Find the maxima, minima, and saddle points of the function f(x,y) = 2xy − x2 − 2y2 + 3x + 4.

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

The partial derivative with respect to $y$ is:

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

1. $2y - 2x + 3 = 0$
2. $2x - 4y = 0$

From equation (2), we can express $x$ in terms of $y$:
$2x = 4y$
$x = 2y$

Substitute this into equation (1):
$2y - 2(2y) + 3 = 0$
$2y - 4y + 3 = 0$
$-2y + 3 = 0$
$-2y = -3$
$y = \frac{3}{2}$

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

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

The final answer is $\boxed{\text{Local maximum at } \left(3, \frac{3}{2}\right) \text{ with value } \frac{17}{2}}$.