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 first-order partial derivatives of the function.
2. Set the first-order partial derivatives to zero to find the critical points.
3. Determine the second-order partial derivatives.
4. Use the second derivative test to classify each critical point as a local maximum, local minimum, or saddle point.

Step 1: First-order Partial Derivatives

The first-order partial derivatives of $f(x, y) = x^4 + y^4 - 4xy$ are:

$f_x(x, y) = \frac{\partial f}{\partial x} = 4x^3 - 4y$

$f_y(x, y) = \frac{\partial f}{\partial y} = 4y^3 - 4x$

Step 2: Set the First-order Partial Derivatives to Zero

To find the critical points, we solve the equations:

$f_x(x, y) = 0 \implies 4x^3 - 4y = 0$

$f_y(x, y) = 0 \implies 4y^3 - 4x = 0$

Simplifying these equations gives:

$x^3 = y$

$y^3 = x$

Solving the Equations

Substitute $y = x^3$ into $y^3 = x$:

$(x^3)^3 = x \implies x^9 = x$

$x(x^8 - 1) = 0$

So, $x = 0$ or $x^8 = 1$.

• If $x = 0$, then $y = 0^3 = 0$. This gives us the critical point $(0, 0)$.

• If $x^8 = 1$, then $x = \pm 1$.

For $x = 1$, $y = 1^3 = 1$, giving the critical point $(1, 1)$.

For $x = -1$, $y = (-1)^3 = -1$, giving the critical point $(-1, -1)$.

Thus, the critical points are: $(0, 0)$, $(1, 1)$, and $(-1, -1)$.

Step 3: Second-order Partial Derivatives

Compute the second-order partial derivatives:

$f_{xx} = \frac{\partial^2 f}{\partial x^2} = 12x^2$

$f_{yy} = \frac{\partial^2 f}{\partial y^2} = 12y^2$

$f_{xy} = \frac{\partial^2 f}{\partial x \partial y} = -4$

Step 4: Second Derivative Test

For a function $f(x, y)$, the second derivative test involves the determinant of the Hessian matrix $H$:

$D(x, y) = f_{xx}(x, y) \cdot f_{yy}(x, y) - (f_{xy}(x, y))^2$

1. At $(0, 0)$:

$f_{xx}(0, 0) = 0, \quad f_{yy}(0, 0) = 0, \quad f_{xy}(0, 0) = -4$

$D(0, 0) = (0)(0) - (-4)^2 = -16$

Since $D(0, 0) < 0$, $(0, 0)$ is a saddle point.

2. At $(1, 1)$:

$f_{xx}(1, 1) = 12 \cdot 1^2 = 12, \quad f_{yy}(1, 1) = 12 \cdot 1^2 = 12, \quad f_{xy}(1, 1) = -4$

$D(1, 1) = (12)(12) - (-4)^2 = 144 - 16 = 128$

Since $D(1, 1) > 0$ and $f_{xx}(1, 1) > 0$, $(1, 1)$ is a local minimum.

3. At $(-1, -1)$:

$f_{xx}(-1, -1) = 12 \cdot (-1)^2 = 12, \quad f_{yy}(-1, -1) = 12 \cdot (-1)^2 = 12, \quad f_{xy}(-1, -1) = -4$

$D(-1, -1) = (12)(12) - (-4)^2 = 144 - 16 = 128$

Since $D(-1, -1) > 0$ and $f_{xx}(-1, -1) > 0$, $(-1, -1)$ is a local minimum.

Conclusion

• Saddle Point: $(0, 0)$
• Local Minima: $(1, 1)$ and $(-1, -1)$
• No Local Maximum is present.

Would you like more details or have any questions?

Related Questions

1. How can we visually identify saddle points on a graph of a function?
2. What is the significance of the Hessian matrix in determining the nature of critical points?
3. Are there functions with both local maxima and minima but no saddle points?
4. How does changing the coefficients in a multivariable function affect its critical points?
5. What are some real-life applications of finding saddle points in multivariable functions?

Tip

When solving for critical points, ensure to verify each solution by substituting back into all original equations to avoid missing any solutions or introducing extraneous ones.