91Ó°ÊÓ

First, we need to calculate the partial derivatives of the function f(x, y) with respect to x and y, which are given by: \( \frac{\partial f}{\partial x} = \frac{\partial}{\partial x}(x^2 + y^3 - 3xy) = 2x - 3y \) \( \frac{\partial f}{\partial y} = \frac{\partial}{\partial y}(x^2 + y^3 - 3xy) = 3y^2 - 3x \)

Now, we set these partial derivatives equal to 0 to find the critical points: \(2x - 3y = 0\) \(3y^2 - 3x = 0\)

Solve these equations simultaneously to get the critical points. From the first equation, we get: \(y = \frac{2}{3}x\) Substitute this into the second equation and we get: \(3\left(\frac{2}{3}x\right)^2 - 3x = 4x^2 - 3x = 0\) \(x(4x - 3) = 0\) Hence, x = 0 or x = 3/4 When x = 0, y = 0 and when x = 3/4, y = 1/2 So the critical points are (0, 0) and (3/4, 1/2).

To classify the critical points, we use the second partial derivative test. We calculate the second partial derivatives and the determinant D: \(D(x, y) = \left(\frac{\partial^2 f}{\partial x^2}\frac{\partial^2 f}{\partial y^2} - \left(\frac{\partial^2 f}{\partial x\partial y}\right)^2\right)\) Calculate the second partial derivatives: \(\frac{\partial^2 f}{\partial x^2} = \frac{\partial}{\partial x}(2x - 3y) = 2\) \(\frac{\partial^2 f}{\partial x\partial y} = \frac{\partial}{\partial y}(2x - 3y) = -3\) \(\frac{\partial^2 f}{\partial y^2} = \frac{\partial}{\partial y}(3y^2 - 3x) = 6y\) Calculating the determinant D: \(D(x, y) = (2)(6y) - (-3)^2 = 12y - 9\) Now we evaluate the determinant D at each critical point: 1. At (0, 0): \(D(0, 0) = 12(0) - 9 = -9 < 0\) Since D < 0, (0, 0) is a saddle point. 2. At (3/4, 1/2): \(D\left(\frac{3}{4}, \frac{1}{2}\right) = 12\left(\frac{1}{2}\right) - 9 = 3 > 0\) Since D > 0 and \(\frac{\partial^2 f}{\partial x^2} > 0\), (\(3/4, 1/2\)) is a relative minimum. In conclusion, the critical point (0, 0) is a saddle point, and the critical point \((\frac{3}{4}, \frac{1}{2})\) is a relative minimum of the function \(f(x, y) = x^2 + y^3 - 3xy\).