Application of Derivatives

Question

Show that the height of the cylinder of maximum volume, which can be inscribed in 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

Given, radius of the sphere is R.
Let r and h be the radius and the height of the inscribed cylinder respectively.
We have:
$straight h equals 2 square root of straight R squared minus straight r squared end root$
Let Volume of cylinder = V
$straight V equals πr squared straight h$
$equals πr squared cross times 2 square root of straight R squared minus straight r squared end root equals 2 πr squared square root of straight R squared minus straight r squared end root$

Differentiating the above function w.r.t r, we have,
$straight V space equals space 2 straight pi squared square root of straight R squared minus straight r squared end root dV over dr equals 4 πr open parentheses square root of straight R squared minus straight r squared end root close parentheses minus fraction numerator 4 πr cubed over denominator 2 square root of straight R squared minus straight r squared end root end fraction space space space space space equals fraction numerator 4 πr open parentheses straight R squared minus straight r squared close parentheses minus 4 πr cubed over denominator 2 square root of straight R squared minus straight r squared end root end fraction dV over dr equals fraction numerator 4 πrR squared minus 4 πr cubed minus 2 πr cubed over denominator 2 square root of straight R squared minus straight r squared end root end fraction space space space space space equals fraction numerator 4 πrR squared minus 6 πr cubed over denominator 2 square root of straight R squared minus straight r squared end root end fraction$
For maxima or minima,
$dV over dr equals 0 space rightwards double arrow space space 4 πrR squared minus 6 πr cubed space equals 0 rightwards double arrow space 6 πr cubed space equals space 4 πR squared rightwards double arrow space straight r squared space equals space fraction numerator 2 straight R squared over denominator 3 end fraction dV over dr space equals space fraction numerator 4 πrR squared minus 6 πr cubed over denominator 2 square root of straight R squared minus straight r squared end root end fraction$
$Now comma space fraction numerator straight d squared straight V over denominator dr squared end fraction space equals 1 half open square brackets fraction numerator square root of straight R squared minus straight r squared end root left parenthesis 4 πR squared minus 18 πr squared right parenthesis minus left parenthesis 4 πrR squared minus 6 πr cubed right parenthesis begin display style fraction numerator left parenthesis negative 2 straight r right parenthesis over denominator 2 square root of straight R squared minus straight r squared end root end fraction end style over denominator left parenthesis straight R squared minus straight r squared right parenthesis end fraction close square brackets$
$equals space 1 half open square brackets fraction numerator left parenthesis straight R squared minus straight r squared right parenthesis left parenthesis 4 πR squared minus 18 πr squared right parenthesis plus straight r left parenthesis 4 πrR squared minus 6 πr cubed right parenthesis over denominator left parenthesis straight R squared minus straight r squared right parenthesis to the power of begin display style 3 over 2 end style end exponent end fraction close square brackets equals 1 half open square brackets fraction numerator 4 πR to the power of 4 minus 22 πr squared straight R squared plus 12 πr to the power of 4 plus 4 πr squared straight R squared over denominator left parenthesis straight R squared minus straight r squared right parenthesis to the power of begin display style 3 over 2 end style end exponent end fraction close square brackets$
Now, when $straight r squared space equals space fraction numerator 2 straight R squared over denominator 3 end fraction comma space fraction numerator straight d squared straight V over denominator dr squared end fraction less than 0.$
$therefore space space Volume space is space the space maximum space when space straight r squared space equals space fraction numerator 2 straight R squared over denominator 3 end fraction.$
When $straight r squared equals space fraction numerator 2 straight R squared over denominator 3 end fraction comma space space straight h space equals space 2 square root of straight R squared minus fraction numerator 2 straight R squared over denominator 3 end fraction end root space equals space 2 square root of straight R squared over 3 end root space equals space fraction numerator 2 straight R over denominator square root of 3 end fraction.$
Hence, the volume of the cylinder is the maximum when the height of the cylinder is $fraction numerator 2 straight R over denominator square root of 3 end fraction.$

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