Final answer:
The first function has a line of critical points with no clear classifications. The second function's critical point at (0, 0) is determined to be a saddle point after evaluating the determinant of the Hessian matrix.
Step-by-step explanation:
For the first function F(x, y) = 3x²y + y² - y² + 2, we notice it simplifies to F(x, y) = 3x²y + 2 because the y² terms cancel each other out. We find the critical points by taking the partial derivatives and setting them to zero. Firstly, we find the partial derivative with respect to x: F_x = 6xy. Setting F_x = 0, we get y = 0 as one condition for critical points.
Secondly, the partial derivative with respect to y is F_y = 3x². Setting F_y = 0 gives us no solution for x, which means the only critical point is at (x, y) = (any real number, 0). However, this critical point is not isolated but forms a line of critical points, so we cannot classify it using the second derivative test.
For the second function F(x, y) = 4 + x³ - 3xy, the critical points are found in a similar way. We find F_x = 3x² - 3y and F_y = -3x. Setting these to zero, we obtain the system of equations: 3x² - 3y = 0 and -3x = 0, which has a solution at (x, y) = (0, 0).
To classify this critical point, we compute the second partial derivatives F_xx, F_yy, and F_xy, and evaluate the determinant of the Hessian matrix at the critical point. The computations yield a determinant that indicates whether it's a local maximum, minimum, or saddle point. In this case, it turns out to be a saddle point since the determinant is negative.