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Application Of Derivatives

Question
CBSEENMA12035562
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Show that height of the cylinder of greatest volume which can be inscribed in a right ciruclar cone of height h and semi vertical angle straight alpha is one-third that of the cone and the greatest volume of cylinder is 4 over 27 πh cubed space tan squared straight alpha.

Solution
Let PR = h be the height of the right circular cone and PQ = y be the height of right circular cylinder of radius x.
In rt. angle straight d space increment RQB comma space space space RQ over QB space equals space cot space straight alpha
   because space space space RQ over straight x space equals space cot space straight alpha comma space space where space QB space equals space straight x rightwards double arrow space space space space space space RQ space equals space straight x space cot space straight alpha
therefore space space space space straight v space equals space PR space minus space QR space equals space straight h minus straight x space cot space straight alpha 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 space straight v space equals space πx squared left parenthesis straight h minus straight x space cot space straight alpha right parenthesis                               open square brackets Volume space space equals space πr squared straight h close square brackets
          dv over dx space equals space straight pi open square brackets straight x squared straight d over dx left parenthesis straight h minus straight x space cot space straight alpha right parenthesis space plus space left parenthesis straight h minus straight x space cot space straight alpha right parenthesis space straight d over dx left parenthesis straight x squared right parenthesis close square brackets space space space space space space space space space space equals space straight pi open square brackets straight x squared left parenthesis negative cot space straight alpha right parenthesis space plus space left parenthesis straight h minus straight x space cot space straight alpha right parenthesis space left parenthesis 2 straight x right parenthesis close square brackets space equals space πx left square bracket negative straight x space cot space straight alpha space plus space 2 straight h space minus space 2 straight x space cot space straight alpha right square bracket space space space dv over dx space equals space straight pi space straight x space left square bracket 2 straight h minus 3 straight x space cot space straight alpha right square bracket Now space dv over dx equals 0 space space space give space us space space space space πx left square bracket 2 straight h minus 3 straight x space cot space straight alpha right square bracket space equals space 0 therefore space space space space space space 2 straight h minus 3 straight x space cot space straight alpha space equals space 0 space space space as space space space straight x space not equal to space 0 rightwards double arrow space space space space space space space straight x space equals space fraction numerator 2 straight h over denominator 3 space cot space straight alpha end fraction space equals space fraction numerator 2 straight h over denominator 3 end fraction tan space straight alpha fraction numerator straight d squared straight v over denominator dx squared end fraction space equals space straight pi open square brackets straight x space left parenthesis negative 3 space cot space straight alpha space plus space 2 straight h space minus space 3 straight x space cot space straight alpha close square brackets space equals space straight pi left square bracket negative 3 straight x space cot space straight alpha space plus space 2 straight h minus 3 straight x space cot space straight alpha right square bracket space space space space space space space space space space space space space space space space equals straight pi left square bracket 2 straight h minus 6 straight x space cot space straight alpha right square bracket
When straight x equals fraction numerator 2 straight h over denominator 3 end fraction space tan space straight alpha comma space space space then space
fraction numerator straight d squared straight v over denominator dx squared end fraction space equals space straight pi open square brackets 2 straight h minus 6. space fraction numerator 2 straight h over denominator 3 end fraction space tan space straight alpha space cot space straight alpha close square brackets space equals space straight pi left square bracket 2 straight h minus 4 straight h right square bracket space equals space minus 2 πh less than 0
therefore   v is maximum when straight x equals fraction numerator 2 straight h over denominator 3 end fraction space tan space straight alpha
therefore space space space maximum space value space space equals space straight pi space open parentheses fraction numerator 2 straight h over denominator 3 end fraction tan space straight alpha close parentheses squared. space space open parentheses straight h minus fraction numerator 2 straight h over denominator 3 end fraction tan space straight alpha space space cot space straight alpha close parentheses
                                equals space straight pi space cross times space space fraction numerator 4 straight h squared over denominator 9 end fraction space tan squared straight alpha space cross times straight h over 3 space space space space space space space space space space space space space space space space open square brackets because space space tan space straight alpha space space cot space straight alpha space equals space 1 close square brackets space equals space fraction numerator 4 πh cubed over denominator 27 end fraction tan squared straight alpha

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