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Application Of Derivatives

Question

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Show that the height of the right circular cylinder of maximum volume that can be inscribed in a given circular cone of height h is $1 third straight h$

Solution

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

angle straight d space RQB comma space space RQ over QB space equals space cot space straight alpha

therefore space space space space space space space RQ over straight x space equals space cot space straight alpha comma space space space where space QB space equals space straight x space space space space space space rightwards double arrow space space space space space RQ space space equals space straight x space cot space straight alpha therefore space space space space space space space space space space straight y 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... 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 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

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 thin space left parenthesis 2 space straight x right parenthesis close square brackets space equals space πx open square brackets negative straight x space cot space straight alpha space plus space 2 space straight h space minus space 2 straight x space cot space straight alpha close square brackets

dv over dx space equals space πx left square bracket 2 straight h space minus space 3 space straight x space cot space straight alpha right square bracket

Now

dv over dx space equals space 0 space space space space gives space us space space space space straight pi space straight x left square bracket 2 straight h minus space 3 space straight x space cot space straight alpha right square bracket space equals 0

therefore space space space space 2 straight h minus 3 straight x space cot space straight alpha space equals space 0 space space as space space straight x space not equal to 0 space space space space space rightwards double arrow space space space space straight x space equals space fraction numerator 2 space 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 space tan space space straight alpha space space space space space space space space 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 right parenthesis space plus space 2 straight h space minus space 3 space 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 space minus space 3 straight x space cot space straight alpha right square bracket space equals space straight pi left square bracket 2 straight h minus 6 straight x space cot space straight alpha right square bracket

When

straight x space equals space fraction numerator 2 straight h over denominator 3 end fraction space tan space straight alpha comma space space then space

fraction numerator straight d squared straight v over denominator dx squared end fraction space equals space straight pi space open square brackets 2 straight h minus 6. space fraction numerator 2 straight h over denominator 3 end fraction 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 space space 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 end fraction space tan space space straight alpha

therefore space space space space space space straight v space equals space straight h minus fraction numerator 2 straight h over denominator 3 end fraction space tan space straight alpha. space space cot space straight alpha

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

equals space straight h minus fraction numerator 2 straight h over denominator 3 end fraction space equals space straight h over 3.

Hence the 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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