To find the local maximum, local minimum, and saddle points of the function $f(x, y) = x^2 + y^4 + 2xy$, we can follow these steps:

Step 1: Find the Partial Derivatives

Let's find the partial derivatives $f_x$ and $f_y$ with respect to $x$ and $y$:

1. Partial Derivative with respect to $x$: $f_x = \frac{\partial}{\partial x}(x^2 + y^4 + 2xy) = 2x + 2y$

2. Partial Derivative with respect to $y$: $f_y = \frac{\partial}{\partial y}(x^2 + y^4 + 2xy) = 4y^3 + 2x$

Step 2: Solve the System of Equations

To find the critical points, set $f_x = 0$ and $f_y = 0$:

1. From $f_x = 0$: $2x + 2y = 0 \Rightarrow x = -y$

2. Substitute $x = -y$ into $f_y = 0$: $4y^3 + 2(-y) = 0$ Simplify the equation: $4y^3 - 2y = 0 \Rightarrow 2y(2y^2 - 1) = 0$

   Therefore, $y = 0$ or $y = \pm \frac{1}{\sqrt{2}}$.

For $y = 0$:

$x = -y = 0$ So, one critical point is $(0, 0)$.

For $y = \pm \frac{1}{\sqrt{2}}$:

$x = -\left(\pm \frac{1}{\sqrt{2}}\right) = \mp \frac{1}{\sqrt{2}}$ This gives two more critical points: $\left( \frac{-1}{\sqrt{2}}, \frac{1}{\sqrt{2}} \right)$ and $\left( \frac{1}{\sqrt{2}}, \frac{-1}{\sqrt{2}} \right)$.

So, the critical points are: $(0, 0), \quad \left( \frac{-1}{\sqrt{2}}, \frac{1}{\sqrt{2}} \right), \quad \left( \frac{1}{\sqrt{2}}, \frac{-1}{\sqrt{2}} \right)$

Step 3: Apply the Second Derivative Test

To classify each critical point, we need the second partial derivatives: $f_{xx} = 2, \quad f_{yy} = 12y^2, \quad f_{xy} = 2$

The Hessian determinant $D = f_{xx}f_{yy} - (f_{xy})^2$:

1. At $(0, 0)$: $f_{xx} = 2, \quad f_{yy} = 0, \quad f_{xy} = 2$ $D = (2)(0) - (2)^2 = -4$ Since $D < 0$, $(0, 0)$ is a saddle point.

2. At $\left( \frac{-1}{\sqrt{2}}, \frac{1}{\sqrt{2}} \right)$: $f_{xx} = 2, \quad f_{yy} = 6, \quad f_{xy} = 2$ $D = (2)(6) - (2)^2 = 8 > 0, \quad f_{xx} > 0$ Since $D > 0$ and $f_{xx} > 0$, this point is a local minimum.

3. At $\left( \frac{1}{\sqrt{2}}, \frac{-1}{\sqrt{2}} \right)$: $f_{xx} = 2, \quad f_{yy} = 6, \quad f_{xy} = 2$ $D = (2)(6) - (2)^2 = 8 > 0, \quad f_{xx} > 0$ This point is also a local minimum.

Summary of Results

• Local Minima: $\left( \frac{-1}{\sqrt{2}}, \frac{1}{\sqrt{2}} \right), \left( \frac{1}{\sqrt{2}}, \frac{-1}{\sqrt{2}} \right)$
• Saddle Point: $(0, 0)$
• Local Maximum: None

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Follow-Up Questions

1. How do we use the Hessian determinant to classify critical points?
2. Can this function have a global maximum or minimum?
3. How would we find critical points if there were more variables?
4. What happens if the Hessian determinant equals zero?
5. Are there cases where a point can be both a saddle point and a local extremum?

Tip: The nature of the critical point depends on the signs of the second derivative test, so always check $D$ and $f_{xx}$ carefully!