given f(x,y) = -2x^2 + 2x - y^2 +4, use the method of Lagrange to find the max and min of f(x,y) subject to the constraint x^2 + y^2 = 4

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 \( (-2, 0) \).

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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.

Given f(x,y) = -2x^2 + 2x - y^2 +4, use the method of Lagrange to find the max and min of f(x,y) subject to the constraint x^2 + y^2 = 4.

Solve the problem of finding the maximum and minimum values of the function f(x, y) = -2x^2 + 2x - y^2 + 4 using the method of Lagrange multipliers. This solution includes step-by-step calculations with the constraint x^2 + y^2 = 4 and explores key concepts in optimization and multivariable calculus.

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