To solve this problem using Lagrange multipliers, we need to maximize and minimize the function $f(x, y) = 2x + y$ subject to the constraint $g(x, y) = x^2 + y^2 = 5$.

Step 1: Set up the Lagrange multiplier equations

We introduce a Lagrange multiplier $\lambda$, and form the Lagrange function:

$\mathcal{L}(x, y, \lambda) = f(x, y) - \lambda \cdot g(x, y)$ $\mathcal{L}(x, y, \lambda) = (2x + y) - \lambda (x^2 + y^2 - 5)$

Step 2: Take partial derivatives and set them equal to zero

1. Partial derivative with respect to $x$: $\frac{\partial \mathcal{L}}{\partial x} = 2 - \lambda \cdot 2x = 0$ $2 = 2\lambda x$ \lambda = \frac{1}{x} \quad \text{(for x \neq 0)}

2. Partial derivative with respect to $y$: $\frac{\partial \mathcal{L}}{\partial y} = 1 - \lambda \cdot 2y = 0$ $1 = 2\lambda y$ $\lambda = \frac{1}{2y}$

3. Partial derivative with respect to $\lambda$: $\frac{\partial \mathcal{L}}{\partial \lambda} = -(x^2 + y^2 - 5) = 0$ $x^2 + y^2 = 5 \quad \text{(this is just the constraint equation)}$

Step 3: Solve the system of equations

From the equations for $\lambda$: $\frac{1}{x} = \frac{1}{2y}$ Cross-multiply to get: $2y = x$

Substitute $x = 2y$ into the constraint equation $x^2 + y^2 = 5$:

$(2y)^2 + y^2 = 5$ $4y^2 + y^2 = 5$ $5y^2 = 5$ $y^2 = 1$ $y = \pm 1$

Now, substitute $y = \pm 1$ into $x = 2y$:

• When $y = 1$, $x = 2(1) = 2$
• When $y = -1$, $x = 2(-1) = -2$

Step 4: Evaluate $f(x, y)$ at these points

• When $(x, y) = (2, 1)$, $f(x, y) = 2(2) + 1 = 4 + 1 = 5$
• When $(x, y) = (-2, -1)$, $f(x, y) = 2(-2) + (-1) = -4 - 1 = -5$

Step 5: Conclusion

The maximum value of $f(x, y)$ is 5, and the minimum value is -5.

Thus, the correct answer is:

c. 5, -5

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Would you like a detailed explanation of any part of the solution? Here are some related questions you might find helpful:

1. How do you determine if a point is a maximum or minimum when using Lagrange multipliers?
2. Can Lagrange multipliers be applied to problems with more than one constraint?
3. How would you approach this problem if the constraint was not a circle, but an ellipse or another shape?
4. What are the geometric interpretations of the Lagrange multiplier method in optimization problems?
5. Can you use Lagrange multipliers to find the extrema of a function on an open domain?

Tip: Always verify that the solution satisfies the constraint equation when using Lagrange multipliers!