## To find the dimensions of the rectangular package with a square end that maximizes the volume while satisfying the limit on the size of packages, we can use calculus and optimization.

Let's assume that the side length of the square end is 'x' inches. Then the length of the rectangular package would be 'x' inches, and the width would be 'w' inches.

Since we want to maximize the volume, we need to optimize the volume function V = x * x * w = x^2 * w. However, we also need to take into account the constraint that the length plus the girth must be 84 inches or less.

The girth of the package is given by 2w, and the length is x. So we have the constraint x + 2w â‰¤ 84.

To proceed with the optimization, we can solve for one of the variables in terms of the other using the constraint equation. Rearranging the constraint equation, we have w â‰¤ (84 - x)/2.

Substituting this expression for w into the volume function, we get V = x^2 * ((84 - x)/2).

Now, we need to find the critical points of the volume function by taking the derivative with respect to x and setting it equal to zero.

dV/dx = 2x * ((84 - x)/2) - x^2/2 = x * (84 - x - x/2) = x * (84 - (3x/2)).

Setting dV/dx = 0, we have x * (84 - (3x/2)) = 0.

This equation gives us two possible solutions for x: x = 0 or (84 - (3x/2)) = 0.

Simplifying the second equation, we get (3x/2) = 84, which gives x = (2/3) * 84 = 56.

Now we can check these critical points and evaluate the volume at these points to find the maximum.

When x = 0, the volume is V = 0 * (84 - 0) = 0.

When x = 56, the volume is V = 56^2 * ((84 - 56)/2) = 56^2 * 14 = 54880 cubic inches.

Since the volume is non-negative, the maximum volume occurs at x = 56 inches.

Thus, the dimensions of the rectangular package with a square end that maximizes the volume are: side length (x) = 56 inches, length (x) = 56 inches, and width (w) = (84 - 56)/2 = 14 inches.