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Application Of Derivatives

Question
CBSEENMA12035538

A square piece of a tin of side 18 cm is to be made into a box without top, by cutting a square from each corner and folding up the flaps to form a box. What should be the side of the square to be cut off, so that the volume of the box is the maximum possible?

Solution
Let x (0 < x < 9) cm be the length of each side of the square which is to be cut from corners of the square tin sheet of each side 18 cm. Let V be the volume of the box formed by folding up the flaps.
      therefore space space space space straight V space equals space straight x left parenthesis 18 minus 2 straight x right parenthesis space left parenthesis 18 minus 2 straight x right parenthesis space equals space straight x left parenthesis 18 minus 2 straight x right parenthesis squared space equals space 4 straight x left parenthesis 9 minus straight x right parenthesis squared space equals space 4 straight x left parenthesis straight x squared minus 18 straight x plus 81 right parenthesis
space space space space space space space space space space space space space equals 4 left parenthesis straight x cubed minus 18 straight x squared plus 81 straight x right parenthesis
space space dV over dx space equals space 4 left parenthesis 3 straight x squared minus 36 straight x plus 81 right parenthesis
space space dV over dx space equals space 0 space space space space space space space space space space space rightwards double arrow space space space space space 4 left parenthesis 3 straight x squared minus 36 straight x plus 81 right parenthesis space equals space 0
rightwards double arrow space space space space straight x squared minus 12 straight x plus 27 space equals space 0 space space space space rightwards double arrow space space space left parenthesis straight x minus 3 right parenthesis thin space left parenthesis straight x minus 9 right parenthesis space equals space 0 space space space space rightwards double arrow space space straight x space equals space 3 comma space 9
rightwards double arrow space space straight x space equals space 3                                                                            open square brackets because space space space straight x space equals space 9 space not an element of left parenthesis 0 comma space 9 right parenthesis close square brackets
       fraction numerator straight d squared straight V over denominator dx squared end fraction space equals space 4 space left parenthesis 6 straight x minus 36 right parenthesis
At 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 4 left parenthesis 18 minus 36 right parenthesis space equals space minus 72 space less than 0
therefore space space space space space space straight V space is space local space maximum space at space straight x space equals space 3
But space straight x space equals space 3 space is space the space only space extreme space point
therefore space space space space space space space space straight V space is space maximum space at space straight x space equals space 3
rightwards double arrow space space space space space side space of space the space square space space equals space 3 space cm.
max. space straight V space equals space 3 space left parenthesis 18 minus 6 right parenthesis squared space equals space 3 space cross times space 144 space space equals 432 space cubic space cm.