+91-8076753736 [email protected]

-->

Application Of Derivatives

Question

Wired Faculty App

Find the volume of the largest cylinder that can be inscribed in a sphere of radius r.

Solution

Let h be the height and R be the base radius of the inscribed cylinder.
In

increment OCA comma

OC squared plus CA squared space equals space OA squared space space space space rightwards double arrow space space space space open parentheses straight h over 2 close parentheses squared plus straight R squared space equals space straight r squared therefore space space space space space straight R squared space equals space straight r squared minus straight h squared over 4 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 cylinder

therefore space space space straight V space equals space πR squared straight h space equals space straight pi open parentheses straight r squared minus straight h squared over 4 close parentheses straight h space equals space straight pi over 4 left parenthesis 4 straight r squared straight h minus straight h cubed right parenthesis space space space space space space space space dV over dh space equals straight pi over 4 left parenthesis 4 straight r squared minus 3 straight h squared right parenthesis space space space space space space space space dV over dh space equals space 0 space space space space rightwards double arrow space space space space space straight pi over 4 left parenthesis 4 straight r squared minus 3 straight h squared right parenthesis space equals space 0 space space space space space rightwards double arrow space space space 4 straight r squared space minus 3 space straight h squared space equals space 0 space space space space space space rightwards double arrow space space space space space space straight h squared space equals space 4 over 3 straight r squared space space space space space space space space space space fraction numerator straight d squared straight V over denominator dh squared end fraction space equals space straight pi over 4 left parenthesis 0 minus 6 straight h right parenthesis space equals space minus fraction numerator 3 πh over denominator 2 end fraction When space straight h space equals space fraction numerator 2 over denominator square root of 3 end fraction straight r comma space space space fraction numerator straight d squared straight V over denominator dh squared end fraction space equals space minus fraction numerator 3 straight pi over denominator 2 end fraction. space fraction numerator 2 straight r over denominator square root of 3 end fraction space equals space minus square root of 3 space straight pi space straight r space less than 0

therefore space space space space space space space straight V space space is space maximum space when space straight h space equals space fraction numerator 2 over denominator square root of 3 end fraction straight r therefore space space space space greatest space volume space space equals space straight pi over 4 open parentheses 4 straight r squared. space space fraction numerator 2 over denominator square root of 3 end fraction straight r space minus space fraction numerator 8 over denominator 3 square root of 3 end fraction straight r cubed close parentheses space space space space equals space straight pi over 4 open parentheses fraction numerator 8 over denominator square root of 3 end fraction straight r cubed space minus space fraction numerator 8 over denominator 3 square root of 3 end fraction straight r cubed close parentheses space equals space straight pi over 4 open parentheses fraction numerator 24 straight r cubed minus 8 straight r cubed over denominator 3 square root of 3 end fraction close parentheses equals straight pi over 4 open parentheses fraction numerator 16 straight r cubed over denominator 3 square root of 3 end fraction close parentheses space equals space fraction numerator 4 πr cubed over denominator 3 square root of 3 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.

Mock Test Series

Mock Test Series

NCERT Sample Papers

Entrance Exams Preparation