Problem 351 - Calculus Volume 3

Question

The sum of the length and the girth (perimeter of a cross-section) of a package carried by a delivery service cannot exceed 108 in. Find the dimensions of the rectangular package of largest volume that can be sent.

Step-by-Step Solution

Answer

The largest volume package has dimensions 36 in (L), 18 in (W), and 18 in (H).

Step 1: Define problem variables

Let's assume the package has length \( L \), width \( W \), and height \( H \). The girth is the perimeter of the cross-section, so for a rectangular package, it's defined as \( 2(W + H) \). According to the problem, the sum of the length and the girth must not exceed 108 inches: \( L + 2(W + H) \leq 108 \).

Step 2: Express length in terms of other dimensions

From the constraint equation \( L + 2(W + H) \leq 108 \), we get \( L = 108 - 2(W + H) \). This expresses the length in terms of the width and height.

Step 3: Write the volume function

The volume \( V \) of the package can be expressed as \( V = L \, \times \, W \, \times \, H = (108 - 2(W + H)) \, \times \, W \, \times \, H \).

Step 4: Simplify the volume function

Expand and simplify the volume function: \[ V = (108 - 2W - 2H)WH = 108WH - 2W^2H - 2WH^2. \] This is the expression we'll maximize.

Step 5: Apply calculus to find critical points

To find the package dimensions that yield the maximum volume, differentiate the volume function with respect to \( W \) and \( H \), then set these partial derivatives to 0 to find critical points. These partial derivatives are:\[ \frac{\partial V}{\partial W} = 108H - 4WH - 2H^2 \]\[ \frac{\partial V}{\partial H} = 108W - 2W^2 - 4WH \]Set each derivative to 0 and solve the system of equations.

Step 6: Solve the system of equations

Solving\begin{align*}108H - 4WH - 2H^2 &= 0 \108W - 2W^2 - 4WH &= 0\end{align*}is complex algebra, but one approach could be substituting the first into the second to get relations like \( W = H \). Solving yields \( W = 18 \) and \( H = 18 \).

Step 7: Find length using the constraint

Using the constraint equation with \( W = 18 \) and \( H = 18 \), solve for \( L \): \( L = 108 - 2(18 + 18) = 108 - 72 = 36 \).

Step 8: Confirm the dimensions maximize the volume

Substitute back into the volume expression and ensure it satisfies maximum conditions, checking second derivatives or other verification strategies if needed. The dimensions maximize the volume.

Key Concepts

Calculus Volume Optimization

When tackling the problem of maximizing volume within given constraints, calculus is an essential tool. Imagine you have a box, and you want it to have the largest possible space inside, or volume, while adhering to certain limitations. In this case, your limit is the total sum of the box's length and girth.
Volume optimization is about finding the best combination of dimensions to achieve this goal. You start by defining the volume of the rectangular package as a function of its dimensions: length (\(L\)), width (\(W\)), and height (\(H\)). For our package, the volume function can be represented as: \[ V = L imes W imes H \].
However, using constraints such as the maximum permissible length and girth, this volume function is expressed in terms of two variables, e.g., \(W\) and \(H\), making it easier to manage with calculus. Once expressed, calculus allows us to find critical points of the volume function—those points are where volume could potentially be maximized.

Partial Derivatives

Partial derivatives come into play when dealing with functions of multiple variables, like in this problem. Since the volume is dependent on width (\(W\)) and height (\(H\)), partial derivatives help us examine how small changes in one dimension affect the volume.
To do this, you first take the partial derivative of volume \(V\) with respect to \(W\), treating \(H\) as a constant. This calculates the rate at which volume changes for a small change in \(W\). Do the same with respect to \(H\), treating \(W\) as constant.
In mathematical terms, these are: \( \frac{\partial V}{\partial W} \) and \( \frac{\partial V}{\partial H} \). Set these derivatives equal to zero to find where volume stops rising—these are called critical points. Solving these equations simultaneously helps pinpoint the exact dimensions where the maximum volume occurs.

Constraint Optimization

The task of maximizing the volume under a restriction is a classic example of constraint optimization. Constraints limit the values that variables can take, guiding you to optimal solutions that satisfy these conditions.
When given a constraint such as "the sum of length and girth should not exceed a certain value," you express one variable (\(L\)) in terms of others (\(W\) and \(H\)), reducing the complexity of the problem. In this context, you have \( L = 108 - 2(W + H) \), which keeps the volume within constraints.
By substituting this expression into the volume function, you change a problem with three variables into one with two, streamlining calculations. Solving such problems typically involves using methods like the Method of Lagrange Multipliers or systematically applying calculus to reach the solution that satisfies all constraints and achieves optimal volume.