The function
f(x, y) = x
3 + y
3 + x
2
y + xy2 − 8x − 8y + 5
has stationary points at some of the following points (x, y) given below. In each
case identify whether the point is stationary, and if so find out if it is a maximum,
minimum or saddle point.
• (2, −2) is:
• (−2, 2) is:
• ( √
2
3
, − √
2
3
) is:• (− √
2
3
, √
2
3
) is:
• ( √
2
3
, √
2
3
) is:
• (− √
2
3
, − √
2
3
) is:
• (2, 2) is:


To determine whether each of the given points is a stationary point of the function \( f(x, y) = x^3 + y^3 + x^2y + xy^2 - 8x - 8y + 5 \), and if so, to classify the nature of the stationary point (i.e., maximum, minimum, or saddle point), we will follow these steps:

### Steps:

1. **Find the partial derivatives** of \( f(x, y) \) with respect to \( x \) and \( y \).
2. **Set the partial derivatives equal to zero** to find the critical points.
3. **Evaluate the second partial derivatives** to use the **Hessian matrix** to classify the stationary points.

### 1. Finding the Partial Derivatives
Compute the first partial derivatives of \( f(x, y) \) with respect to \( x \) and \( y \).

\[
f_x(x, y) = \frac{\partial f}{\partial x} = 3x^2 + 2xy + y^2 - 8
\]

\[
f_y(x, y) = \frac{\partial f}{\partial y} = 3y^2 + 2xy + x^2 - 8
\]

### 2. Stationary Points
A stationary point occurs where both \( f_x(x, y) = 0 \) and \( f_y(x, y) = 0 \).

Let's evaluate each given point to check whether it is a stationary point.

1. **(2, -2):**
   - \( f_x(2, -2) = 3(2)^2 + 2(2)(-2) + (-2)^2 - 8 = 12 - 8 + 4 - 8 = 0 \)
   - \( f_y(2, -2) = 3(-2)^2 + 2(2)(-2) + (2)^2 - 8 = 12 - 8 + 4 - 8 = 0 \)

   This is a stationary point.

2. **(-2, 2):**
   - \( f_x(-2, 2) = 3(-2)^2 + 2(-2)(2) + (2)^2 - 8 = 12 - 8 + 4 - 8 = 0 \)
   - \( f_y(-2, 2) = 3(2)^2 + 2(-2)(2) + (-2)^2 - 8 = 12 - 8 + 4 - 8 = 0 \)

   This is a stationary point.

3. **\( \left(\frac{\sqrt{2}}{3}, -\frac{\sqrt{2}}{3}\right) \):**
   - \( f_x\left(\frac{\sqrt{2}}{3}, -\frac{\sqrt{2}}{3}\right) \) and \( f_y\left(\frac{\sqrt{2}}{3}, -\frac{\sqrt{2}}{3}\right) \) need to be evaluated numerically. After simplifications, we find these are not stationary points.

4. **\( \left(-\frac{\sqrt{2}}{3}, \frac{\sqrt{2}}{3}\right) \):**
   - Similar to the previous point, evaluating numerically shows these are not stationary points.

5. **\( \left(\frac{\sqrt{2}}{3}, \frac{\sqrt{2}}{3}\right) \):**
   - Similarly, this is not a stationary point.

6. **\( \left(-\frac{\sqrt{2}}{3}, -\frac{\sqrt{2}}{3}\right) \):**
   - Again, this is not a stationary point.

7. **(2, 2):**
   - \( f_x(2, 2) = 3(2)^2 + 2(2)(2) + (2)^2 - 8 = 12 + 8 + 4 - 8 = 16 \) (not zero)
   - \( f_y(2, 2) = 3(2)^2 + 2(2)(2) + (2)^2 - 8 = 12 + 8 + 4 - 8 = 16 \) (not zero)

   This is not a stationary point.

### 3. Classifying Stationary Points (Using Hessian)
To classify the stationary points at \((2, -2)\) and \((-2, 2)\), we calculate the second partial derivatives to form the Hessian matrix.

\[
f_{xx} = \frac{\partial^2 f}{\partial x^2} = 6x + 2y, \quad f_{yy} = \frac{\partial^2 f}{\partial y^2} = 6y + 2x, \quad f_{xy} = \frac{\partial^2 f}{\partial x \partial y} = 2x + 2y
\]

For \((2, -2)\):

\[
H(2, -2) = \begin{bmatrix} f_{xx}(2, -2) & f_{xy}(2, -2) \\ f_{xy}(2, -2) & f_{yy}(2, -2) \end{bmatrix} = \begin{bmatrix} 8 & 0 \\ 0 & -8 \end{bmatrix}
\]

The determinant \( \text{det}(H) = 8 \times (-8) - 0^2 = -64 \), indicating a **saddle point**.

For \((-2, 2)\):

\[
H(-2, 2) = \begin{bmatrix} f_{xx}(-2, 2) & f_{xy}(-2, 2) \\ f_{xy}(-2, 2) & f_{yy}(-2, 2) \end{bmatrix} = \begin{bmatrix} -8 & 0 \\ 0 & 8 \end{bmatrix}
\]

The determinant \( \text{det}(H) = -8 \times 8 - 0^2 = -64 \), indicating another **saddle point**.

### Summary of Results:
- (2, -2) is: **Saddle point**
- (-2, 2) is: **Saddle point**
- \( \left(\frac{\sqrt{2}}{3}, -\frac{\sqrt{2}}{3}\right) \) is: **Not a stationary point**
- \( \left(-\frac{\sqrt{2}}{3}, \frac{\sqrt{2}}{3}\right) \) is: **Not a stationary point**
- \( \left(\frac{\sqrt{2}}{3}, \frac{\sqrt{2}}{3}\right) \) is: **Not a stationary point**
- \( \left(-\frac{\sqrt{2}}{3}, -\frac{\sqrt{2}}{3}\right) \) is: **Not a stationary point**
- (2, 2) is: **Not a stationary point**

Would you like further details or have any questions?

---

**Related Questions:**
1. What is a saddle point in the context of multivariable calculus?
2. How is the Hessian matrix used to classify stationary points?
3. What are critical points, and how are they found in multivariable functions?
4. Can a function have more than one type of stationary point?
5. What are the conditions for a point to be a local maximum or minimum?
6. How do we interpret the eigenvalues of the Hessian matrix in classification?
7. What role do second partial derivatives play in optimization problems?
8. How does the behavior of a function around a stationary point determine its nature?

**Tip:** Always check the second partial derivatives when classifying stationary points to ensure accurate identification of maxima, minima, or saddle points.

Learn how to identify stationary points and classify their nature using multivariable calculus techniques. Explore examples involving partial derivatives, Hessian matrices, and the second derivative test. Suitable for advanced college-level students and enthusiasts of calculus.

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