Problem 21 - Calculus Early Transcendentals: Pearson New International Edition

Question

Find the three-dimensional vector with length 9 , the sum of whose components is a maximum.

Step-by-Step Solution

Answer

\( \langle 3\sqrt{3}, 3\sqrt{3}, 3\sqrt{3} \rangle \)

Step 1: Set up the vector components

Let's denote the three-dimensional vector as \( \mathbf{v} = \langle x, y, z \rangle \). We need to maximize the sum \( x + y + z \).

Step 2: Apply the condition on vector length

The problem states the vector's length is 9. Hence, we use the formula for the magnitude of the vector: \( \sqrt{x^2 + y^2 + z^2} = 9 \). Simplifying, we square both sides to get \( x^2 + y^2 + z^2 = 81 \).

Step 3: Use symmetry to maximize the sum of components

Since we want to maximize \( x + y + z \) and the equation is symmetrical, assume \( x = y = z \). Now, substitute into \( x^2 + y^2 + z^2 = 81 \) to get \( 3x^2 = 81 \).

Step 4: Solve for the component value

Solving \( 3x^2 = 81 \), divide both sides by 3: \( x^2 = 27 \). Taking the square root, we find \( x = \sqrt{27} = 3\sqrt{3} \).

Step 5: Substitute back to find the vector

Since \( x = y = z = 3\sqrt{3} \), the vector is \( \mathbf{v} = \langle 3\sqrt{3}, 3\sqrt{3}, 3\sqrt{3} \rangle \). This vector has length 9 and the maximum sum of components.

Key Concepts

Vector Magnitude

In understanding three-dimensional vectors, learning to calculate vector magnitude is fundamental. The magnitude of a vector gives us the length of that vector in space, much like how the length of a line segment measures how far it stretches. For a vector denoted as \( \mathbf{v} = \langle x, y, z \rangle \), the magnitude is calculated using the formula:

• \( \|\mathbf{v}\| = \sqrt{x^2 + y^2 + z^2} \)

This formula is derived from the Pythagorean theorem in three dimensions. Imagine the vector as the diagonal of a box in 3D space formed by its components on the x, y, and z axes. By squaring each component, adding them, and taking the square root, we find this diagonal's length, which is the magnitude of the vector.
For example, if you need a vector with a specific magnitude, like 9 in this case, you should ensure that \( x^2 + y^2 + z^2 = 81 \) since the square of the magnitude (9) is 81. This equation is key in ensuring the vector has the desired length.

Symmetry in Vectors

Symmetry in vectors often simplifies problems involving vectors, especially when maximizing or minimizing expressions. Symmetry suggests that all parts of the vector are equal or behave similarly.
For an equation concerning vectors with a symmetric property, like the sum of components \( x + y + z \), you can assume equal values for each component. Here, assuming \( x = y = z \) exploits symmetry effectively.

• This assumption converts the problem from dealing with three unknowns into a single unknown, simplifying calculations significantly.

In the given problem, the vector is set up such that all its components are equal, i.e., \( x = y = z \). By doing so, the problem depends on solving for \( x \), which then applies to all components. Solving \( 3x^2 = 81 \) gives \( x = 3\sqrt{3} \), and thus the vector components are equal, simplifying the task of finding the solution.

Maximizing Vector Component Sum

Maximizing the sum of vector components involves strategic selection and manipulation of the components. In practical terms, you must comply with any given constraints, like a fixed magnitude, while maximizing the target sum expression.
To maximize \( x + y + z \) when \( x^2 + y^2 + z^2 = 81 \), observe that a symmetric distribution allows for a maximum sum.

• By setting each component to \( x = y = z \), you ensure an equally distributed vector, which is advantageous in maximizing their sum given the constraint.

Solving \( 3x^2 = 81 \), we find the value of the sum-maximizing component \( x = 3\sqrt{3} \). Substituting these into your vector \( \mathbf{v} = \langle 3\sqrt{3}, 3\sqrt{3}, 3\sqrt{3} \rangle \), confirms that the vector not only meets the magnitude requirement but also maximizes the sum of its components, thereby achieving the optimization goal.