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Application Of Derivatives

Question
CBSEENMA12035540

An open box with a square base is to be made out of a given quantity of sheet of area c2. Show that the maximum volume of the box is fraction numerator straight c cubed over denominator 6 square root of 3 end fraction.

Solution
Let x be the side of the square base of the open box and y be its height. Let V denote the volume of the box.
    therefore space space space space straight V space equals space straight x squared straight y                                               ...(1)
Also surface area  = c2
rightwards double arrow space space space space space straight x squared plus 4 xy space equals space straight c squared space space space space space space rightwards double arrow space straight y space equals space fraction numerator straight c squared minus straight x squared over denominator 4 straight x end fraction
therefore space space space space from space left parenthesis 1 right parenthesis comma space space space space space straight V space equals straight x squared open parentheses fraction numerator straight c squared minus straight x squared over denominator 4 space straight x end fraction close parentheses
therefore space space space space space space straight V space equals space 1 fourth straight x left parenthesis straight c squared minus straight x squared right parenthesis space equals space 1 fourth left parenthesis straight c squared straight x minus straight x cubed right parenthesis
space space space space space space space dV over dx equals 1 fourth left parenthesis straight c squared minus 3 straight x squared right parenthesis
space space space space space space space space space space fraction numerator straight d squared straight V over denominator dx squared end fraction space equals space 1 fourth left parenthesis negative 6 space straight x right parenthesis space equals space minus fraction numerator 3 straight x over denominator 2 end fraction
For V to be maximum or minimum,
                            dV over dx space equals space 0
therefore space space space space space 1 fourth left parenthesis straight c squared minus 3 straight x squared right parenthesis space equals space 0
rightwards double arrow space space space 3 straight x squared space equals straight c squared space space space space space space space space space space space space space space space space rightwards double arrow space space space straight x squared space equals straight c squared over 3
therefore space space space space straight x space equals space fraction numerator straight c over denominator square root of 3 end fraction space as space straight x space cannot space be space negative
For space straight x space equals space fraction numerator straight c over denominator square root of 3 end fraction comma space space space fraction numerator straight d squared straight V over denominator dx squared end fraction space equals space minus 3 over 2. space fraction numerator straight c over denominator square root of 3 end fraction space equals space minus fraction numerator square root of 3 over denominator 2 end fraction space straight c space less than 0
therefore space space space straight V space is space maximum space when space straight x space equals space fraction numerator straight c over denominator square root of 3 end fraction
Maximum space value space of space straight V space equals space 1 fourth open parentheses straight c squared fraction numerator straight c over denominator square root of 3 end fraction minus fraction numerator straight c cubed over denominator 3 square root of 3 end fraction close parentheses
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 space space space space space space space space space space equals space 1 fourth open parentheses fraction numerator straight c cubed over denominator square root of 3 end fraction minus fraction numerator straight c cubed over denominator 3 square root of 3 end fraction close parentheses space equals space 1 fourth fraction numerator 3 straight c cubed minus straight c cubed over denominator 3 square root of 3 end fraction space equals 1 fourth cross times fraction numerator 2 straight c cubed over denominator 3 square root of 3 end fraction space equals fraction numerator straight c cubed over denominator 6 square root of 3 end fraction cubic space units.

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