We are asked to find the maximum and minimum values of the function $f(x, y) = 3x - 2y$, subject to the constraint $x^2 + y^2 = 13$. This is a standard optimization problem that can be solved using the method of Lagrange multipliers.

Step 1: Define the Lagrange Multiplier Equation

We introduce the Lagrange multiplier $\lambda$ and set up the system of equations by defining the Lagrange function $\mathcal{L}(x, y, \lambda)$: $\mathcal{L}(x, y, \lambda) = f(x, y) - \lambda (g(x, y) - c)$ where $g(x, y) = x^2 + y^2$ and the constraint is $g(x, y) = 13$.

So, the Lagrange function becomes: $\mathcal{L}(x, y, \lambda) = (3x - 2y) - \lambda (x^2 + y^2 - 13)$

Step 2: Compute the Gradient

Now, take the partial derivatives of $\mathcal{L}(x, y, \lambda)$ with respect to $x$, $y$, and $\lambda$, and set them equal to zero: $\frac{\partial \mathcal{L}}{\partial x} = 3 - 2\lambda x = 0 \quad \Rightarrow \quad 3 = 2\lambda x \quad \Rightarrow \quad \lambda = \frac{3}{2x}$ $\frac{\partial \mathcal{L}}{\partial y} = -2 - 2\lambda y = 0 \quad \Rightarrow \quad -2 = 2\lambda y \quad \Rightarrow \quad \lambda = -\frac{1}{y}$ $\frac{\partial \mathcal{L}}{\partial \lambda} = -(x^2 + y^2 - 13) = 0 \quad \Rightarrow \quad x^2 + y^2 = 13$

Step 3: Solve the System of Equations

Now, we have two expressions for $\lambda$: $\lambda = \frac{3}{2x} \quad \text{and} \quad \lambda = -\frac{1}{y}$ Set them equal to each other: $\frac{3}{2x} = -\frac{1}{y} \quad \Rightarrow \quad 3y = -2x \quad \Rightarrow \quad y = -\frac{2}{3}x$

Step 4: Substitute into the Constraint

Substitute $y = -\frac{2}{3}x$ into the constraint $x^2 + y^2 = 13$: $x^2 + \left(-\frac{2}{3}x\right)^2 = 13 \quad \Rightarrow \quad x^2 + \frac{4}{9}x^2 = 13 \quad \Rightarrow \quad \frac{13}{9}x^2 = 13$ $x^2 = 9 \quad \Rightarrow \quad x = 3 \quad \text{or} \quad x = -3$

Step 5: Find the Corresponding $y$-values

For $x = 3$: $y = -\frac{2}{3}(3) = -2$ For $x = -3$: $y = -\frac{2}{3}(-3) = 2$

Step 6: Evaluate $f(x, y)$ at the Critical Points

Now, evaluate the function $f(x, y) = 3x - 2y$ at the points $(3, -2)$ and $(-3, 2)$:

• At $(3, -2)$: $f(3, -2) = 3(3) - 2(-2) = 9 + 4 = 13$
• At $(-3, 2)$: $f(-3, 2) = 3(-3) - 2(2) = -9 - 4 = -13$

Step 7: Conclusion

• The maximum value of $f(x, y)$ is $\boxed{13}$.
• The minimum value of $f(x, y)$ is $\boxed{-13}$.

Would you like a deeper explanation or details? Here are five related questions:

1. How does the method of Lagrange multipliers work for functions with more than two variables?
2. Can the Lagrange multiplier method be used with inequality constraints?
3. How do you visualize the function and constraint in this problem geometrically?
4. What are alternative methods for constrained optimization problems besides Lagrange multipliers?
5. What happens if there is no solution to the system of equations derived from the Lagrange multiplier method?

Tip: When solving with Lagrange multipliers, always verify the constraint is satisfied at the critical points to ensure correctness.