Application of Derivatives

Question

Show that the right circular cylinder of given surface and maximum volume is such that its height is equal to the diameter of the base.

Solution

Let r be the radius of base of circular cylinder and h be its height. Let V be the volume and S be total surface area.
$therefore space space space space straight S space equals space 2 πrh plus 2 πr squared space space space space space rightwards double arrow space space space space 2 πrh space equals space straight S minus 2 πr squared space space space space space space$
   $straight h space equals space fraction numerator 1 over denominator 2 πr end fraction left parenthesis straight S minus 2 πr squared right parenthesis$ ...(1)
$straight V space equals space πr squared straight h space equals πr squared. space space fraction numerator 1 over denominator 2 πr end fraction left parenthesis straight S minus 2 πr squared right parenthesis$    $open square brackets because space of space left parenthesis 1 right parenthesis close square brackets$
$therefore space space space space space straight V space equals space 1 half straight r left parenthesis straight S minus 2 πr squared right parenthesis space equals space 1 half left parenthesis Sr minus 2 πr cubed right parenthesis space where space straight S space is space constant.$
   $dV over dr space equals space 1 half left parenthesis straight S minus 6 πr squared right parenthesis$
Now, $dV over dr space equals space 0 space space space gives space us space 1 half left parenthesis straight S minus 6 πr squared right parenthesis space space equals 0$
$therefore space space space straight S minus 6 πr squared space equals space 0 space space space space space space space space space space space space space space space space space space space space space space space space rightwards double arrow space 2 πrh plus 2 πr squared minus 6 πr squared space equals space 0 therefore space space space 2 πrh space equals space 4 πr squared space space space space space space space space space space space space space space space space space space space space space space space space space rightwards double arrow space space space straight h space equals space 2 straight r space space space space space space space space fraction numerator straight d squared straight V over denominator dr squared end fraction space equals space 1 half left parenthesis 0 minus 12 πr right parenthesis space equals negative 6 πr When space straight r space equals straight h over 2 comma space fraction numerator straight d squared straight V over denominator dr squared end fraction space equals space minus 6 straight pi. space straight h over 2 space equals space minus 3 πh less than 0$
$therefore$ V is maximum when h = 2r i.e., height of the cylinder is equal to the diameter of the base.

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Application Of Derivatives

Show that the right circular cylinder of given surface and maximum volume is such that its height is equal to the diameter of the base.

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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