Application of Derivatives

Question

Show that the height of a cylinder of maximum volume that can be inscribed in a sphere of radius R is a sphere of radius R is $fraction numerator 2 straight R over denominator square root of 3 end fraction.$ Also, find the maximum volume.

Solution

Let h be the height and r be the base radius of the inscribed cylinder in a sphere of radius R.
In rt. angled $increment OCA comma$
$OC squared plus CA squared space space equals OA squared therefore space space space space straight h squared over 4 plus straight r squared space equals space straight R squared therefore space space space space space straight r squared space equals space straight R squared minus straight h squared over 4 space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space... left parenthesis 1 right parenthesis$
Let V be the volume of the 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 minus straight h cubed right parenthesis space space space space space space space dV over dh space equals space straight pi over 4 left parenthesis 4 straight R squared minus 3 straight h squared right parenthesis and space fraction numerator straight d squared straight V over denominator dh squared end fraction space equals space straight pi over 4 left parenthesis negative 6 straight h right parenthesis space equals space minus fraction numerator 3 πh over denominator 2 end fraction For space straight V space to space be space maxima space or space minima. space dV over dx space equals space 0 therefore 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 space rightwards double arrow space space space space space space space space space 4 straight R squared minus 3 straight h squared space equals space 0 space space space space space space space space rightwards double arrow space space space space space space straight h squared space equals space fraction numerator 4 straight R squared over denominator 3 end fraction therefore space space space straight h space equals space fraction numerator 2 straight R over denominator square root of 3 end fraction When space straight h space equals space fraction numerator 2 straight R over denominator square root of 3 end fraction 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 space 0 therefore space space space space space straight V 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 Max space volume space equals space straight pi over 4 open parentheses 4 straight R squared cross times fraction numerator 2 over denominator square root of 3 end fraction straight R. 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 cross times fraction numerator 8 straight R cubed over denominator 3 square root of 3 end fraction left parenthesis 3 minus 1 right parenthesis space equals fraction numerator 4 πR cubed over denominator 3 square root of 3 end fraction$

For Daily Free Study Material Join wiredfaculty