Critical Points & Steepest Ascent/Descent

🎯 Learning Objectives

• Find critical points of $f(x,y)$ by solving $\nabla f = \mathbf{0}$
• Classify critical points using the second derivative test (Hessian)
• Determine the direction of steepest ascent and descent
• Use the gradient to find maxima and minima

📋 Prerequisites

• Gradient (Week 9) — $\nabla f$ must be zero at interior extrema
• Tangent Planes (this week) — horizontal tangent plane at critical point
• Partial Derivatives (Week 9) — computing first and second derivatives

1. Intuition: Flat Spots on a Surface

• Local maximum (peak)
• Local minimum (valley)
• Saddle point (mountain pass — max in one direction, min in another) (Diagram)

2. Finding Critical Points

Definition (Critical Point). A point $\mathbf{a}$ is a critical point of $f$ if $\nabla f(\mathbf{a}) = \mathbf{0}$ or $\nabla f$ does not exist. For $f(x,y)$, solve:

Example 1: Finding critical points

$f(x,y) = x^2 + y^2 - 4x + 6y + 13$.

$f_x = 2x - 4 = 0 \Rightarrow x = 2$. $f_y = 2y + 6 = 0 \Rightarrow y = -3$.

Critical point: $(2,-3)$. $f(2,-3) = 4 + 9 - 8 - 18 + 13 = 0$. Example 2: Multiple critical points

$f(x,y) = x^3 - 3x + y^2$.

$f_x = 3x^2 - 3 = 0 \Rightarrow x = \pm 1$. $f_y = 2y = 0 \Rightarrow y = 0$.

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

3. Classification: Second Derivative Test

Theorem (Second Derivative Test for $f(x,y)$). Let $(a,b)$ be a critical point of $f$. Define:

where $H$ is the Hessian matrix.

• If $D > 0$ and $f_{xx} > 0$: local minimum
• If $D > 0$ and $f_{xx} < 0$: local maximum
• If $D < 0$: saddle point
• If $D = 0$: test is inconclusive Example 3: Classifying Example 1

$f(x,y) = x^2 + y^2 - 4x + 6y + 13$, critical point $(2,-3)$.

$f_{xx} = 2$, $f_{yy} = 2$, $f_{xy} = 0$. $D = 2\cdot2 - 0^2 = 4 > 0$, $f_{xx} = 2 > 0$ ⇒ local minimum. Example 4: Classifying Example 2

$f(x,y) = x^3 - 3x + y^2$.

$f_{xx} = 6x$, $f_{yy} = 2$, $f_{xy} = 0$.

At $(1,0)$: $D = 6(1)\cdot2 - 0 = 12 > 0$, $f_{xx}=6>0$ ⇒ local min. At $(-1,0)$: $D = 6(-1)\cdot2 - 0 = -12 < 0$ ⇒ saddle point. Example 5: Saddle point

$f(x,y) = x^2 - y^2$ at $(0,0)$.

$f_x = 2x = 0$, $f_y = -2y = 0$ ⇒ $(0,0)$ is critical. $f_{xx}=2$, $f_{yy}=-2$, $f_{xy}=0$. $D = 2(-2) - 0 = -4 < 0$ ⇒ saddle point.

Indeed: $f(x,0) = x^2$ has min at $x=0$; $f(0,y) = -y^2$ has max at $y=0$.

4. Classifying Critical Points in 3 Variables

• All eigenvalues positive ⇒ local minimum
• All eigenvalues negative ⇒ local maximum
• Mixed signs ⇒ saddle point
• Any zero eigenvalue ⇒ inconclusive

5. Global vs. Local Extrema

6. Edge Cases & Gotchas

7. Common Pitfalls

❌ Pitfall 1: Forgetting to check $f_{xx}$ sign when $D > 0$

❌ Pitfall 2: Missing critical points where derivative doesn't exist

❌ Pitfall 3: Assuming all critical points are extrema

8. Formula Summary Table

9. 📝 Practice Questions

Q1: Find critical point

$f(x,y) = x^2 + 3y^2 - 2x + 12y + 10$.

Solution: $f_x = 2x - 2 = 0 \Rightarrow x = 1$. $f_y = 6y + 12 = 0 \Rightarrow y = -2$. Critical point: $(1,-2)$. Q2: Classify critical point

Classify the critical point from Q1.

