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.