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Application Of Derivatives

Question
CBSEENMA12035541

An open topped box is to be constructed by removing equal squares from each corner of a 3 metre by 8 metre rectangular sheet of aluminum and folding up the sides. Find the volume of the largest such box.

 

Solution
Let x metre be the length of a side of the removed squares. Then the height of the box is x, length is (8 – 2x) and breadth is (3 – 2 x).
Let V be the volume of the box.
   therefore space space space space space straight V space space equals straight x space left parenthesis 3 minus 2 straight x right parenthesis thin space left parenthesis 8 minus 2 straight x right parenthesis space equals space 4 straight x cubed minus 22 straight x squared plus 24 straight x
space space space space dV over dx space equals space
and space fraction numerator straight d squared straight V over denominator dx squared end fraction space equals space
For V to be maximum or minimum.
                         dV over dx space equals space 0 space space space space space space space space space space space rightwards double arrow space 12 straight x squared minus 44 straight x plus 24 space equals space 0
rightwards double arrow space space space 3 straight x squared minus 11 straight x plus 6 space equals space 0 space space space space space space space space rightwards double arrow space space space straight x space equals space fraction numerator 11 plus-or-minus square root of 121 minus 72 end root over denominator 6 end fraction
rightwards double arrow space space space space space space straight x space equals space fraction numerator 11 plus-or-minus square root of 49 over denominator 6 end fraction space equals space fraction numerator 11 plus-or-minus 7 over denominator 6 end fraction space equals 3 comma space 2 over 3
Rejecting x = 3 as breadth cannot be negative, we get, straight x space equals 2 over 3
At space space straight x space equals space 2 over 3 comma space dV over dx squared space equals space 24 space open parentheses 2 over 3 close parentheses minus 44 minus 16 minus 4 space equals space 28 less than 0
therefore space space space space space space straight V space space is space maximum space when space straight x space equals space 2 over 3
therefore space space space space space space space volume space of space largest space box space space equals open parentheses 2 over 3 close parentheses space open parentheses 3 minus 4 over 3 close parentheses space open parentheses 8 minus 4 over 3 close parentheses space equals space 2 over 3 cross times 5 over 3 cross times 20 over 3 space equals space 200 over 27 straight m cubed.

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