Answer:
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
