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: Compute first partial derivatives

$$\frac{\partial f}{\partial x} = 2x(2-y)$$

$$\frac{\partial f}{\partial y} = -x^2 - 3y^2 + 6y + 9$$

Step 2: Find critical points

Set $\frac{\partial f}{\partial x} = 0$: $$2x(2-y) = 0$$ $$x = 0 \text{ or } y = 2$$

Set $\frac{\partial f}{\partial y} = 0$: $$-x^2 - 3y^2 + 6y + 9 = 0$$ $$x^2 = -3y^2 + 6y + 9$$

Case 1: If $x = 0$: $$0 = -3y^2 + 6y + 9$$ $$3y^2 - 6y - 9 = 0$$ $$y^2 - 2y - 3 = 0$$ $$(y-3)(y+1) = 0$$ $$y = 3 \text{ or } y = -1$$

Critical points: $(0, 3)$ and $(0, -1)$

Case 2: If $y = 2$: $$x^2 = -3(4) + 6(2) + 9 = -12 + 12 + 9 = 9$$ $$x = \pm 3$$

Critical points: $(3, 2)$ and $(-3, 2)$

Step 3: Apply second derivative test

Compute second partial derivatives: $$f_{xx} = 2(2-y)$$ $$f_{yy} = -6y + 6$$ $$f_{xy} = -2x$$

Hessian determinant: $D = f_{xx}f_{yy} - (f_{xy})^2$

Test each critical point:

(A) $(0, -1)$: $$f_{xx} = 2(2-(-1)) = 6$$ $$f_{yy} = -6(-1) + 6 = 12$$ $$f_{xy} = 0$$ $$D = 6 \cdot 12 - 0 = 72 > 0$$

Since $D > 0$ and $f_{xx} > 0$: local minimum, not a saddle point 

(B) $(0, 3)$: $$f_{xx} = 2(2-3) = -2$$ $$f_{yy} = -6(3) + 6 = -12$$ $$f_{xy} = 0$$ $$D = (-2)(-12) - 0 = 24 > 0$$

Since $D > 0$ and $f_{xx} < 0$: local maximum, not a saddle point 

(C) $(3, 2)$: $$f_{xx} = 2(2-2) = 0$$ $$f_{yy} = -6(2) + 6 = -6$$ $$f_{xy} = -2(3) = -6$$ $$D = 0 \cdot (-6) - (-6)^2 = -36 < 0$$

Since $D < 0$: saddle point 

(D) $(-3, 2)$: $$f_{xx} = 2(2-2) = 0$$ $$f_{yy} = -6(2) + 6 = -6$$ $$f_{xy} = -2(-3) = 6$$ $$D = 0 \cdot (-6) - 6^2 = -36 < 0$$

Since $D < 0$: saddle point 

Answer: (C) and (D) are saddle points