Find the maxima, minima, and saddle points of the function f(x,y) = x2 + xy + y2 + 3x - 3y + 4.

The given function is $f(x,y) = x^2 + xy + y^2 + 3x - 3y + 4$.
The partial derivative with respect to $x$ is:

$$f_x = \frac{\partial}{\partial x}(x^2 + xy + y^2 + 3x - 3y + 4) = 2x + y + 3$$

The partial derivative with respect to $y$ is:

$$f_y = \frac{\partial}{\partial y}(x^2 + xy + y^2 + 3x - 3y + 4) = x + 2y - 3$$

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

1. $2x + y + 3 = 0$
2. $x + 2y - 3 = 0$

From equation (1), we can express $y$ in terms of $x$:
$y = -2x - 3$

Substitute this into equation (2):
$x + 2(-2x - 3) - 3 = 0$
$x - 4x - 6 - 3 = 0$
$-3x - 9 = 0$
$-3x = 9$
$x = -3$

Now substitute $x = -3$ back into the expression for $y$:
$y = -2(-3) - 3 = 6 - 3 = 3$
The only critical point is $(-3,3)$.

$$f_{xx} = \frac{\partial}{\partial x}(2x + y + 3) = 2$$ $$f_{yy} = \frac{\partial}{\partial y}(x + 2y - 3) = 2$$ $$f_{xy} = \frac{\partial}{\partial y}(2x + y + 3) = 1$$

• At $(-3,3)$: $$D(-3,3) = 3$$ Since $D(-3,3) > 0$, we check $f_{xx}(-3,3)$: $$f_{xx}(-3,3) = 2$$ Since $f_{xx}(-3,3) > 0$, the point $(-3,3)$ is a local minimum. The value of the function at this point is: $$f(-3,3) = (-3)^2 + (-3)(3) + (3)^2 + 3(-3) - 3(3) + 4$$ $$f(-3,3) = 9 - 9 + 9 - 9 - 9 + 4$$ $$f(-3,3) = -5$$

The final answer is $\text{Local minimum at } (-3,3) \text{ with value } -5$.