Question

Find the dimensions of the rectangular package of largest volume subject to the constraint that the sum of the length and the girth cannot exceed 108 inches (see figure). (Hint: Maximize \(V=x y z\) subject to the constraint \(x+2 y+2 z=108\).)

Short answer

The dimensions of the rectangular package that would give the largest volume under the given constraint are: x = 36 inches, y = 18 inches and z = 27 inches.

Step-by-step solution

Understand the Variables and Constraint

For a rectangular box, the volume \(V\) is computed as \(V = x \cdot y \cdot z\). The constraint is that the sum of length (\(x\)) and girth (which is twice the sum of width and height, \(\(2y + 2z)\)) should not exceed 108 inches. Mathematically, this constraint is expressed as \(x + 2y + 2z = 108\).

Express the Volume in Terms of a Single Variable

From the constraint, we can express either \(y\) or \(z\) in terms of \(x\) (or vice versa). For simplicity, express \(y\) in terms of \(x\) and \(z\) as follows: \(y = \frac{108 - x - 2z}{2}\). Substitute this into the volume equation to get \(V = x \cdot \frac{(108 - x- 2z)}{2} \cdot z = \frac{x(108 - x - 2z)z}{2}\).

Find the Maximum Volume

To find the maximum volume, differentiate the volume equation with respect to \(z\), equate the derivative to zero and solve for \(z\): \( \frac{dV}{dz} = \frac{x(108 - 3x - 4z)}{2} = 0\). Solving this equation yields \(z = \frac{27 - x}{2}\). Substitute this back into the constraint equation \(x + 2y + 2z = 108\) to find the corresponding values of \(x\) and \(y\). Using these values gives us the dimensions that maximize the volume.

Verify the Solution

Verify that the solution obtained in Step 3 indeed provides a maximum by applying the second derivative test. If the second derivative of the volume is less than zero, then \(z = \frac{27 - x}{2}\) is a maximum. This ensures that the solution is not a minimum or an inflection point.

Key concepts

Constrained Optimization

Constrained optimization is a fundamental concept in calculus and applied mathematics, where the goal is to optimize (either maximize or minimize) a function subject to certain restrictions, or constraints. In practical terms, it's about finding the best solution within a defined set of rules.

For example, when designing a package, we might seek to maximize the volume while staying within material usage limits, such as the total length of the packaging material. This constraint would form part of our optimization problem.

In the given exercise, we are tasked with finding the maximum volume of a rectangular package while adhering to a constraint on its dimensions. By setting up equations for the volume and the constraints, we turn a real-world problem into an optimization task that can be solved using calculus.

Volume Maximization

Volume maximization is a specific type of optimization problem where the objective is to find the shape dimensions that provide the largest volume. These problems are common in manufacturing and design since materials and space usually have constraints.

Our problem requires maximizing the volume of a rectangular box given a limit to the sum of its length and girth. The volume of a rectangular box is the product of its length, width, and height (\( V = x \times y \times z \)). To tackle such a problem, we often start by expressing the volume in terms of a single variable—this simplification lets us use the tools of calculus to find the maximum volume.

Derivatives

Derivatives are at the heart of calculus, often thought of as mathematical tools that measure change. When dealing with functions, a derivative gives us the rate at which one variable changes with respect to another. It's fundamental for solving optimization problems.

In our rectangular package problem, we first express the volume in terms of one variable and then take the derivative with respect to that variable. By setting this derivative equal to zero, we find critical points which are potential candidates for our maximum volume. This process, known as taking the derivative, is crucial because it guides us to the values we seek for optimization.

Second Derivative Test

Once we've found potential candidates for our optimized solution using derivatives, we need to confirm whether these points indeed represent maximum (or minimum) values. The second derivative test is a method used to verify our findings.

If the second derivative of our volume function is negative at a critical point, it indicates that we have a local maximum—just what we're looking for in volume maximization problems. In the context of our package problem, applying the second derivative test to the volume function with respect to the variable ensures that we've indeed found the dimensions that maximize the volume under the given constraint.