Problem 49 - Calculus Early Transcendentals

Question

Find the dimensions of a rectangular box of maximum volume such that the sum of the lengths of its 12 edges is a constant \(c .\)

Step-by-Step Solution

Answer

The dimensions of maximum volume are \(x = y = z = \frac{c}{12}\).

Step 1: Understanding the Problem

We need to find the dimensions of a rectangular box with maximum volume, given the constraint that the sum of the lengths of its edges is constant at \(c\). The box has 12 edges and these edges are determined by the three dimensions: length \(x\), width \(y\), and height \(z\).

Step 2: Expressing Edge Sum Constraint

The sum of the lengths of the 12 edges of the box is \(4x + 4y + 4z = c\). We can simplify this equation to \(x + y + z = \frac{c}{4}\), which expresses the constraint condition.

Step 3: Formulating the Volume Function

The volume \(V\) of the rectangular box is given by \(V = x \, y \, z\). Our objective is to maximize this volume, \(V\), subject to the constraint \(x + y + z = \frac{c}{4}\).

Step 4: Using the Method of Lagrange Multipliers

Set up the Lagrange function: \( \mathcal{L}(x, y, z, \lambda) = xyz + \lambda \left( \frac{c}{4} - (x+y+z) \right)\). Partial derivatives with respect to \(x\), \(y\), and \(z\) will give us equations: \( yz = \lambda \), \( xz = \lambda \), and \( xy = \lambda \).

Step 5: Solving the Equations Simultaneously

From \( yz = xz = xy = \lambda \), we infer \(x = y = z\), since the equations are symmetric and assuming all dimensions are positive and non-zero. Substitute \(x = y = z\) into the constraint equation \(x + y + z = \frac{c}{4}\), we get \(3x = \frac{c}{4}\). Thus, \(x = \frac{c}{12}\), and as \(x = y = z\), we have \( y = \frac{c}{12}\) and \( z = \frac{c}{12}\).

Step 6: Finding the Maximum Volume

Plug back the values \(x = y = z = \frac{c}{12}\) into the volume equation, \(V = xyz = \left(\frac{c}{12}\right)^3 = \frac{c^3}{1728}\).

Step 7: Conclusion of the Optimization

The dimensions of the box that yield the maximum volume, given the constraint, are \( x = y = z = \frac{c}{12} \). The volume achieved is \(\frac{c^3}{1728}\).

Key Concepts

Lagrange Multipliers

When faced with optimization problems in calculus, Lagrange multipliers provide an elegant method for finding the maxima or minima of a function subject to constraints. This technique is particularly useful when dealing with multivariable functions. In our scenario, the goal is to maximize the volume of a rectangular box with a fixed total edge length.

The essence of Lagrange multipliers is to introduce an auxiliary variable, denoted as \( \lambda \), which helps in solving the problem with constraints. By setting up a new function that combines the original function and the constraint, we can calculate partial derivatives to find the points at which our function, subject to the constraint, reaches its maximum or minimum value. The formula consists of creating the Lagrangian:

• \( \mathcal{L}(x, y, z, \lambda) = f(x, y, z) + \lambda (g(x, y, z) - c) \)

In our exercise, the original function would be the volume \( xyz \) and the constraint is \( x+y+z = \frac{c}{4} \), leading to the Lagrangian: \( \mathcal{L}(x, y, z, \lambda) = xyz + \lambda \left(\frac{c}{4} - (x+y+z)\right) \). Calculating the partial derivatives gives us the necessary equations to solve for the variables.

Volume of Rectangular Box

The volume of a rectangular box, also known as a rectangular prism, can be expressed in a straightforward manner. It is simply the product of its three dimensions: length \( x \), width \( y \), and height \( z \). Thus, the formula for calculating the volume \( V \) is given by:

• \( V = x \cdot y \cdot z \)

This formula plays a central role in optimization problems that aim to determine maximum volume with specific constraints.

Understanding the physical meaning of this formula helps to visualize problems more concretely. A larger product of \( x, y, \) and \( z \) implies a larger box. Hence, the challenge of maximizing volume under certain conditions translates into finding values of \( x, y, \) and \( z \) that make the most effective use of the given constraints.

Constraints in Calculus

Constraints in calculus optimization problems act as conditions that our solution must satisfy. In many real-world situations, we don't have complete freedom to choose the values of variables; instead, they are subject to these limitations. Known simply as constraints, they define the feasible region for solutions.

In the context of this exercise, the constraint represents the fixed sum of the edges of the box. This constraint is expressed mathematically as \( x + y + z = \frac{c}{4} \). Such constraints necessitate the use of methods like Lagrange multipliers.

• The equation forms a boundary within which the box dimensions must lie.
• The constraint ensures our solution is realistic and feasible given the conditions.

Solving within these constraints gives us dimensions of a box that aligns with the given edge total while optimizing the desired property, such as volume.

Maximization Problems

Maximization problems in calculus involve finding the largest possible value of a function, given certain restrictions or parameters. Often encountered in fields ranging from economics to engineering, these problems build on our understanding of how variables interact under constraints to achieve optimal results.

The problem we're dealing with is a classic example of a maximization problem. Here, we aim to achieve the maximum volume of the box while adhering to a constraint on the sum of the edges.

• The objective is clearly defined: maximize \( V = x \cdot y \cdot z \).
• The constraint provides a boundary condition: \( x + y + z = \frac{c}{4} \).

Through the application of Lagrange multipliers, we solved the conditions to find a cube-shaped box where \( x = y = z = \frac{c}{12} \). This configuration yields the largest possible volume for the box, given the edge-length constraint.