Question

For the function 𝑓∶ℝ×ℝ→ℝ defined by 𝑓(𝑥, 𝑦)=2𝑥²-𝑥𝑦-3𝑦 ²-3𝑥+7𝑦 , the point (1,1) is

• A. a point of local maximum
• B. a point of local minimum
• C. a saddle point
• D. NOT a critical point

Answer 1

The Correct Option is C

To determine the nature of the point \((1, 1)\) for the function \(f(x, y) = 2x^2 - xy - 3y^2 - 3x + 7y\), we will use the second derivative test for functions of two variables.

First, find the first partial derivatives: 

• \(f_x = \frac{\partial f}{\partial x} = 4x - y - 3\).
• \(f_y = \frac{\partial f}{\partial y} = -x - 6y + 7\).

Setting these partial derivatives to zero, we find critical points.

Solving \(4x - y - 3 = 0\) and \(-x - 6y + 7 = 0\):

• From \(4x - y - 3 = 0\), we have \(y = 4x - 3\).
• Substitute \(y = 4x - 3\) into \(-x - 6y + 7 = 0\):
  • \(-x - 6(4x - 3) + 7 = 0\)
  • \(-x - 24x + 18 + 7 = 0\)
  • \(-25x + 25 = 0 \Rightarrow x = 1\)
• Then \(y = 4(1) - 3 = 1\)

So, the point \((1, 1)\) is a critical point.

Next, find the second order partial derivatives:

• \(f_{xx} = \frac{\partial^2 f}{\partial x^2} = 4\).
• \(f_{yy} = \frac{\partial^2 f}{\partial y^2} = -6\).
• \(f_{xy} = f_{yx} = \frac{\partial^2 f}{\partial x \partial y} = -1\).

The second derivative test for functions of two variables involves calculating the determinant of the Hessian matrix at the critical point:

Hessian Matrix, \(H = \begin{bmatrix} f_{xx} & f_{xy} \\ f_{yx} & f_{yy} \end{bmatrix}\)
\(= \begin{bmatrix} 4 & -1 \\ -1 & -6 \end{bmatrix}\)

Calculate the determinant of this matrix:

• \(D = f_{xx}f_{yy} - (f_{xy})^2 = (4)(-6) - (-1)^2 = -24 - 1 = -25\)

Since \(D < 0\), the point \((1, 1)\) is a saddle point.

Thus, the correct answer is: a saddle point.