Question

The value of the determinant of the Hessian of \( f(x, y) = x^2 + y^2 + xy - 8x - 7y \) at its stationary point is

• A. \( 2 \)
• B. \( 3 \)
• C. \( 4 \)
• D. \( 5 \)

Hint:

Find the stationary point, then look at the second partial derivatives. For a quadratic function these do not change with position.

Answer 1

The Correct Option is B

Step 1: Find the stationary point.
A stationary point is where both first partial derivatives are zero.
\[ f_x = \frac{\partial f}{\partial x} = 2x + y - 8 \]
\[ f_y = \frac{\partial f}{\partial y} = 2y + x - 7 \]
Set both equal to zero: \( 2x + y - 8 = 0 \) and \( x + 2y - 7 = 0 \).
From the first equation, \( y = 8 - 2x \). Put this into the second equation: \( x + 2(8-2x) - 7 = 0 \), so \( x + 16 - 4x - 7 = 0 \), giving \( -3x + 9 = 0 \), so \( x = 3 \).
Then \( y = 8 - 2(3) = 2 \). The stationary point is \( (3, 2) \).

Step 2: Write the Hessian matrix.
The Hessian collects the second partial derivatives.
\[ f_{xx} = 2, \quad f_{yy} = 2, \quad f_{xy} = f_{yx} = 1 \]
\[ H = \begin{pmatrix} f_{xx} & f_{xy} \\ f_{yx} & f_{yy} \end{pmatrix} = \begin{pmatrix} 2 & 1 \\ 1 & 2 \end{pmatrix} \]
Since \( f \) is a plain quadratic in \( x \) and \( y \), these second derivatives are constants, so the Hessian is the same at every point, not just at \( (3,2) \).

Step 3: Compute the determinant.
\[ \det(H) = f_{xx}f_{yy} - (f_{xy})^2 = (2)(2) - (1)^2 = 4 - 1 = 3 \]
Option (C) 4 is what you get if you multiply the diagonal terms and forget to subtract the square of the off diagonal term. Options (A) 2 and (D) 5 come from similar slips in combining the four entries.

Final Answer:
The determinant of the Hessian at the stationary point is 3. \[ \boxed{3} \]