
Find the local maxima, local minima, and saddle points (if any) of the function 
f(x,y) = x^4 + y^4 - 4xy

To find the local maxima, local minima, and saddle points of the function \( f(x, y) = x^4 + y^4 - 4xy \), we will follow these steps:

1. **Find the partial derivatives** \( f_x \) and \( f_y \).
2. **Find the critical points** by setting the partial derivatives equal to zero.
3. **Use the second derivative test** to classify the critical points.

### Step 1: Compute the Partial Derivatives
Let's start by finding the first-order partial derivatives of the function with respect to \( x \) and \( y \):

\[
f_x(x, y) = \frac{\partial}{\partial x}(x^4 + y^4 - 4xy) = 4x^3 - 4y
\]
\[
f_y(x, y) = \frac{\partial}{\partial y}(x^4 + y^4 - 4xy) = 4y^3 - 4x
\]

### Step 2: Find the Critical Points
To find the critical points, set the partial derivatives equal to zero:

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

We can simplify these equations as follows:

\[
x^3 = y \quad \text{(from equation 1)}
\]
\[
y^3 = x \quad \text{(from equation 2)}
\]

Now, substitute \( y = x^3 \) into \( y^3 = x \):

\[
(x^3)^3 = x \implies x^9 = x
\]

This gives us:

\[
x(x^8 - 1) = 0
\]

So, \( x = 0 \) or \( x^8 = 1 \).

For \( x^8 = 1 \), \( x = 1 \) or \( x = -1 \).

Thus, the possible values of \( x \) are \( x = 0 \), \( x = 1 \), or \( x = -1 \). 

### Corresponding \( y \) Values:
- If \( x = 0 \), then \( y = 0^3 = 0 \).
- If \( x = 1 \), then \( y = 1^3 = 1 \).
- If \( x = -1 \), then \( y = (-1)^3 = -1 \).

The critical points are: \( (0, 0) \), \( (1, 1) \), and \( (-1, -1) \).

### Step 3: Use the Second Derivative Test
We now use the second derivative test to classify these points. First, calculate the second-order partial derivatives:

\[
f_{xx} = \frac{\partial}{\partial x}(4x^3 - 4y) = 12x^2
\]
\[
f_{yy} = \frac{\partial}{\partial y}(4y^3 - 4x) = 12y^2
\]
\[
f_{xy} = \frac{\partial}{\partial y}(4x^3 - 4y) = -4
\]

The **Hessian determinant** \( D(x, y) \) is given by:

\[
D(x, y) = f_{xx} f_{yy} - (f_{xy})^2 = (12x^2)(12y^2) - (-4)^2 = 144x^2 y^2 - 16
\]

#### Classifying the Critical Points:

1. **At \( (0, 0) \)**:
   \[
   D(0, 0) = 144(0)(0) - 16 = -16 \quad (\text{Negative, so } (0,0) \text{ is a saddle point})
   \]

2. **At \( (1, 1) \)**:
   \[
   D(1, 1) = 144(1)(1) - 16 = 128 \quad (\text{Positive, so we check } f_{xx} = 12 > 0, \text{ so } (1,1) \text{ is a local minimum})
   \]

3. **At \( (-1, -1) \)**:
   \[
   D(-1, -1) = 144(1)(1) - 16 = 128 \quad (\text{Positive, so we check } f_{xx} = 12 > 0, \text{ so } (-1,-1) \text{ is a local minimum})
   \]

### Conclusion:

- **Local minima** at \( (1, 1) \) and \( (-1, -1) \).
- **Saddle point** at \( (0, 0) \).
- **No local maxima**.

Would you like more details or have any questions?

Here are some related questions to expand on this:
1. How does the second derivative test work for classifying critical points?
2. What is the significance of the Hessian matrix in multivariable calculus?
3. How do saddle points differ from local minima and maxima?
4. Can there be cases where the second derivative test is inconclusive?
5. How can this method be extended to functions of more than two variables?

**Tip:** When the Hessian determinant is negative, it indicates that the critical point is a saddle point.

Learn how to find local maxima, minima, and saddle points of the function f(x, y) = x^4 + y^4 - 4xy using multivariable calculus techniques. Detailed step-by-step solution with the second derivative test. Suitable for advanced college-level students and enthusiasts of multivariable calculus.

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