Let $f(x, y)=x^{2}-y^{3}-30 x y$
List the critical points:
The critical point at the origin is a Select an answer $\checkmark$
The critical point with negative $y$ value is a Select an answer $v$
Answer
\(\boxed{\text{The critical points of the function } f(x, y)=x^{2}-y^{3}-30 x y \text{ are at } (-2250, -150) \text{ and } (0, 0). \text{ The critical point at } (-2250, -150) \text{ is a saddle point and the critical point at } (0, 0) \text{ is a local minimum.}}\)
Step 1 ：Define the function \(f(x, y)=x^{2}-y^{3}-30 x y\).
Step 2 ：Find the partial derivatives of the function with respect to \(x\) and \(y\), which are \(f_x = 2x - 30y\) and \(f_y = -30x - 3y^2\) respectively.
Step 3 ：Set these partial derivatives equal to zero to find the critical points. The critical points are \((-2250, -150)\) and \((0, 0)\).
Step 4 ：Calculate the second partial derivatives of the function, which are \(f_{xx} = 2\), \(f_{yy} = -6y\), and \(f_{xy} = -30\).
Step 5 ：Use these second partial derivatives to find the determinant of the Hessian matrix at each critical point. If the determinant is positive and the second partial derivative with respect to \(x\) is positive, the point is a local minimum. If the determinant is positive and the second partial derivative with respect to \(x\) is negative, the point is a local maximum. If the determinant is negative, the point is a saddle point.
Step 6 ：Classify the critical points. The critical point at \((-2250, -150)\) is a saddle point and the critical point at \((0, 0)\) is a local minimum.
Step 7 ：\(\boxed{\text{The critical points of the function } f(x, y)=x^{2}-y^{3}-30 x y \text{ are at } (-2250, -150) \text{ and } (0, 0). \text{ The critical point at } (-2250, -150) \text{ is a saddle point and the critical point at } (0, 0) \text{ is a local minimum.}}\)
