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

3: 2

step-by-step explanation:

shorter piece is 2: 3, so just reverse it, and you get 3: 2