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Application Of Derivatives

Question
CBSEENMA12035547

Show that the surface area of a closed cuboid with square base and given volume is minimum when it is a cube.

Solution

Let x, x, y be the length, width and height of the cuboid such that base is a square of side x.
therefore space space space straight V space equals space straight x squared straight y                                                                   ...(1)
where V is volume of cuboid.
Let S be surface area of cuboid.
therefore space space space straight S space equals space straight x squared plus straight x squared plus 4 xy space space space open square brackets because space space straight S space equals area space of space base space plus space area space of space top space plus area space of space four space walls close square brackets
space space space space space space space space space space space
           space equals space 2 straight x squared plus 4 straight x. space straight V over straight x squared                                                    open square brackets because space space of space left parenthesis 1 right parenthesis close square brackets
therefore space space space space straight S space equals space 2 straight x squared plus fraction numerator 4 straight V over denominator straight x end fraction space space space space rightwards double arrow space space space dS over dx space equals 4 straight x minus fraction numerator 4 straight V over denominator straight x squared end fraction
                 dS over dx space equals 0 space space space space rightwards double arrow space space 4 straight x minus fraction numerator 4 straight V over denominator straight x squared end fraction space equals space 0 space space space space space rightwards double arrow space space space straight x cubed space equals space straight V space space space space rightwards double arrow space space straight x space equals space straight V to the power of 1 third end exponent
fraction numerator straight d squared straight S over denominator dx squared end fraction space equals space 4 plus fraction numerator 8 straight V over denominator straight x cubed end fraction
When straight x space equals straight V to the power of 1 third end exponent. space fraction numerator straight d squared straight S over denominator dx squared end fraction space equals space 4 plus fraction numerator 8 straight V over denominator straight V end fraction space equals space 4 plus 8 space equals space 12 space greater than 0
therefore    S is minimum when straight x equals straight V to the power of 1 third end exponent
When straight x space equals space straight V to the power of 1 third end exponent comma space space space space space straight y space equals space straight V over straight x squared space equals space straight V over straight V to the power of begin display style 2 over 3 end style end exponent space equals space straight V to the power of 1 third end exponent space equals space straight x
therefore space space space space space straight S space is space minimum space when space length comma space width space and space height space are space straight x space each
therefore space space space space straight S space is space minimum space when space cuboid space is space straight a space cube. space