Question

two squares of length x are cut out of adjacent corners of an 18 inch by 18 inch piece of cardboard and two rectangles of length 9 and width x are cut out of the other two corners of the cardboard (see figure below) the resulting piece of cardboard is then folded along the dashed lines to form a closed box. Find the dimensions and volume of the largest box that can be formed in this way.

Answer 1

• 3 in × 6 in × 12 in
• 216 in³

Explanation:

You want the dimensions and volume of a cuboid that can be made from an 18 in square of cardboard when an x-inch square is cut from two adjacent corners, and an x-inch by 9-inch rectangle is cut from the other two corners.

Dimensions

The longest side of the cuboid will have length (18 -2x). The second longest side will have length (18 -9 -x) = (9 -x). The shortest side will have length x.

Volume

The volume of the cuboid is the product of these side lengths:

V = (18 -2x)(9 -x)(x) = 2x(9 -x)²

Maximum volume

The value of x that maximizes volume will be the value that makes the derivative with respect to x be zero.

V' = 2(9 -x)² -4x(9 -x) = (9 -x)(2(9 -x) -4x) = 2(9 -x)(9 -3x)

V' = 6(9 -x)(3 -x)

The derivative will be zero when its factors are zero, at x=9 and x=3. The value x=9 gives zero volume.

The value x=3 gives a volume of ...

V = 2·3(9 -3)² = 6³ = 216 . . . . . cubic inches

The dimensions are ...

• x = 3
• 9 -x = 9 -3 = 6
• 18 -2x = 18 -6 = 12

The dimensions for maximum volume are 3 inches by 6 inches by 12 inches. The maximum volume is 216 cubic inches.