Problem 29 - Calculus of a Single Variable

Question

Maximum Volume A rectangular package to be sent by a postal service can have a maximum combined length and girth (perimeter of a cross section) of 108 inches (see figure). Find the dimensions of the package of maximum volume that can be sent. (Assume the cross section is square.)

Step-by-Step Solution

Answer

The dimensions of the package for maximum volume that can be sent by the postal service should be 18 inches by 18 inches by 36 inches.

Step 1: Set up the equations

Let x represent one side length of the square cross-section (which is also the width and the height), and let y represent the length of the package. According to the problem, the girth of the package (which is 4x in this case because the cross-section is square) plus the length cannot exceed 108 inches. So we get the first equation: \(4x + y = 108\). The volume V of the package is given by \(V = x^2y\), since it's a rectangular box and the cross section is a square.

Step 2: Express the volume in terms of one variable

In order to maximize the volume, we need to express it in terms of one variable. From the first equation \(4x + y = 108\), we can express y in terms of x, getting \(y = 108 - 4x\). Substitute this into the volume equation, we get \(V = x^2(108-4x) = 108x^2 - 4x^3\). This is the function we will maximize.

Step 3: Find the maximum volume

To find the maximum volume, we first find the derivative of the volume function: \(V' = 216x - 12x^2\). Setting the derivative equal to zero, we find \(216x - 12x^2 = 0\). Solving this quadratic equation for x, we obtain \(x = 18\) inches as the solution that makes sense in this context (the other solution \(x = 0\) is discarded because it doesn't make sense for the dimensions of a box). Substituting \(x = 18\) back into the equation \(4x + y = 108\), we get \(y = 36\) inches. So the optimal dimensions for the package are 18 inches by 18 inches by 36 inches.