What is the maximum value of the rectangular box (Calculus)

Question

Find the dimensions of the open rectangular box of maximum volume that can be made from a sheet of cardboard 11in. by 7in. by cutting congruent squares from the corners and folding up the sides. Then find the volume.

Answer 1

By folding up the sides, we get an "open top" box with length (11 - 2x), a width of (7 - 2x) and a height of x. So, since volume is length times width times height, we can say V(x) = (11 - 2x) (7 - 2x)(x) or, multiplying it out, we get V(x) = 4x³ - 36x² + 77x

If we want to find local extremes for the volume, we take the first derivative and set it equal to zero.

V '(x) = 12x² - 72x + 77

12x² - 72x + 77 = 0 solves to x = 1.393 and x = 4.607

We know that x = 4.607 is not a real answer because the width is only 7 so the biggest square cutout must be smaller than 3.5 so the answer must be 1.393. To check, let's plug in a few values and see how the volume varies.

V(0.5) = 30

V(1) = 45

V(1.393) = 48.217

V(1,5) = 48

V(2) = 42

This checks out so when x = 1.393, the box dimensions are 8.214 inches x 4.214 inches x 1.393 inches yielding again a volume of 48.217 cubic inches