-->

Application Of Derivatives

Question
CBSEENMA12035554

An open tank with a square base and vertical sides is to be constructed from a metal sheet so as to hold a given quantity of water, show that the cost of the material will be least when the depth of the tank is half of its width.

Solution
Let x be sides of the square base and h be the depth of the given tank of volume V.
therefore space space space space straight V space equals space straight x squared straight h space space space space space space space rightwards double arrow space space space space straight h space space equals space space straight V over straight x squared                                                    ...(1)
Surface area of tanki = left parenthesis straight x squared plus 4 xh right parenthesis space sq. space units
Let Rs. p be the cost per square unit of material and C be the total cost.
therefore space space space straight C space equals space left parenthesis straight x squared plus 4 xh right parenthesis straight p space equals space open parentheses straight x squared plus fraction numerator 4 straight V over denominator straight x end fraction close parentheses space straight p. space where space straight p space is space constant. space space space space space space left square bracket because space of space left parenthesis 1 right parenthesis right square bracket
space space space space space dC over dx equals straight p open parentheses 2 straight x minus fraction numerator 4 straight V over denominator straight x squared end fraction close parentheses
Now comma space dC over dx equals 0 space space gives space us
space space
          straight p open parentheses 2 straight x minus fraction numerator 4 straight V over denominator straight x squared end fraction close parentheses space equals 0 space space space space space space or space space space space 2 straight x minus fraction numerator 4 straight V over denominator straight x squared end fraction space equals 0
therefore space space space straight x cubed minus 2 straight V space equals space 0 space space space space space space space rightwards double arrow space space space straight x cubed space equals space 2 thin space straight V                                                ...(2)
space fraction numerator straight d squared straight C over denominator dx squared end fraction space equals space straight p open parentheses 2 plus fraction numerator 8 space straight V over denominator straight x cubed end fraction close parentheses space space equals straight p space open parentheses 2 plus 8 over straight x cubed. space straight x cubed over 2 close parentheses space space space space space space space space space space space space space open square brackets because space space of space left parenthesis 2 right parenthesis close square brackets
space space space space space space space space space space space equals space straight p left parenthesis 2 plus 4 right parenthesis space equals space 6 straight p greater than 0
therefore space space space space space space straight C space is space minimum.
Now space space space straight h space equals space straight V over straight x squared space equals space fraction numerator begin display style straight x cubed over 2 end style over denominator straight x squared end fraction space equals space straight x over 2
therefore space space space depth space of space tank space equals space half space of space its space width.