MinMax07 Dimensions of a bricklike box.

Postal rates increase when the girth (once around) plus the length of a package exceeds 84 inches. What are the dimensions of a "bricklike" shaped box with square ends to provide maximum volume?

The defining equation is the volume which in this case is the area of the end, x², times the length, y: V = x²y

The constraint is that the girth, 4x, plus the length, y, is limited to 84: 4x + y = 84

The simplest way to write the V =ź equation in one variable is to solve the constraint equation for y: y = 84 - 4x and substitute for y in the defining equation.

V = x²(84 - 4x) = 84x² - 4x³

The first derivative of V is V' = 168x - 12x² and setting V' = 0, 12x(14 - x) = 0 produces two values of x where the slope of V vs. x is zero: x = 0 and x = 14. The value x = 0 produces a zero volume, about as minimal as you can get, so x = 14 is a good bet for maximum volume.

The second derivative of V is V'' = 168 - 24x . Evaluating V'' at x = 14 is V''(14) = 168 - 24(14) = 168 - 336 = -168 verifying our suspicion that x = 14 produced the maximum volume.

Going back to the constraint equation solved for y, the corresponding y dimension is y = 84 - 4(14) = 84 - 56 = 28.

A box with a square end 14 inches on a side and length 28 produces the maximum volume within the girth and length restrictions.