A grocer wants to make fruit boxesfrom square cardboard 36 inches long and 36 inches wide as shown.
The shaded areas are cut away and the rest is folded along the dashed lines. The sides are folded up and stuck together using the four flaps marked F. The lid has two flaps, marked L, which are not glued.
Calculate the volume of the finished box. What is the largest volume possible if the 4 inches can be changed to any measurement?

You want to know the volume of the box that can be formed from the pattern shown, and the maximum volume that can be had by changing the 4-inch dimension.

Volume

The volume of the box is the product of its length, width, and height. The height of the box is shown as 4 inches.

The length is shown as the difference between 36 inches and twice the 4-inch dimension:

length = 36 -2(4) = 28 . . . . inches

The width is shown as half the difference between 36 inches and twice the 4-inch dimension. That will make it be half the length.

width = (1/2)(36 -2(4)) = 14 . . . . inches

Then the volume of the box from the pattern shown is ...

V = LWH = (28 in)(14 in)(4 in) = 1568 in³

Maximum volume

Using the above observations regarding dimensions, we can write the volume in terms of the height of the box (x) as ...

height = x

length = 36 -2x

width = 18 -x

volume = (x)(36 -2x)(18 -x) = 2x³ -72x² +648x

The maximum volume will be found at the value of x that makes the derivative of volume be zero:

V' = 6x² -144x +648 = 0

x² -24x +108 = 0 . . . . . . . . divide by 6

(x -6)(x -18) = 0 . . . . . . . . factor

Values of x that make the derivative zero are x=6 and x=18. The original equation is a cubic with a positive leading coefficient, so we know these values correspond to a local maximum and a local minimum, respectively.

The largest volume possible is obtained with a corner cut of 6 inches. That volume is ....

V = (6)(36 -2·6)(18 -6) = 6(24)(12) = 1728 . . . . cubic inches

The largest volume possible is 1728 in³.