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Application Of Derivatives

Question
CBSEENMA12035537

A square piece of 30 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 volume of the box is maximum?

Solution
Let x (0 < x < 15) be the length of each side of the square which is to be cut from corners of the square tin sheet of each side 30 cm. Let V be the volume of the open box formed by folding up the flaps.
 therefore space space space space straight V space equals space straight x space left parenthesis 30 minus 2 straight x right parenthesis thin space left parenthesis 30 minus 2 straight x right parenthesis space equals space straight x left parenthesis 30 minus 2 straight x right parenthesis squared space equals space 4 straight x left parenthesis 15 minus straight x right parenthesis squared
space space space space space space space space space space space space equals space 4 straight x left parenthesis straight x squared minus 30 straight x plus 225 right parenthesis space equals space 4 left parenthesis straight x cubed minus 30 straight x squared plus 225 straight x right parenthesis
space space space space dV over dx space equals space 4 left parenthesis 3 straight x squared minus 60 straight x plus 225 right parenthesis
space space space space dV over dx space equals space 0 space space space space rightwards double arrow space space space space 4 left parenthesis 3 straight x squared minus 60 straight x plus 225 right parenthesis space equals space 0 space space space space space space rightwards double arrow space space space space space straight x squared minus 20 straight x plus 75 space equals space 0
rightwards double arrow space space space space space space left parenthesis straight x minus 5 right parenthesis thin space left parenthesis straight x minus 15 right parenthesis space equals space 0 space space space space rightwards double arrow space space space space space straight x space equals space 5 comma space space 15 space space space space space space space rightwards double arrow space space space space space straight x space equals space 5 space space space as space space straight x space equals space 15 space not an element of space left parenthesis 0 comma space 15 right parenthesis
space space space space space space space space fraction numerator straight d squared straight V over denominator dx squared end fraction space equals space 4 left parenthesis 6 straight x minus 60 right parenthesis
At space space space straight x space equals space 5 comma space space space space space fraction numerator straight d squared straight V over denominator dx squared end fraction space equals space 4 space space left parenthesis 30 minus 60 right parenthesis space equals space minus 120 space less than 0
therefore space space space space space space space space space space space straight V space space is space maximum space at space straight x space equals space 5 space space space space space space space space space space space space space space space space space space space space space space space
But x = 5 is the only extreme point
therefore space space space space space space straight V space is space maximum space at space straight x space equals space 5
therefore space space space space side space of space the space square space space equals space 5 space cm.