To find the maximum and minimum of the function $f(x, y) = -2x^2 + 2x - y^2 + 4$ subject to the constraint $g(x, y) = x^2 + y^2 - 4 = 0$ (i.e., the circle of radius 2 centered at the origin), we use the Lagrange multiplier method.

Step 1: Set up the Lagrange multiplier system

The Lagrange multiplier method requires us to solve the system of equations obtained by equating the gradients of the function $f(x, y)$ and the constraint function $g(x, y)$.

We need to solve: $\nabla f(x, y) = \lambda \nabla g(x, y)$ where $\lambda$ is the Lagrange multiplier.

Compute the gradients:

• $\nabla f(x, y) = \left( \frac{\partial f}{\partial x}, \frac{\partial f}{\partial y} \right) = (-4x + 2, -2y)$
• $\nabla g(x, y) = \left( \frac{\partial g}{\partial x}, \frac{\partial g}{\partial y} \right) = (2x, 2y)$

So, we have the system of equations: $(-4x + 2, -2y) = \lambda (2x, 2y)$

Step 2: Set up the system of equations

This gives us the following system:

1. $-4x + 2 = \lambda 2x$
2. $-2y = \lambda 2y$
3. $g(x, y) = x^2 + y^2 = 4$ (the constraint equation)

Step 3: Solve the system

From equation (2):

$-2y = \lambda 2y$ If $y \neq 0$, dividing by $2y$, we get $\lambda = -1$.

If $y = 0$, substitute into the constraint $x^2 + y^2 = 4$, which gives $x^2 = 4$, so $x = \pm 2$.

Case 1: $y \neq 0$ (i.e., $\lambda = -1$)

Substitute $\lambda = -1$ into equation (1): $-4x + 2 = -2x$ Simplifying: $-4x + 2 = -2x \quad \Rightarrow \quad -2x = -2 \quad \Rightarrow \quad x = 1$

Now, substitute $x = 1$ into the constraint $x^2 + y^2 = 4$: $1^2 + y^2 = 4 \quad \Rightarrow \quad y^2 = 3 \quad \Rightarrow \quad y = \pm \sqrt{3}$

Case 2: $y = 0$

As mentioned earlier, if $y = 0$, then $x = \pm 2$.

Step 4: Evaluate the function at the critical points

Now, we evaluate $f(x, y)$ at the critical points:

• $(x, y) = (1, \sqrt{3})$: $f(1, \sqrt{3}) = -2(1)^2 + 2(1) - (\sqrt{3})^2 + 4 = -2 + 2 - 3 + 4 = 1$

• $(x, y) = (1, -\sqrt{3})$: $f(1, -\sqrt{3}) = -2(1)^2 + 2(1) - (-\sqrt{3})^2 + 4 = 1$

• $(x, y) = (2, 0)$: $f(2, 0) = -2(2)^2 + 2(2) - (0)^2 + 4 = -8 + 4 + 4 = 0$

• $(x, y) = (-2, 0)$: $f(-2, 0) = -2(-2)^2 + 2(-2) - (0)^2 + 4 = -8 - 4 + 4 = -8$

Step 5: Conclusion

The maximum value of $f(x, y)$ is $1$ at the points $(1, \sqrt{3})$ and $(1, -\sqrt{3})$.

The minimum value of $f(x, y)$ is $-8$ at the point $(-2, 0)$.

---

Would you like more details or clarifications on any step?

Follow-up Questions:

1. How does the method of Lagrange multipliers handle inequality constraints?
2. Can you explain the significance of the Lagrange multiplier $\lambda$ in optimization problems?
3. What if the constraint were an ellipse instead of a circle—how would that change the solution?
4. How would we apply this method in three-dimensional problems?
5. Could the method of Lagrange be applied if $f(x, y)$ had more complicated terms (e.g., involving trigonometric functions)?

Tip: The Lagrange multiplier method is particularly useful when the constraint is non-linear, as it simplifies finding extrema while keeping the constraint intact.