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Application Of Derivatives

Question
CBSEENMA12035539
Wired Faculty App

A square piece of tin of side 24 cm. is to be made into a box without top by cutting a square from each corner arid 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 maximum ? Also, find this maximum volume.

Solution
Let x (0 < x < 12) 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 open box formed by folding up the flaps.
therefore space space space space straight V space equals space straight x left parenthesis 24 minus 2 straight x right parenthesis space left parenthesis 24 minus 2 straight x right parenthesis space equals space straight x left parenthesis 24 minus 2 straight x right parenthesis squared space space space space space space space space space space space space space equals 4 straight x left parenthesis 12 minus straight x right parenthesis squared space space equals space 4 straight x left parenthesis straight x squared minus 24 straight x plus 144 right parenthesis space equals space 4 left parenthesis straight x cubed minus 24 straight x squared plus 144 straight x right parenthesis space space space dV over dx space equals space 4 left parenthesis 3 straight x squared minus 48 straight x plus 144 right parenthesis space space space space dV over dx space equals space 0 space space space space space rightwards double arrow space space space space 4 left parenthesis 3 straight x squared minus 48 straight x plus 144 right parenthesis space space space rightwards double arrow space space space straight x squared minus 16 straight x plus 48 space equals space 0 rightwards double arrow space space space space left parenthesis straight x minus 4 right parenthesis thin space left parenthesis straight x minus 12 right parenthesis space equals space 0 space space space space space space rightwards double arrow space space straight x space equals space 4 comma space space space 12 space space space space rightwards double arrow space space straight x space equals space 4 space as space straight x space equals space 12 space not an element of left parenthesis 0 comma space 12 right parenthesis space
                   fraction numerator straight d squared straight V over denominator dx squared end fraction space equals space 4 left parenthesis 6 straight x minus 48 right parenthesis
At space straight x space equals 4 comma space space space space fraction numerator straight d squared straight V over denominator dx squared end fraction space equals space 4 left parenthesis 24 minus 48 right parenthesis space equals space minus 96 thin space less than 0 therefore space space space space space space space straight V space is space local space maximum space at space straight x space equals space 4 But space straight x space equals 4 space is space the space only space extreme space point therefore space space space space straight V space is space maximum space at space straight x space equals space 4 therefore space space space side space of space the space square space space equals space 4 space cm. and space max. space value space equals space 4 space cross times 4 cross times left parenthesis 12 minus 4 right parenthesis squared space equals space 1024 space cubic space cm. space
                   

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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