find maximum and minimum value of x^2-2xy+2y^2-2x+2y+1

To find the maximum and minimum values of the function \( f(x, y) = x^2 - 2xy + 2y^2 - 2x + 2y + 1 \), we need to follow these steps:

### Step 1: Find the Partial Derivatives

First, find the first-order partial derivatives of \( f(x, y) \) with respect to \( x \) and \( y \).

\[
f_x = \frac{\partial f}{\partial x} = 2x - 2y - 2
\]
\[
f_y = \frac{\partial f}{\partial y} = -2x + 4y + 2
\]

### Step 2: Set the Partial Derivatives to Zero

To find the critical points, set the partial derivatives equal to zero:

\[
2x - 2y - 2 = 0
\]
\[
-2x + 4y + 2 = 0
\]

#### Solving the first equation:

\[
2x - 2y = 2 \implies x - y = 1 \quad \text{(Equation 1)}
\]

#### Solving the second equation:

\[
-2x + 4y = -2 \implies -x + 2y = -1 \quad \text{(Equation 2)}
\]

Now solve these two equations simultaneously.

From Equation 1, \( x = y + 1 \).

Substitute \( x = y + 1 \) into Equation 2:

\[
-(y + 1) + 2y = -1
\]
\[
-y - 1 + 2y = -1 \implies y - 1 = -1 \implies y = 0
\]

Now, substitute \( y = 0 \) into Equation 1:

\[
x = 0 + 1 = 1
\]

So, the critical point is \( (x, y) = (1, 0) \).

### Step 3: Determine the Nature of the Critical Point

To determine whether the critical point is a maximum, minimum, or saddle point, we examine the second-order partial derivatives:

\[
f_{xx} = \frac{\partial^2 f}{\partial x^2} = 2
\]
\[
f_{yy} = \frac{\partial^2 f}{\partial y^2} = 4
\]
\[
f_{xy} = \frac{\partial^2 f}{\partial x \partial y} = -2
\]

Now compute the Hessian determinant \( D \):

\[
D = f_{xx}f_{yy} - (f_{xy})^2 = (2)(4) - (-2)^2 = 8 - 4 = 4
\]

Since \( D > 0 \) and \( f_{xx} > 0 \), the function has a **local minimum** at the critical point \( (1, 0) \).

### Step 4: Calculate the Function Value at the Critical Point

Finally, substitute \( x = 1 \) and \( y = 0 \) into the original function:

\[
f(1, 0) = (1)^2 - 2(1)(0) + 2(0)^2 - 2(1) + 2(0) + 1 = 1 - 2 + 1 = 0
\]

So, the minimum value of the function is **0** at \( (x, y) = (1, 0) \).

### Step 5: Consider Behavior at Infinity for Maximum Value

Since the function is a quadratic function in both \( x \) and \( y \), and the leading terms are positive, the function goes to infinity as \( x \) or \( y \) tends to infinity. Hence, the **maximum value** is **infinity**.

### Summary
- **Minimum value:** 0 at \( (x, y) = (1, 0) \)
- **Maximum value:** \( \infty \)

### Do you want details or have any questions?
Here are five relative questions:
1. How do you determine if a critical point is a maximum, minimum, or saddle point?
2. What happens if \( f(x, y) \) had negative leading coefficients?
3. Can you solve for the critical points without partial derivatives?
4. How does the Hessian matrix help in analyzing critical points?
5. What is the significance of the discriminant in quadratic functions?

**Tip:** When analyzing functions of two variables, always check the nature of the critical points using the Hessian determinant to ensure correct identification of maxima, minima, or saddle points.

Discover how to find the maximum and minimum values of the function f(x, y) = x^2 - 2xy + 2y^2 - 2x + 2y + 1 with detailed steps and solutions. Learn about partial derivatives, critical points, and the Hessian determinant. Suitable for advanced undergraduate mathematics.

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