To solve for the extrema of $f(x, y) = 2 + x - x^2 - y^2$ under the constraint $x^2 = y^2 = 1$ using Lagrange multipliers, we can follow these steps:

Step 1: Formulate the Lagrange Multiplier Equations

We are given the objective function $f(x, y) = 2 + x - x^2 - y^2$ and the constraint $g(x, y) = x^2 - 1 = 0$ (since both $x^2 = 1$ and $y^2 = 1$).

The method of Lagrange multipliers involves finding the gradients of the objective function $f$ and the constraint $g$, and setting them proportional. That is, we need to solve:

$\nabla f(x, y) = \lambda \nabla g(x, y)$

Step 2: Compute the Gradients

First, compute the gradient of $f(x, y)$:

$\nabla f(x, y) = \left( \frac{\partial f}{\partial x}, \frac{\partial f}{\partial y} \right) = \left( 1 - 2x, -2y \right)$

Next, compute the gradient of $g(x, y)$:

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

Step 3: Set up the System of Equations

Using the Lagrange multiplier method:

$1 - 2x = \lambda \cdot 2x$ $-2y = \lambda \cdot 2y$ And we also have the constraint $x^2 = 1$.

Step 4: Solve the System of Equations

Equation 1:

$1 - 2x = 2 \lambda x$ Rearrange: $1 = 2x + 2 \lambda x$ $1 = 2x(1 + \lambda)$ Thus: $\lambda = \frac{1}{2x} - 1$

Equation 2:

$-2y = 2 \lambda y$ For $y \neq 0$, divide both sides by $2y$: $-1 = \lambda$ Thus, $\lambda = -1$.

Substitute $\lambda = -1$ into Equation 1:

$1 = 2x(1 - 1)$ This simplifies to: $1 = 0$ which is a contradiction. Therefore, we must have $y = 0$.

Step 5: Evaluate the Constraint with $y = 0$

If $y = 0$, from the constraint $x^2 = 1$, we have two possibilities: $x = 1$ or $x = -1$.

Step 6: Compute the Objective Function at these Points

1. When $x = 1$ and $y = 0$: $f(1, 0) = 2 + 1 - 1^2 - 0^2 = 2 + 1 - 1 = 2$

2. When $x = -1$ and $y = 0$: $f(-1, 0) = 2 + (-1) - (-1)^2 - 0^2 = 2 - 1 - 1 = 0$

Step 7: Conclusion

The extrema of $f(x, y)$ subject to the constraint $x^2 = y^2 = 1$ occur at:

• A maximum at $(x, y) = (1, 0)$ with $f(1, 0) = 2$
• A minimum at $(x, y) = (-1, 0)$ with $f(-1, 0) = 0$

Would you like a more detailed breakdown of the steps, or have any questions?

Here are 5 related questions you can explore next:

1. How do the Lagrange multiplier equations change for other types of constraints (e.g., equality or inequality)?
2. What would the solution look like if the constraint was $x^2 + y^2 = 1$?
3. How can we interpret the Lagrange multiplier $\lambda$ geometrically?
4. What happens if there are more than two variables in the optimization problem?
5. How can we verify the solutions using the second derivative test?

Tip: Lagrange multipliers are a powerful method for constrained optimization, but always be mindful of the constraint conditions—sometimes, they can lead to situations where no valid solutions exist!