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Application Of Derivatives

Question

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An open box with a square base is to be made out of a given iron sheet of area 27 square m. Show that the maximum volume of the box is 13.5 cubic m.

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 space space space space space space space space space straight V space equals space straight x squared straight y

...(1)
Also surface area = 27 square metres

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therefore space space space from space left parenthesis 1 right parenthesis comma space space straight V space equals space straight x squared space open parentheses fraction numerator 27 minus straight x squared over denominator 4 straight x end fraction close parentheses therefore space space space space space space space space straight V space equals space 1 fourth straight x left parenthesis 27 minus straight x squared right parenthesis space equals space 1 fourth left parenthesis 27 minus straight x cubed right parenthesis space space space space space dV over dx space equals 1 fourth left parenthesis 27 minus 3 straight x squared right parenthesis 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 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 space space space space rightwards double arrow space space space space space 1 fourth left parenthesis 27 minus 3 straight x squared right parenthesis space equals 0 space space rightwards double arrow space space space 27 minus 3 straight x squared space equals 0

rightwards double arrow space space space space space straight x squared space equals space 9 space space space space space rightwards double arrow space space straight x space equals space 3 space as space straight x space cannot space be space negative space space space space space space space For space straight x space equals space 3 comma space space fraction numerator straight d squared straight V over denominator dx squared end fraction space equals space minus 3 over 2 cross times 3 space equals space minus 9 over 2 less than 0 therefore space space space space straight V space is space maximum space when space straight x space equals space 3 therefore space space space space maximum space value space of space straight V space equals space 1 fourth left parenthesis 27 space cross times 3 space minus space 27 right parenthesis space equals space 1 fourth cross times 54 space equals space 13.5 space cubic space metres.

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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