) Find and classify all the stationary points of f(x, y) = x
2 − 4xy + y
3 + 4y .

To find and classify the stationary points of the function \( f(x, y) = x^2 - 4xy + y^3 + 4y \), we follow these steps:

### 1. Find the First Partial Derivatives
We need to compute the partial derivatives of \( f(x, y) \) with respect to \( x \) and \( y \), and then solve the system of equations obtained by setting these derivatives equal to zero.

#### Partial derivative with respect to \( x \):
\[
f_x = \frac{\partial}{\partial x}(x^2 - 4xy + y^3 + 4y) = 2x - 4y
\]

#### Partial derivative with respect to \( y \):
\[
f_y = \frac{\partial}{\partial y}(x^2 - 4xy + y^3 + 4y) = -4x + 3y^2 + 4
\]

### 2. Solve the System of Equations
The stationary points are the points where both \( f_x = 0 \) and \( f_y = 0 \). This gives us the system of equations:

\[
2x - 4y = 0 \quad \text{(1)}
\]
\[
-4x + 3y^2 + 4 = 0 \quad \text{(2)}
\]

#### Solving equation (1) for \( x \):
\[
x = 2y
\]

#### Substitute \( x = 2y \) into equation (2):
\[
-4(2y) + 3y^2 + 4 = 0
\]
\[
-8y + 3y^2 + 4 = 0
\]

This simplifies to:
\[
3y^2 - 8y + 4 = 0
\]

We can solve this quadratic equation using the quadratic formula:
\[
y = \frac{-(-8) \pm \sqrt{(-8)^2 - 4(3)(4)}}{2(3)}
\]
\[
y = \frac{8 \pm \sqrt{64 - 48}}{6} = \frac{8 \pm \sqrt{16}}{6} = \frac{8 \pm 4}{6}
\]

So we get two solutions for \( y \):
\[
y = \frac{8 + 4}{6} = 2 \quad \text{and} \quad y = \frac{8 - 4}{6} = \frac{4}{6} = \frac{2}{3}
\]

#### Find \( x \) corresponding to these \( y \)-values:
Using \( x = 2y \):

- When \( y = 2 \), \( x = 2(2) = 4 \)
- When \( y = \frac{2}{3} \), \( x = 2\left(\frac{2}{3}\right) = \frac{4}{3} \)

Thus, the stationary points are:
\[
(4, 2) \quad \text{and} \quad \left(\frac{4}{3}, \frac{2}{3}\right)
\]

### 3. Classify the Stationary Points
To classify the stationary points, we use the **second derivative test**. This involves calculating the second partial derivatives and using the **Hessian determinant**.

#### Second partial derivatives:

- \( f_{xx} = \frac{\partial}{\partial x}(2x - 4y) = 2 \)
- \( f_{yy} = \frac{\partial}{\partial y}(-4x + 3y^2 + 4) = 6y \)
- \( f_{xy} = \frac{\partial}{\partial y}(2x - 4y) = -4 \)

The Hessian determinant \( H \) is given by:
\[
H = f_{xx} f_{yy} - (f_{xy})^2
\]
We calculate this for both points.

#### For the point \( (4, 2) \):

- \( f_{xx} = 2 \)
- \( f_{yy} = 6(2) = 12 \)
- \( f_{xy} = -4 \)

Thus, the Hessian is:
\[
H = (2)(12) - (-4)^2 = 24 - 16 = 8
\]

Since \( H > 0 \) and \( f_{xx} > 0 \), the point \( (4, 2) \) is a **local minimum**.

#### For the point \( \left(\frac{4}{3}, \frac{2}{3}\right) \):

- \( f_{xx} = 2 \)
- \( f_{yy} = 6\left(\frac{2}{3}\right) = 4 \)
- \( f_{xy} = -4 \)

Thus, the Hessian is:
\[
H = (2)(4) - (-4)^2 = 8 - 16 = -8
\]

Since \( H < 0 \), the point \( \left(\frac{4}{3}, \frac{2}{3}\right) \) is a **saddle point**.

### 4. Conclusion
The function \( f(x, y) = x^2 - 4xy + y^3 + 4y \) has:

- A **local minimum** at \( (4, 2) \)
- A **saddle point** at \( \left(\frac{4}{3}, \frac{2}{3}\right) \)

Would you like more details on any specific part of the solution?

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Here are 5 related questions:
1. How do you find the global minimum or maximum of a function?
2. What is the significance of the Hessian matrix in classifying stationary points?
3. Can a point be both a local maximum and a saddle point?
4. How does the second derivative test work in higher dimensions?
5. What happens if the Hessian determinant is zero?

**Tip**: Always check the sign of the Hessian determinant and the second derivative to classify stationary points correctly.

Find and classify all the stationary points of f(x, y) = x^2 − 4xy + y^3 + 4y.

Learn how to find and classify the stationary points of the function f(x, y) = x^2 - 4xy + y^3 + 4y using partial derivatives and the second derivative test. This solution involves solving quadratic equations and determining the Hessian determinant for local minimum and saddle points. Suitable for undergraduate students studying multivariable calculus.

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