91Ó°ÊÓ

We need to find the first partial derivatives of \(g(x, y)\), so we will differentiate it with respect to x and y: \(g_x(x,y) = \frac{\partial g}{\partial x} = -2x - 2y + 1\) \(g_y(x,y) = \frac{\partial g}{\partial y} = 2y - 2x - 4\)

In order to locate the critical points, we will set the first partial derivatives equal to zero: \(-2x - 2y + 1 = 0\) (1) \(2y - 2x - 4 = 0\) (2) Now we will solve the system of equations. We can add equation (1) and equation (2) to eliminate x: \((-2x - 2y + 1) + (2y - 2x - 4) = 0 \Rightarrow -4x -3 = 0\) By solving for x, we get: \(x = \frac{3}{4}\). Substituting this value back into equation (1), we get: \(-2\cdot\frac{3}{4} - 2y + 1 = 0\) Solving for y, we obtain: \(y = \frac{1}{4}\). Therefore, the critical point is at \((\frac{3}{4}, \frac{1}{4})\).

Now, we will find the second partial derivatives of \(g(x, y)\): \(g_{xx}(x, y) = -2\) \(g_{xy}(x, y) = -2\) \(g_{yy}(x, y) = 2\) We will evaluate the second partial derivatives at the critical point \((\frac{3}{4}, \frac{1}{4})\), but notice that they are all constants. Therefore, the values are: \(g_{xx} = -2\) \(g_{xy} = -2\) \(g_{yy} = 2\) Now, let's compute the Hessian determinant: \(D(x, y) = g_{xx}g_{yy} - g_{xy}^2 = (-2)(2) - (-2)^2 = -4 - 4 = -8\) Since \(D(x, y) < 0\), the second partial derivative test tells us that the critical point \((\frac{3}{4}, \frac{1}{4})\) is a saddle point. To summarize, the function \(g(x, y)\) has one critical point at \((\frac{3}{4}, \frac{1}{4})\), and it is classified as a saddle point.