Question

Let $f : \mathbb{R}^2 \to \mathbb{R}$ be defined by $f(x, y) = x^2(2 - y) - y^3 + 3y^2 + 9y$, where $(x, y) \in \mathbb{R}^2$. Which of the following is/are saddle point(s) of $f$?
 

• A. $(0, -1)$
• B. $(0, 3)$
• C. $(3, 2)$
• D. $(-3, 2)$

Hint:

To identify saddle points, use the second derivative test: if $D < 0$, the point is a saddle.

Answer 1

The Correct Option is C, D

Step 1: Find the critical points.
Compute partial derivatives: \[ f_x = 2x(2 - y), f_y = -x^2 - 3y^2 + 6y + 9. \] Set both to zero: From $f_x = 0 $\Rightarrow$ x = 0$ or $y = 2$. Case 1: $x = 0$ Then $f_y = -3y^2 + 6y + 9 = 0 $\Rightarrow$ y^2 - 2y - 3 = 0 $\Rightarrow$ y = 3, -1.$ Thus, points $(0,3)$ and $(0,-1)$. Case 2: $y = 2$ Then $f_y = -x^2 - 3(2)^2 + 6(2) + 9 = -x^2 + 9 = 0 $\Rightarrow$ x = \pm 3.$ Thus, points $(3,2)$ and $(-3,2)$.

Step 2: Compute second partial derivatives.
\[ f_{xx} = 2(2 - y), f_{yy} = -6y + 6, f_{xy} = -2x. \]

Step 3: Use second derivative test.
Determinant $D = f_{xx} f_{yy} - (f_{xy})^2$. At $(0, -1)$: $f_{xx} = 6, \ f_{yy} = 12, \ f_{xy} = 0 $\Rightarrow$ D = 72 > 0$, and $f_{xx} > 0$ → Local minimum. At $(0, 3)$: $f_{xx} = -2, \ f_{yy} = -12, \ f_{xy} = 0 $\Rightarrow$ D = 24 > 0$, and $f_{xx} < 0$ → Local maximum. At $(3, 2)$ and $(-3, 2)$: $f_{xx} = 0, \ f_{yy} = -6, \ f_{xy} = -6$ → $D = -36 < 0$, indicating saddle points.

Step 4: Conclusion.
\[ \boxed{(0, -1) \text{ and } (0, 3) \text{ are saddle points.}} \]