Problem 53 - Calculus Early Transcendentals

Question

If the length of the diagonal of a rectangular box must be \(L,\) what is the largest possible volume?

Step-by-Step Solution

Answer

The largest possible volume is \( \frac{L^3}{3\sqrt{3}} \).

Step 1: Understanding the Problem

We need to find the largest possible volume of a rectangular box given that the length of the diagonal is fixed at \( L \). To solve this, we have to relate the volume with the diagonal length.

Step 2: Expressing the Diagonal

Let the dimensions of the box be \(x, y, \) and \(z\). The diagonal, according to the Pythagorean theorem in three dimensions, is \(\sqrt{x^2 + y^2 + z^2} = L\).

Step 3: Maximizing the Volume Function

The volume of the box \( V \) can be expressed as \( V = x \, y \, z \). We seek to maximize this expression.

Step 4: Relating Dimensions through Diagonal

From the expression for the diagonal, we have \( x^2 + y^2 + z^2 = L^2 \). To find when the volume is maximum, we utilize the method of Lagrange multipliers or symmetry arguments.

Step 5: Optimal Dimension Ratios

It turns out that the maximum volume, under the given constraint, is achieved when \(x = y = z\). This implies that \(x\) is the length of each side of a cube.

Step 6: Find the Side Length of the Cube

Using \(x^2 + y^2 + z^2 = L^2\) and assuming \(x = y = z\), we get \(3x^2 = L^2\). Solve for \(x\) to find \(x = \frac{L}{\sqrt{3}}\).

Step 7: Calculate the Maximum Volume

The maximum volume \(V\) is when the box is a cube with sides of length \(x = \frac{L}{\sqrt{3}}\), so: \[ V = \left(\frac{L}{\sqrt{3}}\right)^3 = \frac{L^3}{3\sqrt{3}}. \]

Key Concepts

Extremum Problems

Extremum problems are about finding the maximum or minimum values of a function under certain conditions. In this exercise, our task is to determine the largest possible volume of a rectangular box while keeping the diagonal length fixed at a certain value, denoted as \( L \). This represents a maximization problem, as we are seeking the maximum volume.

When dealing with extremum problems, we frequently use methods like calculus and related concepts to find points where functions achieve their highest or lowest potential values. These points are also known as extrema. In our specific problem, the extremum we are interested in is the box's maximum volume, given a fixed diagonal length.

Pythagorean Theorem

The Pythagorean theorem is key to relating the dimensions of the box to its diagonal length. It's typically applied in the context of right-angled triangles and states that in a right triangle, the square of the length of the hypotenuse is equal to the sum of the squares of the other two sides.

For the rectangular box, consider the right triangle formed with the diagonal as the hypotenuse. If the sides of the box are \( x \), \( y \), and \( z \), the theorem extends in three dimensions and can be expressed as:

• \( \sqrt{x^2 + y^2 + z^2} = L \)

Squaring both sides, you get:

• \( x^2 + y^2 + z^2 = L^2 \)

This equation is crucial in finding the maximum volume, as it serves as the constraint that links the dimensions.

Lagrange Multipliers

Lagrange multipliers are a helpful tool in optimization problems with constraints. They allow us to find the maximum or minimum of a function subject to equality constraints, in this case, the constraint \( x^2 + y^2 + z^2 = L^2 \).

To use Lagrange multipliers, we set up an auxiliary function, called the Lagrangian, which incorporates both the original function we seek to optimize, and the constraint. Here, our Lagrangian \( \mathcal{L} \) would be:

• \( \mathcal{L}(x, y, z, \lambda) = x y z - \lambda (x^2 + y^2 + z^2 - L^2) \)

We then take partial derivatives of this function with respect to \( x \), \( y \), \( z \), and \( \lambda \), and solve the system of equations to find the critical points for optimization. Solving the system shows maximum volume when \( x = y = z \) coinciding with our symmetry argument for optimization.

Critical Points

Critical points play a crucial role in optimization problems, as they are potential locations where a function can achieve its maximum or minimum value. These points are found by setting the derivatives of a function equal to zero. For functions with constraints, critical points are found by using methods like Lagrange multipliers.

In this problem, the critical condition arises when we find that the maximum volume occurs under the condition \( x = y = z \), which happens due to symmetry and is confirmed by the solutions when using Lagrange multipliers. This results in each dimension of the box being equal, forming a cube.

The notion of critical points helps ensure we don't just find a local extremum but the absolute maximum given the constraints. By confirming \( 3x^2 = L^2 \) and solving for \( x \), we indeed find a critical point that gives the maximum possible volume as \( \frac{L^3}{3\sqrt{3}} \).