Find the maxima, minima, and saddle points of the function f(x,y) = x2 + xy + y2 + 3x + 2y + 5.

The given function is $f(x,y) = x^2 + xy + y^2 + 3x + 2y + 5$.
The partial derivative with respect to $x$ is:

The partial derivative with respect to $y$ is:

(Note: The $f_y$ in your work was $x+2$, which is incorrect for the given function. We will use the correct $f_y = x+2y+2$.)

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

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

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

Substitute this into equation (2):
$x + 2(-2x - 3) + 2 = 0$
$x - 4x - 6 + 2 = 0$
$-3x - 4 = 0$
$-3x = 4$
$x = -\frac{4}{3}$

Now substitute $x = -\frac{4}{3}$ back into the expression for $y$:
$y = -2\left(-\frac{4}{3}\right) - 3$
$y = \frac{8}{3} - \frac{9}{3}$
$y = -\frac{1}{3}$
The only critical point is $\left(-\frac{4}{3}, -\frac{1}{3}\right)$.

• At $\left(-\frac{4}{3}, -\frac{1}{3}\right)$: $$D\left(-\frac{4}{3}, -\frac{1}{3}\right) = 3$$ Since $D > 0$, we check $f_{xx}$: $$f_{xx}\left(-\frac{4}{3}, -\frac{1}{3}\right) = 2$$ Since $f_{xx} > 0$, the critical point $\left(-\frac{4}{3}, -\frac{1}{3}\right)$ is a local minimum.

The final answer is $\boxed{\text{Local minimum at } \left(-\frac{4}{3}, -\frac{1}{3}\right) \text{ with value } \frac{8}{3}}$.