Find the critical points and classify them for the function f(x,y) = 2x2 + 3xy + 4y2 - 5x + 2y.

The given function is:

$$f(x,y) = 2x^2 + 3xy + 4y^2 - 5x + 2y$$

Differentiate $f(x,y)$ with respect to $x$ (treating $y$ as a constant):

$$f_x = \frac{\partial}{\partial x}(2x^2 + 3xy + 4y^2 - 5x + 2y)$$ $$f_x = 4x + 3y - 5$$

Differentiate $f(x,y)$ with respect to $y$ (treating $x$ as a constant):

$$f_y = \frac{\partial}{\partial y}(2x^2 + 3xy + 4y^2 - 5x + 2y)$$ $$f_y = 3x + 8y + 2$$

Set $f_x = 0$:

$$4x + 3y - 5 = 0 \quad (1)$$

Set $f_y = 0$:

$$3x + 8y + 2 = 0 \quad (2)$$

From equation (1), we have $4x + 3y = 5$.
From equation (2), we have $3x + 8y = -2$.

To solve this system of linear equations, multiply equation (1) by 3 and equation (2) by 4:

$$3(4x + 3y) = 3(5) \implies 12x + 9y = 15 \quad (3)$$ $$4(3x + 8y) = 4(-2) \implies 12x + 32y = -8 \quad (4)$$

Subtract equation (3) from equation (4):

$$(12x + 32y) - (12x + 9y) = -8 - 15$$ $$23y = -23$$ $$y = -1$$

Substitute $y = -1$ into equation (1):

$$4x + 3(-1) - 5 = 0$$ $$4x - 3 - 5 = 0$$ $$4x - 8 = 0$$ $$4x = 8$$ $$x = 2$$

The only critical point is $(2, -1)$.

$$f_{xx} = \frac{\partial}{\partial x}(4x + 3y - 5) = 4$$ $$f_{yy} = \frac{\partial}{\partial y}(3x + 8y + 2) = 8$$ $$f_{xy} = \frac{\partial}{\partial y}(4x + 3y - 5) = 3$$

The discriminant is given by the formula:

$$D(x,y) = f_{xx}f_{yy} - (f_{xy})^2$$

Substitute the partial derivatives:

$$D(x,y) = (4)(8) - (3)^2$$ $$D(x,y) = 32 - 9$$ $$D(x,y) = 23$$

For the critical point $(2, -1)$:

$$D(2, -1) = 23$$

Since $D(2, -1) = 23 > 0$, we need to check $f_{xx}(2, -1)$.

$$f_{xx}(2, -1) = 4$$

Since $f_{xx}(2, -1) = 4 > 0$ and $D(2, -1) > 0$, the point $(2, -1)$ is a local minimum.

Calculate the function value at $(2, -1)$:

$$f(2, -1) = 2(2)^2 + 3(2)(-1) + 4(-1)^2 - 5(2) + 2(-1)$$ $$f(2, -1) = 2(4) - 6 + 4(1) - 10 - 2$$ $$f(2, -1) = 8 - 6 + 4 - 10 - 2$$ $$f(2, -1) = 2 + 4 - 10 - 2$$ $$f(2, -1) = 6 - 10 - 2$$ $$f(2, -1) = -4 - 2$$ $$f(2, -1) = -6$$

Summary of results:

• Local minima: There is a local minimum at $\boxed{(2,-1)}$ with a value of $f(2,-1) = \boxed{-6}$.
• Local maxima: There are no local maxima.
• Saddle points: There are no saddle points.