Maxima

Maxima

1. What We Seek

For $z = f(x,y)$, a local maximum at $(a,b)$ means $f(a,b) ≥ f(x,y)$ for all $(x,y)$ near $(a,b)$ (inside some disc). Global maximum is ≥ all domain points. This topic covers local maxima without constraints (Lagrange will add constraints).

2. Necessary Condition — Stationary (Critical) Points

If $f$ differentiable and $(a,b)$ is interior local maxima/minima, then:

i.e. tangent plane horizontal. Points satisfying both are stationary / critical points. On boundaries, maxima can occur with non-zero gradient — but interior needs $∇f = 0$.

Finding them: solve simultaneous $f_x=0, f_y=0$. Expect 1-4 solutions for polynomial $f$.

3. Second-Derivative Test — The Discriminant

Let:

is the Hessian determinant $D = f_{xx} f_{yy} - f_{xy}² = det [[A,B],[B,C]]$.

$D$ and $A$	Conclusion at $(a,b)$
$D > 0$ and $A < 0$	Local maximum (concave down both ways)
$D > 0$ and $A > 0$	Local minimum
$D < 0$	Saddle point (not maxima nor minima — rises one direction, falls another)
$D = 0$	Test fails — need higher-order or direct inspection

Derivation sketch: second-order Taylor: $Δf ≈ 1/2[ A h² +2B hk + C k²]$ is a quadratic form; $D$ decides its sign definite.

4. Worked Examples

(a) f(x,y)= x^3 + y^3 -3x y:
    f_x=3x^2-3y=0 ⇒ y=x^2 ; f_y=3y^2-3x=0 ⇒ 3x^4-3x=0 ⇒ x=0 or 1,13? Actually x(x^3-1)=0 ⇒ (0,0) and (1,1)
    At (0,0): A=6x=0, B=-3, C=6y=0 ⇒ D= -9 <0 ⇒ SADDLE ✓
    At (1,1): A=6, B=-3, C=6 ⇒ D=36-9=27>0, A>0 ⇒ MINIMUM (value -1) (no maximum) ✓

(b) f(x,y)= x^4 + y^4 -2x^2 +4x y -2y^2:
    f_x=4x^3-4x+4y=0, f_y=4y^3+4x-4y=0 ⇒ adding: 4x^3+4y^3=0 ⇒ y=-x ⇒ 4x^3-4x-4x=0 ⇒ x=0,±√2
    Points (0,0), (√2,-√2), (-√2,√2); evaluate D and A to classify ⇒ (0,0) saddle, other two minima ✓

(c) f(x,y)= sin x sin y sin(x+y) (0<x,y<π): stationary at x=y=π/3 ⇒ A<0, D>0 ⇒ MAXIMUM value 3√3/8 ✓

(d) Need to also check boundary for global: inside disc x^2+y^2 ≤1, maximum may be on circle — use Lagrange later.

5. Saddle vs Maxima — Geometric Picture

At $D<0$, cross-sections $x=a$ and $y=b$ give opposite concavities — horse saddle $z=x²-y²$. Maxima is a peak ($A<0$), minima a bowl ($A>0$).

6. Exam Points

• Solve $f_x=0, f_y=0$ to find stationary points — 3 marks, almost certain.
• Compute $D = f_{xx}f_{yy}-f_{xy}²$ and classify as maxima/minima/saddle — 7 marks, very frequent.
• Show $D=0$ case needs separate check — 3 marks.
• Worked polynomial $x^3+y^3-3xy$ -type classification — 7-10 marks, almost certain.