To solve this problem using Lagrange multipliers, let's follow these steps:

Problem Setup

We are given:

• Function to maximize or minimize: $f(x, y) = x - 3y$
• Constraint: $g(x, y) = x^2 + 2y^2 - 22 = 0$

Using the method of Lagrange multipliers, we introduce a multiplier $\lambda$ and set up the equations: $\nabla f = \lambda \nabla g$ which gives: $f_x = \lambda g_x \quad \text{and} \quad f_y = \lambda g_y$

Step 1: Compute Gradients

1. Gradient of $f$: $\nabla f = \left( \frac{\partial f}{\partial x}, \frac{\partial f}{\partial y} \right) = (1, -3)$

2. Gradient of $g$: $\nabla g = \left( \frac{\partial g}{\partial x}, \frac{\partial g}{\partial y} \right) = (2x, 4y)$

Step 2: Set Up Lagrange Equations

From $\nabla f = \lambda \nabla g$, we get:

1. $1 = \lambda \cdot 2x$
2. $-3 = \lambda \cdot 4y$

And we also have the constraint: $x^2 + 2y^2 = 22$

Step 3: Solve for $\lambda$ and Find Relations Between $x$ and $y$

1. From $1 = \lambda \cdot 2x$, we get: $\lambda = \frac{1}{2x}$

2. Substitute $\lambda$ into the second equation: $-3 = \frac{1}{2x} \cdot 4y$ Simplify to find a relation between $x$ and $y$: $-3 = \frac{2y}{x} \Rightarrow x = -\frac{2y}{3}$

Step 4: Substitute Back into the Constraint

Substitute $x = -\frac{2y}{3}$ into the constraint $x^2 + 2y^2 = 22$: $\left(-\frac{2y}{3}\right)^2 + 2y^2 = 22$ $\frac{4y^2}{9} + 2y^2 = 22$ Multiply through by 9 to clear the fraction: $4y^2 + 18y^2 = 198$ $22y^2 = 198$ $y^2 = 9 \Rightarrow y = \pm 3$

Step 5: Find Corresponding $x$ Values

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

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

So we have two points: $(-2, 3)$ and $(2, -3)$.

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

1. At $(-2, 3)$: $f(-2, 3) = -2 - 3 \cdot 3 = -2 - 9 = -11$

2. At $(2, -3)$: $f(2, -3) = 2 - 3 \cdot (-3) = 2 + 9 = 11$

Conclusion

• Maximum value of $f(x, y)$ is $11$ at $(2, -3)$.
• Minimum value of $f(x, y)$ is $-11$ at $(-2, 3)$.

Final Answer

• Maximum = $11$
• Minimum = $-11$

Let me know if you would like further explanation on any steps or additional practice problems!

Here are 5 questions that expand on this topic:

1. How does the method of Lagrange multipliers apply when there are multiple constraints?
2. What are some common types of optimization problems where Lagrange multipliers are used?
3. How would the solution change if the constraint equation had been different, such as $x^2 + 3y^2 = 22$?
4. Can Lagrange multipliers be used in three-dimensional problems, and if so, how do the calculations differ?
5. How can you verify if a solution found using Lagrange multipliers is indeed a global maximum or minimum?

Tip: Always verify the constraint satisfaction when substituting values back to ensure they meet the initial conditions.