Question

The minimum value of the function \( f(x, y) = x^2 + xy + y^2 - 3x - 6y + 11 \) is ............

Hint:

For quadratic functions in two variables, use partial derivatives and the determinant \( D = f_{xx}f_{yy} - f_{xy}^2 \) to classify extrema.

Answer 1

Correct Answer: 2

Step 1: Find partial derivatives. 
\[ \frac{\partial f}{\partial x} = 2x + y - 3, \frac{\partial f}{\partial y} = x + 2y - 6. \]

Step 2: Solve for critical point. 
\[ 2x + y - 3 = 0 \text{and} x + 2y - 6 = 0. \] From first, \( y = 3 - 2x. \) Substitute into second: \[ x + 2(3 - 2x) - 6 = 0 \Rightarrow -3x = 0 \Rightarrow x = 0, y = 3. \]

Step 3: Second derivative test. 
\[ f_{xx} = 2, \; f_{yy} = 2, \; f_{xy} = 1. \] \[ D = f_{xx}f_{yy} - f_{xy}^2 = 4 - 1 = 3 > 0, \, f_{xx} > 0. \] Hence, it is a minimum.

Step 4: Minimum value. 
\[ f(0,3) = 0 + 0 + 9 - 0 - 18 + 11 = 2. \] Correction (re-evaluate using substitution): \[ f(0,3) = (0)^2 + (0)(3) + (3)^2 - 3(0) - 6(3) + 11 = 9 - 18 + 11 = 2. \]

Final Answer: \[ \boxed{2} \]