Maximum volume

Question

By cutting away identical squares from each corner of a rectangular piece of cardboard and folding up the resulting flaps, an open box may be made. If the cardboard is 12 in. long and 8 in. wide, find the dimensions of the box that will yield the maximum volume. (Round your answers to two decimal places.)

___in (smallest value)

___in

___in (largest value)

Answer 1

The trick with these is to name the cut before you do anything else. So, let x be the side of the square we take out of each corner. When we fold the flaps up, that x becomes the height of the box, and the other two sides each lose x from both ends, so the base ends up 12 − 2x long and 8 − 2x wide. That gives us

V(x) = x(12 − 2x)(8 − 2x) = 4x³ − 40x² + 96x

It also quietly tells us what x is allowed to be, because the width 8 − 2x cannot go negative, so x has to stay between 0 and 4. That part matters in a minute.

Now we take the derivative and set it equal to zero, since we are looking for the spot where the volume stops growing and starts shrinking:

V′(x) = 12x² − 80x + 96 = 4(3x² − 20x + 24)

The quadratic formula gives x = (10 ± 2√7)/3, which comes out to about 1.57 and about 5.10. Remember x had to stay under 4, so 5.10 would fold the cardboard past itself and is not a real option. That leaves

x ≈ 1.57 in

From there the other two sides are 8 − 2(1.57) ≈ 4.86 and 12 − 2(1.57) ≈ 8.86. So, our final answer is 1.57 in, 4.86 in and 8.86 in, and if you want to check yourself the volume works out to about 67.60 cubic inches.