Question

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

Answer 1

It is given that the box is to be made by a square piece of tin of side 18 cm.

Let the side of the tin that have to be cut is x cm.

Length and breadth after cutting will be 18−x−x or 18−2x.

Height will remain same that is x.

Let, V be the volume of the box.

V( x )=length×breadth×height =( 18−2x )( 18−2x )( x ) = ( 18−2x ) 2 ( x )

Differentiate the function with respect to x,

V ′ ( x )= ( 18−2x ) 2 −4x( 18−2x ) =( 18−2x )( 18−2x−4x ) =( 18−2x )( 18−6x ) (1)

Put V ′ ( x )=0,

( 18−2x )( 18−6x )=0 x=3,9

At x=9, length after cutting will be 18−2×9=0 which is not possible, so, x=3 is the only value.

Differentiate equation (1) with respect to x,

V ″ ( x )=( 18−6x )( −2 )+( 18−2x )( −6 ) =−36+12x−108+12x =24x−144 =−24( 6−x )

Substitute x=3,

V ″ ( 3 )=−24( 6−3 ) =−24( 3 ) =−72 <0

This shows that V ″ ( x ) is negative, so, x=3 is the point of maxima.

Therefore, 3 cm of the cube side should be cut off so that the volume of the box will be maximum.