Find the local maximum and minimum values and saddle point(s) of the function. You are encouraged to use a calculator or computer to graph the function with a domain and viewpoint that reveals all the important aspects of the function. (Enter your answers as comma-separated lists. If an answer does not exist, enter DNE.)

*f*(*x*, *y*) = *x*2 + *y*4 + 2*xy*

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

Would you like further details or a graph of the function? 

### 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!

Find the local maximum and minimum values and saddle point(s) of the function. You are encouraged to use a calculator or computer to graph the function with a domain and viewpoint that reveals all the important aspects of the function. (Enter your answers as comma-separated lists. If an answer does not exist, enter DNE.) f(x, y) = x^2 + y^4 + 2xy

Learn how to find local maxima, minima, and saddle points for the multivariable function f(x, y) = x^2 + y^4 + 2xy. This solution uses partial derivatives and the Hessian determinant to classify critical points. Suitable for college-level multivariable calculus students.

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