Finding the Stationary Point

To find the stationary point, we first compute the first partial derivatives of $f(x,y) = x^2 + y^2 + xy - 8x - 7y$ and set them equal to zero.

• Partial derivative with respect to x: $ \frac{\partial f}{\partial x} = \frac{\partial}{\partial x}(x^2 + y^2 + xy - 8x - 7y) = 2x + y - 8 $ Setting this to zero gives the equation: $ 2x + y - 8 = 0 \implies 2x + y = 8 \quad (1) $
• Partial derivative with respect to y: $ \frac{\partial f}{\partial y} = \frac{\partial}{\partial y}(x^2 + y^2 + xy - 8x - 7y) = 2y + x - 7 $ Setting this to zero gives the equation: $ x + 2y - 7 = 0 \implies x + 2y = 7 \quad (2) $

Now, we solve the system of linear equations (1) and (2):

1. Multiply equation (1) by 2: $ 4x + 2y = 16 \quad (3) $
2. Subtract equation (2) from equation (3): $ (4x + 2y) - (x + 2y) = 16 - 7 $ $ 3x = 9 \implies x = 3 $
3. Substitute $x=3$ into equation (1): $ 2(3) + y = 8 \implies 6 + y = 8 \implies y = 2 $

The stationary point is $(3, 2)$.

Calculating the Hessian Matrix

Next, we find the second partial derivatives to form the Hessian matrix.

• $ f_{xx} = \frac{\partial^2 f}{\partial x^2} = \frac{\partial}{\partial x}(2x + y - 8) = 2 $
• $ f_{yy} = \frac{\partial^2 f}{\partial y^2} = \frac{\partial}{\partial y}(x + 2y - 7) = 2 $
• $ f_{xy} = \frac{\partial^2 f}{\partial x \partial y} = \frac{\partial}{\partial x}(x + 2y - 7) = 1 $ (Note: $f_{yx} = \frac{\partial^2 f}{\partial y \partial x} = \frac{\partial}{\partial y}(2x + y - 8) = 1$, so $f_{xy} = f_{yx}$)

The Hessian matrix $H$ is:

Since the second derivatives are constant, the Hessian matrix is the same at all points, including the stationary point $(3, 2)$.

Determinant of the Hessian at the Stationary Point

Finally, we calculate the determinant of the Hessian matrix evaluated at the stationary point $(3, 2)$.

Determinant:

The value of the determinant of the Hessian at the stationary point is 3.