Let x represent the edge length of the square end. Then the area of the end of the package is x², and the length of the package is 126 -4x. The overall volume is ... V = x²×(126 -4x) V = -4x³ +126x²
The volume will be a maximum where the derivative of volume with respect to x is zero. V' = -12x² +252x V' = -12x(x -21) This will be zero for x = 0 and for x = 21
The maximum volume of the package is V = (21 in)²(126 in -4×21 in) V = 18,522 in³