straight x right parenthesis space 2 straight x close square brackets space equals space straight pi space open square brackets negative square root of 3 straight x squared plus 2 hx minus 2 square root of 3 straight x squared close square brackets space space space space space space space space space space equals space straight pi open square brackets 2 hx minus 3 square root of 3 space straight x squared close square brackets space space equals space straight pi space straight x left parenthesis 2 straight h minus 3 square root of 3 straight x right parenthesis Now space dv over dx space equals 0 space space gives space us space space space πx left parenthesis 2 straight h minus 3 square root of 3 space straight x right parenthesis space equals space 0 therefore space space space space space 2 straight h minus 3 square root of 3 space straight x space equals space 0 space space space space space as space space straight x not equal to 0 space space space space space space rightwards double arrow space space space straight x space equals space fraction numerator 2 straight h over denominator 3 square root of 3 end fraction fraction numerator straight d squared straight v over denominator dx squared end fraction space equals space straight pi left parenthesis 2 straight h minus 6 square root of 3 space straight x right parenthesis

When space straight x space equals space fraction numerator 2 straight h over denominator square root of 3 end fraction. space fraction numerator straight d squared straight v over denominator dx squared end fraction space equals straight pi space open parentheses 2 straight h minus 6 square root of 3. space fraction numerator 2 straight h over denominator 3 square root of 3 end fraction close parentheses space equals space straight pi space left parenthesis 2 straight h minus 4 straight h right parenthesis space space space space space space space space space space space space space space space space space space equals negative 2 πh space less than space 0 therefore space space space space straight v space is space maximum space when space straight x space equals space fraction numerator 2 straight h over denominator 3 square root of 3 end fraction Max. space value space of space straight v space equals space straight pi space open parentheses fraction numerator 2 straight h over denominator 3 square root of 3 end fraction close parentheses squared space space space open parentheses straight h minus square root of 3 cross times fraction numerator 2 over denominator 3 square root of 3 end fraction close parentheses space equals space straight pi space cross times space fraction numerator 4 straight h squared over denominator 27 end fraction cross times straight h over 3 equals space space fraction numerator 4 πh cubed over denominator 81 end fraction

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