Question

A rectangular sheet of tin 45 cm by 24 cm is to be made into a box without top, by cutting off square from each corner and folding up the flaps. 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 a box is to be made by a rectangular sheet of tin of side 35 cm by 24 cm.

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

Length of the box after cutting will be 45−x−x or 45−2x.

Breadth of the box after cutting will be 24−x−x or 24−2x.

Height will remain same that is x.

Let, V( x ) be the volume of the box.

V( x )=length×breadth×height =( 45−2x )( 24−2x )( x ) =x( 1080−90x−48x+4 x 2 ) =4 x 3 −138 x 2 +1080x

Differentiate the function with respect to x,

V ′ ( x )=12 x 2 −276x+1080 =12( x 2 −23x+90 ) =12( x−18 )( x−5 ) (1)

Put V ′ ( x )=0,

12( x−18 )( x−5 )=0 x=5,18

When x=18, then breadth will be 24−2×18=−12. Breadth can’t be negative, so, x=5 is the only value.

Differentiate equation (1) with respect to x,

V ″ ( x )=12( x−5 )( 1 )+12( x−18 )( 1 ) V ″ ( 5 )=12( 5−5 )+12( 5−18 ) =−156

This shows that the function is negative, so, x=5 is the point of maxima.

Therefore, the side of square be cut off by 5c m so that the volume of box is maximum.