To find the local maxima, local minima, and saddle points of the function f(x, y) = 2x^2 - 4xy + 3y^2 - 8x + 3y + 5, we can calculate the critical points and use the second derivative test.
First, let's find the partial derivatives:
∂f/∂x = 4x - 4y - 8
∂f/∂y = -4x + 6y + 3
To find the critical points, we set both partial derivatives equal to zero:
4x - 4y - 8 = 0
-4x + 6y + 3 = 0
Solving these equations simultaneously, we get:
x = 1
y = -1
Now, let's calculate the second partial derivatives:
∂^2f/∂x^2 = 4
∂^2f/∂y^2 = 6
∂^2f/∂x∂y = -4
Using these second partial derivatives, we can calculate the discriminant to determine the nature of the critical point:
D = (∂^2f/∂x^2) * (∂^2f/∂y^2) - (∂^2f/∂x∂y)^2
Plugging in the values, we have:
D = (4)(6) - (-4)^2
D = 24 - 16
D = 8
Since D is positive and (∂^2f/∂x^2) is positive, the critical point (1, -1) corresponds to a local minimum.
Therefore, the correct choices are:
OA. A local minimum occurs at (1, -1). The local minimum value is unknown without further calculation.
OB. There are no local maxima.
OA. A saddle point occurs at (1, -1).