Application of Derivatives

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Question

Of all the closed cylindrical cans (right circular), of a given volume of 100 cubic
centimetres, find the dimensions of the can which has the minimum surface
area?
Or
Of all the closed cylindrical tin cans (right circular) which enclose a given volume of 100 cubic cm., which has the minimum surface area?

Solution

Let V be the volume of cylinder whose radius is r; height is h and surface area is S.

therefore space space space space space space straight V space equals space πr squared straight h comma space space space space or space space space space 100 space equals space πr squared straight h

therefore space space space space space space straight h equals space 100 over πr squared

...(1)
Now,

straight S space equals space 2 πr squared space plus space 2 πrh space equals space 2 πr squared plus 2 πr. space 100 over πr squared

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

therefore space space space space straight S space equals space 2 πr squared plus 200 over straight r space space space space space space space dS over dr space equals space 4 πr minus 200 over straight r squared Now space space dS over dr space equals 0 space space give space us space space space space space space 4 πr minus 200 over straight r squared space equals space 0 comma space space space space space rightwards double arrow space space 4 πr cubed minus 200 space equals space 0 space space space space space rightwards double arrow space space straight r cubed space equals space 50 over straight pi space space space rightwards double arrow space space space space straight r space equals space open parentheses 50 over straight pi close parentheses to the power of 1 third end exponent space space space space space space fraction numerator straight d squared straight S over denominator dr squared end fraction space equals space 4 straight pi space plus 400 over straight r cubed space space space space space space space space space space

When space space straight r space equals space open parentheses 50 over straight pi close parentheses to the power of 1 third end exponent comma space space space space space fraction numerator straight d squared straight S over denominator dr squared end fraction space equals space 4 straight pi plus 400 over 50 cross times straight pi space equals space 12 straight pi greater than 0 therefore space space space space straight S space is space minimum space when space straight r space equals space open parentheses 50 over straight pi close parentheses to the power of 1 third end exponent