f(x,y,z) = xyz, -1<=x,y,z<= 1

find extreme points

To find the extreme points of the function \( f(x, y, z) = xyz \) within the region \(-1 \leq x, y, z \leq 1\), we need to evaluate \( f(x, y, z) \) at both the critical points and the boundary of the region.

### Step 1: Critical Points

The function \( f(x, y, z) = xyz \) is a simple product of variables, and its partial derivatives are given by:

- \( f_x = \frac{\partial}{\partial x}(xyz) = yz \)
- \( f_y = \frac{\partial}{\partial y}(xyz) = xz \)
- \( f_z = \frac{\partial}{\partial z}(xyz) = xy \)

For critical points, we set the partial derivatives equal to zero:

\[
yz = 0, \quad xz = 0, \quad xy = 0
\]

The solutions to these equations are:

1. \( y = 0 \) or \( z = 0 \)
2. \( x = 0 \) or \( z = 0 \)
3. \( x = 0 \) or \( y = 0 \)

The possible combinations of \((x, y, z)\) that satisfy these equations are:

- \( (x, y, z) = (0, y, z) \)
- \( (x, y, z) = (x, 0, z) \)
- \( (x, y, z) = (x, y, 0) \)

These points lie on the planes \( x = 0\), \( y = 0\), or \( z = 0 \), which means the critical points are at the origin and the intersections of the planes, specifically: \( (0, 0, z)\), \( (0, y, 0)\), and \( (x, 0, 0)\).

### Step 2: Boundary Points

Since the region is bounded by \(-1 \leq x, y, z \leq 1\), we need to consider the function at the corners of the cube, i.e., where \( x, y, z \) take on the extreme values \(-1\) or \(1\).

We evaluate the function \( f(x, y, z) = xyz \) at these points:

1. \( f(1, 1, 1) = 1 \cdot 1 \cdot 1 = 1 \)
2. \( f(-1, 1, 1) = -1 \cdot 1 \cdot 1 = -1 \)
3. \( f(1, -1, 1) = 1 \cdot -1 \cdot 1 = -1 \)
4. \( f(1, 1, -1) = 1 \cdot 1 \cdot -1 = -1 \)
5. \( f(-1, -1, 1) = -1 \cdot -1 \cdot 1 = 1 \)
6. \( f(-1, 1, -1) = -1 \cdot 1 \cdot -1 = 1 \)
7. \( f(1, -1, -1) = 1 \cdot -1 \cdot -1 = 1 \)
8. \( f(-1, -1, -1) = -1 \cdot -1 \cdot -1 = -1 \)

### Conclusion

The extreme values of the function \( f(x, y, z) = xyz \) within the region \(-1 \leq x, y, z \leq 1\) are:

- Maximum value: \( 1 \) at points \((1, 1, 1)\), \((-1, -1, 1)\), \((-1, 1, -1)\), and \((1, -1, -1)\).
- Minimum value: \( -1 \) at points \((-1, 1, 1)\), \((1, -1, 1)\), \((1, 1, -1)\), and \((-1, -1, -1)\).

Would you like more details or have any questions about this?

Here are some related questions to consider:

1. How do you determine critical points for multivariable functions?
2. What is the significance of boundary points in optimization problems?
3. Can you explain why critical points are found where partial derivatives are zero?
4. What techniques are used to evaluate functions over bounded regions?
5. How does the geometry of a region affect the search for extreme values?
6. Are there any methods to visualize functions in three dimensions?
7. What role does symmetry play in simplifying optimization problems?
8. How would the problem change if the constraints on \(x\), \(y\), and \(z\) were different?

**Tip:** Always check both critical points and boundary points when finding extreme values over a bounded region.

Learn how to find extreme points of the function f(x, y, z) = xyz within the bounded region -1 <= x, y, z <= 1. This problem involves multivariable calculus concepts, including partial derivatives and optimization within a bounded region.

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