Find the maxima, minima, and saddle points of the function f(x,y) = x3 + 3xy + y3.

The given function is $f(x,y) = x^3 + 3xy + y^3$.
The partial derivative with respect to $x$ is:

$$f_x = \frac{\partial}{\partial x}(x^3 + 3xy + y^3) = 3x^2 + 3y$$

The partial derivative with respect to $y$ is:

$$f_y = \frac{\partial}{\partial y}(x^3 + 3xy + y^3) = 3x + 3y^2$$

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

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

Substitute $y = -x^2$ from equation (1) into equation (2):

$$x + (-x^2)^2 = 0$$ $$x + x^4 = 0$$ $$x(1 + x^3) = 0$$

This equation yields two possible values for $x$:

• $x = 0$
• $1 + x^3 = 0 \implies x^3 = -1 \implies x = -1$

Now, find the corresponding $y$ values using $y = -x^2$:

• If $x = 0$, then $y = -(0)^2 = 0$. Critical point: $(0,0)$.
• If $x = -1$, then $y = -(-1)^2 = -1$. Critical point: $(-1,-1)$.

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

$$f_{xx} = \frac{\partial}{\partial x}(3x^2 + 3y) = 6x$$ $$f_{yy} = \frac{\partial}{\partial y}(3x + 3y^2) = 6y$$ $$f_{xy} = \frac{\partial}{\partial y}(3x^2 + 3y) = 3$$

• For the critical point $(0,0)$:

  $$D(0,0) = 36(0)(0) - 9 = -9$$

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

• For the critical point $(-1,-1)$:

  $$D(-1,-1) = 36(-1)(-1) - 9 = 36 - 9 = 27$$

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

  $$f_{xx}(-1,-1) = 6(-1) = -6$$

  Since $f_{xx}(-1,-1) < 0$, the point $(-1,-1)$ is a local maximum. The function value at this point is:

  $$f(-1,-1) = (-1)^3 + 3(-1)(-1) + (-1)^3 = -1 + 3 - 1 = 1$$

The final answer is:

• Saddle point at $(0,0)$
• Local maximum at $(-1,-1)$ with value 1