Find the dimensions of the open rectangular box of maximum volume that can be made from a sheet of cardboard 41 in. by 22 in. by cutting congruent squares from the corners and folding up the sides. Then find the volume.

V (max) = 1872.42 in³

Step-by-step explanation:

Let call " x " the side of the cut square in each corner, then the sides of rectangular base (L and W) will be

L = 41 - 2*x and

W = 22 - 2*x x = the height of the open box

Volume of the box is

V (b) = L*W*x

And volume of the box as a function of x is:

V (x) = (41 - 2*x) * (22 - 2*x) * x

V (x) = (902 - 82*x - 44*x + 4*x²) * x ⇒ V (x) = (902 - 126*x + 4x²) * x

V (x) = 902*x - 126*x² + 4x³

Taking derivatives on both sides of the equation we get:

V' (x) = 902 - 252*x + 12 * x²

V' (x) = 12*x² - 252*x + 902

V' (x) = 0 ⇒ 12*x² - 252*x + 902 = 0

A second degree equation solving for x (dividing by 2)

6*x² - 126 * x + 451 = 0

x₁,₂ = (126 ± √ 15876 - 10824) / 12

x₁,₂ = (126 ± 71,08) / 12

x₁ = 16 in (we dismiss this value since 2x > than 22 there is not feasible solution for the problem)

x₂ = 54.92/12 ⇒ x₂ = 4,58 in

Then maximum volume is:

V (max) = 31.84 * 12.84 * 4,58

V (max) = 1872.42 in³