Solution: $f_{xx}=2$, $f_{yy}=6$, $f_{xy}=0$. $D = 2(6) - 0 = 12 > 0$, $f_{xx}=2 > 0$ ⇒ local minimum. $f(1,-2) = 1 + 12 - 2 - 24 + 10 = -3$. Q3: Saddle point

$f(x,y) = xy$. Find and classify critical points.

Solution: $f_x = y = 0$, $f_y = x = 0$ ⇒ $(0,0)$. $f_{xx}=0$, $f_{yy}=0$, $f_{xy}=1$. $D = 0\cdot0 - 1^2 = -1 < 0$ ⇒ saddle point. Indeed $f(x,x)=x^2$ has min but $f(x,-x) = -x^2$ has max at $(0,0)$. Q4: Multiple critical points

$f(x,y) = x^3 + y^3 - 3xy$. Find and classify all critical points.

Solution: $f_x = 3x^2 - 3y = 0 \Rightarrow y = x^2$. $f_y = 3y^2 - 3x = 0 \Rightarrow y^2 = x \Rightarrow x^4 = x \Rightarrow x(x^3-1)=0$. $x=0$ ⇒ $y=0$. $x=1$ ⇒ $y=1$.

Critical points: $(0,0)$ and $(1,1)$.

$f_{xx}=6x$, $f_{yy}=6y$, $f_{xy}=-3$.

At $(0,0)$: $D = 0\cdot0 - (-3)^2 = -9 < 0$ ⇒ saddle. At $(1,1)$: $D = 6\cdot6 - (-3)^2 = 36-9=27>0$, $f_{xx}=6>0$ ⇒ local min. $f(1,1)=1+1-3=-1$. Q5: Global extremum

Find global minimum of $f(x,y) = x^2 + y^2$ on $\mathbb{R}^2$.

Solution: $f_x=2x=0$, $f_y=2y=0$ ⇒ $(0,0)$. $D=2\cdot2-0=4>0$, $f_{xx}>0$ ⇒ local min. Since $f(x,y)\geq 0$ and $f(0,0)=0$, this is also the global minimum. No global maximum (unbounded above). Q6: Boundary check

Find max and min of $f(x,y)=x+y$ on the unit disk $x^2+y^2\leq 1$.

Solution: Interior: $\nabla f=(1,1)\neq(0,0)$, so no interior critical points. Boundary $x^2+y^2=1$: Use Lagrange multipliers or parameterise. $f(x,y) = x+y$ on circle radius 1. Maximum of $x+y$ occurs when $(x,y)=(1/\sqrt{2}, 1/\sqrt{2})$, value $\sqrt{2}$. Minimum at $(-1/\sqrt{2}, -1/\sqrt{2})$, value $-\sqrt{2}$. Q7: $D=0$ case

$f(x,y) = x^4 + y^4$. Find and classify $(0,0)$.

Solution: $\nabla f = (4x^3, 4y^3) = (0,0)$ ⇒ $(0,0)$. $f_{xx}=12x^2$, $f_{yy}=12y^2$, $f_{xy}=0$. At $(0,0)$: $f_{xx}=f_{yy}=0$, $D=0$ ⇒ inconclusive. By inspection: $f(x,y) = x^4+y^4 \geq 0$ and $f(0,0)=0$, so it's a local (and global) minimum. Q8: Steepest ascent

From $(1,1)$ on $f(x,y)=x^2+y^2$, in which direction does $f$ increase most rapidly?

Solution: $\nabla f = (2x,2y)$, $\nabla f(1,1) = (2,2)$. Direction of steepest ascent: $\frac{(2,2)}{\sqrt{8}} = \left(\frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}}\right)$. Q9: No critical points

Show $f(x,y)=e^{x+y}$ has no critical points.

Solution: $\nabla f = (e^{x+y}, e^{x+y}) \neq (0,0)$ since $e^{x+y} > 0$ for all $x,y$. Q10: Second derivative test in 3D

$f(x,y,z) = x^2 + y^2 + z^2$. Find and classify critical points.

Solution: $\nabla f = (2x,2y,2z) = (0,0,0)$ ⇒ $(0,0,0)$. Hessian:

. Eigenvalues: $2,2,2$ (all positive) ⇒ local minimum. $f(0,0,0)=0$ is global minimum.

🔗 Cross-References

• Next topic: Hessian Matrix & Local Extrema
• Week 11 (Hessian): Detailed classification using eigenvalues
• BSMA1001 (Maths 1): Single-variable critical points
• BSCS3004 (Deep Learning): Gradient descent optimisation, saddle points in high dimensions Join Discord Previous10.1 Tangent PlanesNext11.1 Hessian Matrix