Find the maxima, minima, and saddle points of the function f(x,y) = 4xy − x4 − y4.

The given function is $f(x,y) = 4xy - x^4 - y^4$.
The partial derivative with respect to $x$ is:

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

The partial derivative with respect to $y$ is:

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

Set $f_x = 0$ and $f_y = 0$:

1. $4y - 4x^3 = 0 \implies y = x^3$
2. $4x - 4y^3 = 0 \implies x = y^3$

Substitute equation (1) into equation (2):

$$x = (x^3)^3$$ $$x = x^9$$ $$x^9 - x = 0$$ $$x(x^8 - 1) = 0$$

This gives three possible values for $x$:

• $x = 0$
• $x^8 - 1 = 0 \implies x^8 = 1 \implies x = \pm 1$

Now, find the corresponding $y$ values using $y = x^3$:

• If $x = 0$, then $y = 0^3 = 0$. Critical point: $(0,0)$.
• If $x = 1$, then $y = 1^3 = 1$. Critical point: $(1,1)$.
• If $x = -1$, then $y = (-1)^3 = -1$. Critical point: $(-1,-1)$. The critical points are $(0,0)$, $(1,1)$, and $(-1,-1)$.

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

$$D(x,y) = (-12x^2)(-12y^2) - (4)^2$$ $$D(x,y) = 144x^2y^2 - 16$$

• At $(0,0)$:

  $$D(0,0) = 144(0)^2(0)^2 - 16 = -16$$

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

• At $(1,1)$:

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

  Since $D(1,1) > 0$, we check $f_{xx}(1,1)$:

  $$f_{xx}(1,1) = -12(1)^2 = -12$$

  Since $f_{xx}(1,1) < 0$, the point $(1,1)$ is a local maximum. The value of the function at this point is $f(1,1) = 4(1)(1) - (1)^4 - (1)^4 = 4 - 1 - 1 = 2$.

• At $(-1,-1)$:

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

  Since $D(-1,-1) > 0$, we check $f_{xx}(-1,-1)$:

  $$f_{xx}(-1,-1) = -12(-1)^2 = -12$$

  Since $f_{xx}(-1,-1) < 0$, the point $(-1,-1)$ is a local maximum. The value of the function at this point is $f(-1,-1) = 4(-1)(-1) - (-1)^4 - (-1)^4 = 4 - 1 - 1 = 2$.

Summary of results:

• The point $(0,0)$ is a saddle point.
• The point $(1,1)$ is a local maximum with value $f(1,1) = 2$.
• The point $(-1,-1)$ is a local maximum with value $f(-1,-1) = 2$.

The final answer is $\boxed{\text{Saddle point at } (0,0), \text{ Local maxima at } (1,1) \text{ and } (-1,-1) \text{ with value } 2}$.