Suppose a package to be mailed using a certain postal service may not measure more than 138 inches in length plus girth.

Question

Suppose a package to be mailed using a certain postal service may not measure more than 138 inches in length plus girth. (Length is the longest dimension and girth is the largest distance around the package, perpendicular to the length.) Find the greatest volume (in inches cubed) of a rectangular box with square base that may be mailed.

I got width, height, and length equal to 46, but the answer was wrong.

How I should solve this problem? I want to get the solution as soon as possible.

Answer 1

We want to recognize since the base is a square, our volume formula, with x being the width of the base and y being the length of the box is: V = x²y.

Now, with the information about the length and girth (perimeter around the box), we get a constraint equation of 4x + y = 138.

Solving for y we obtain: y = 138 - 4x.

From here, we plug this value into our volume equation to get:

V(x) = x²(138 - 4x) = 138x² - 4x³.

Next, we want to maximize this equation, so using the power rule we get:

V'(x) = 276x - 12x².

Setting to zero we obtain:

0 = 276x - 12x²

0 = 12x(23 - x).

Thus, x = 0 (which is not possible) or x = 23.

To verify this is a maximum: notice V'(22) = 264 and V'(24) = -288, thus at x = 23, our derivative is zero and changes from positive to negative, therefore we have obtained a maximum value on V(x).

To finish, we plug x = 23 into our constraint equation:

4(23) + y = 138

92 + y = 138

y = 138 - 92 = 46.

Thus, the max volume is V = (23²)(46) = 24,334 in³.