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Application Of Derivatives

Question
CBSEENMA12035565
Wired Faculty App

Prove that the volume of the largest cone that can be inscribed in a sphere of radius R is 8 over 27 of the volume of the sphere.
Or
Find the volume of the largest cone that can be inscribed in a sphere of radius r.

Solution

Let R be the radius of cone, r be the radius of sphere and OD = x
∴   height of cone i.e. h = x + r
Now x2 + R2 = r2 ⇒ R2 = r2 – x2 ...(1)
Let V be the volume of cone.

therefore space space space space space straight V space equals space 1 third πR squared straight h space equals space 1 third straight pi left parenthesis straight r squared minus straight x squared right parenthesis thin space left parenthesis straight x plus straight r right parenthesis space space space space space space space space space space space space space space equals space straight pi over 3 left parenthesis straight x plus straight r right parenthesis squared space left parenthesis straight r minus straight x right parenthesis dV over dx space equals space straight pi over 3 open square brackets straight x plus straight r right parenthesis squared space left parenthesis negative 1 right parenthesis space plus space left parenthesis straight r minus straight x right parenthesis. space 2 left parenthesis straight x plus straight r right parenthesis close square brackets space space space space space space space space space space space equals space straight pi over 3 left parenthesis straight x plus straight r right parenthesis open square brackets negative straight x minus straight r plus 2 straight r minus 2 straight x close square brackets space equals space straight pi over 3 left parenthesis straight x plus straight r right parenthesis thin space left parenthesis straight r minus 3 straight x right parenthesis dV over dx space equals space 0 space space space space space space space rightwards double arrow space space space space straight pi over 3 left parenthesis straight x plus straight r right parenthesis thin space left parenthesis straight r minus 3 straight x right parenthesis space equals 0 space space space space rightwards double arrow space space space space straight x space equals space minus straight r comma space space straight r over 3
Now x = – r is not possible as x cannot be negative.
therefore space space space space straight x space equals space straight r over 3
space fraction numerator straight d squared straight V over denominator dx squared end fraction space equals space straight pi over 3 open square brackets left parenthesis straight x plus straight r right parenthesis. space left parenthesis negative 3 right parenthesis space plus space left parenthesis straight r minus 3 straight x right parenthesis. space 1 close square brackets space equals space straight pi over 3 open square brackets negative 3 straight x minus 3 straight r plus straight r minus 3 straight x close square brackets space equals space minus fraction numerator 2 straight pi over denominator 3 end fraction left parenthesis straight r plus 3 straight x right parenthesis At space space straight x space equals space straight r over 3 comma space space fraction numerator straight d squared straight V over denominator dx squared end fraction space equals space minus fraction numerator 2 straight pi over denominator 3 end fraction left square bracket straight r plus straight r right square bracket space equals space minus fraction numerator 4 straight pi over denominator 3 end fraction straight r less than 0 therefore space space space space space space volume space straight V space is space maximum space when space straight x space equals space straight r over 3 therefore space space space space greatest space volume space space equals space straight pi over 3 open square brackets open parentheses straight r over 3 plus straight r close parentheses squared space space space open parentheses straight r minus straight r over 3 close parentheses close square brackets space equals space straight pi over 3 cross times fraction numerator 16 space straight r squared over denominator 9 end fraction cross times space fraction numerator 2 straight r over denominator 3 end fraction space equals fraction numerator 32 πr cubed over denominator 81 end fraction

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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