Question

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.

Answer 1

V(max) = 1872.42 in³

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