We are tasked with finding the maximum and minimum values of the function $f(x, y) = xy$ subject to the constraint $4x^2 + y^2 = 8$ using the method of Lagrange multipliers.

Step 1: Setting up the problem using Lagrange multipliers

The method of Lagrange multipliers involves introducing a new variable, $\lambda$ (the Lagrange multiplier), and solving the system of equations formed by:

1. $\nabla f(x, y) = \lambda \nabla g(x, y)$
2. The constraint equation $g(x, y) = 0$, where $g(x, y) = 4x^2 + y^2 - 8$.

So, our function $f(x, y) = xy$ and the constraint is $g(x, y) = 4x^2 + y^2 - 8$.

Gradient of $f(x, y)$:

$\nabla f(x, y) = \left( \frac{\partial}{\partial x}, \frac{\partial}{\partial y} \right) f(x, y) = (y, x)$

Gradient of $g(x, y)$:

$\nabla g(x, y) = \left( \frac{\partial}{\partial x}, \frac{\partial}{\partial y} \right) g(x, y) = (8x, 2y)$

Step 2: Setting up the system of equations

Using $\nabla f(x, y) = \lambda \nabla g(x, y)$, we get the system: y = \lambda (8x) \tag{1} x = \lambda (2y) \tag{2}

Additionally, we have the constraint: 4x^2 + y^2 = 8 \tag{3}

Step 3: Solving the system

From equation (2):

x = \lambda (2y) \implies \lambda = \frac{x}{2y} \quad \text{(assuming y \neq 0)}

Substitute $\lambda$ into equation (1):

$y = \frac{x}{2y} (8x) = \frac{8x^2}{2y} \implies y^2 = 4x^2$ Thus, $y = 2x$ or $y = -2x$.

Case 1: $y = 2x$

Substitute $y = 2x$ into the constraint equation $4x^2 + y^2 = 8$: $4x^2 + (2x)^2 = 8 \implies 4x^2 + 4x^2 = 8 \implies 8x^2 = 8 \implies x^2 = 1 \implies x = \pm 1$ When $x = 1$, $y = 2$; and when $x = -1$, $y = -2$.

Case 2: $y = -2x$

Substitute $y = -2x$ into the constraint equation: $4x^2 + (-2x)^2 = 8 \implies 4x^2 + 4x^2 = 8 \implies 8x^2 = 8 \implies x^2 = 1 \implies x = \pm 1$ When $x = 1$, $y = -2$; and when $x = -1$, $y = 2$.

Step 4: Evaluating $f(x, y) = xy$ at the critical points

We now have four critical points: $(1, 2), (-1, -2), (1, -2), (-1, 2)$.

• At $(1, 2)$, $f(1, 2) = 1 \times 2 = 2$.
• At $(-1, -2)$, $f(-1, -2) = (-1) \times (-2) = 2$.
• At $(1, -2)$, $f(1, -2) = 1 \times (-2) = -2$.
• At $(-1, 2)$, $f(-1, 2) = (-1) \times 2 = -2$.

Step 5: Conclusion

• Maximum value: The maximum value of $f(x, y) = xy$ is $\boxed{2}$, and it occurs at 2 points: $(1, 2)$ and $(-1, -2)$.

• Minimum value: The minimum value of $f(x, y) = xy$ is $\boxed{-2}$, and it occurs at 2 points: $(1, -2)$ and $(-1, 2)$.

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Do you need further details or explanations?

Here are some related questions to expand your understanding:

1. How does the method of Lagrange multipliers handle problems with multiple constraints?
2. What are the conditions for a critical point to be a local maximum or minimum in constrained optimization?
3. Can Lagrange multipliers be applied to three-variable functions with two constraints?
4. How do we interpret the Lagrange multiplier $\lambda$ geometrically?
5. How can we solve constrained optimization problems numerically if they cannot be solved algebraically?

Tip: Always verify the solution by checking if the critical points satisfy both the objective function and the constraint.