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Application Of Derivatives

Question

Wired Faculty App

Show that the height of a closed right circular cylinder of given volume and least surface area is equal to its diameter.

Solution

Let r be the radius of base of circular cylinder and h be its height. Let V be the volume and S be the total surface area.

therefore space space space space space straight V space equals space πr squared straight h space space space space space space space space space space space space space space rightwards double arrow space space space straight h space equals straight V over πr squared

...(1)
Also,

straight S space equals space 2 πrh plus 2 πr squared space equals space 2 πr. space straight V over πr squared plus 2 πr squared

open square brackets because space space of space left parenthesis 1 right parenthesis close square brackets

therefore space space straight S space equals fraction numerator 2 space straight V over denominator straight r end fraction plus 2 πr squared therefore space space space dS over dr space equals negative fraction numerator 2 straight V over denominator straight r squared end fraction plus 4 πr Now space dS over dr equals 0 space space space rightwards double arrow space space space space fraction numerator negative 2 straight V over denominator straight r squared end fraction plus 4 space straight pi space straight r space equals space 0 space space space space space space space space rightwards double arrow space space space space 2 space straight V space equals space 4 πr cubed

rightwards double arrow space space space 2 πr squared straight h space equals space 4 πr cubed

open square brackets because space space space straight V space equals space πr squared straight h close square brackets

rightwards double arrow space space space straight r space equals space straight h over 2

Now,

fraction numerator straight d squared straight S over denominator dr squared end fraction space equals space fraction numerator 4 straight V over denominator straight r squared end fraction plus 4 straight pi

When space straight r space equals straight h over 2 comma space fraction numerator straight d squared straight S over denominator dr squared end fraction space equals space fraction numerator 4 straight V over denominator straight h cubed end fraction cross times 8 plus 4 straight pi space equals space fraction numerator 32 space straight V over denominator straight h cubed end fraction plus 4 straight pi space greater than 0 therefore space space space straight S space is space minimum space when space straight r space equals space straight h over 2 space straight i. straight e. comma space straight h space equals space 2 straight r space straight i. straight e. comma space height space equals space diameter.

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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