Lagrange's Multiplier Method

Lagrange's Multiplier Method

1. Why Constraints Need a New Method

Unconstrained interior needs $∇f=0$, but if $(x,y)$ must satisfy constraint $g(x,y)=0$ (e.g., lie on circle $x²+y²=1$, or sum $x+y=10$), the extremum typically occurs where $∇f≠0$ — along the constraint curve. Lagrange's method finds such constrained extrema without solving $g$ for one variable.

2. Statement — Lagrange Function

To extremise $f(x,y)$ subject to $g(x,y)=0$, form:

Solve simultaneously:

i.e. $f_x + λ g_x =0, f_y + λ g_y =0, g=0$. Geometrically, at optimum $∇f$ is parallel to $∇g$ ($∇f = -λ ∇g$), meaning $f$-contour touches $g$-curve tangentially.

For three variables with one constraint $g(x,y,z)=0$, same with $L(x,y,z,λ)$. With two constraints $g=0, h=0$, use two multipliers: $L = f + λ g + μ h$.

3. Working Steps (exam recipe)

1. Write $L$ and three equations $L_x=0, L_y=0, g=0$.
2. Eliminate $λ$ (often ratio $f_x/g_x = f_y/g_y$ if denominators ≠0) to relate $x,y$.
3. Use $g=0$ to find candidate points.
4. Evaluate $f$ at candidates (and compare) — for proof of max/min, either argue $f$ bounded or check-bordered Hessian/second variation sign if required.

4. Worked Examples

(a) Extremise f= x y subject to x + y = 10:
    L= x y + λ(x+y-10): L_x= y+λ=0 ⇒ λ=-y ; L_y= x+λ=0 ⇒ λ=-x ⇒ x=y ⇒ 2x=10 ⇒ x=5,y=5, λ=-5
    f=25 is MAXIMUM (product maximal when sum fixed and numbers equal; endpoints 0,10 give 0) ✓

(b) Find extrema of f= x^2 + y^2 subject to x^2 + y^2 + x y = 3? Actually constraint g= x+y-? Choose circle:
    Simpler: f= x^2 + y^2 on circle g= x^2 + y^2 + x? Better classic: f= x^2 + y^2 with x+y=1 ⇒ L_x=2x+λ=0, L_y=2y+λ=0 ⇒ x=y=1/2 ⇒ min 1/2 ✓

(c) Rectangular box with given surface area? Classic: maximise volume V=x y z subject to S=2(xy+yz+zx)-constant=0 ⇒ gives cube x=y=z ✓

(d) Distance from origin to line x+2y=5? Minimise f= x^2+y^2 (squared distance) subject to g= x+2y-5=0:
    L_x=2x+λ=0 ⇒ x=-λ/2 ; L_y=2y+2λ=0 ⇒ y=-λ ; g: -λ/2 -2λ=5 ⇒ -5λ/2=5 ⇒ λ=-2 ⇒ x=1, y=2
    Min distance √5 ; check second: Hessian positive ⇒ minimum ✓

(e) f= x^2 + y^2 subject to g= x^4 + y^4 +2x^2 y^2? Actually circle: use polar to verify.

5. Sufficiency — When Asked

Second-order condition via bordered Hessian $B = det [[0,g_x,g_y],[g_x, L_{xx},L_{xy}],[g_y,L_{xy},L_{yy}]]$: at candidate, $B>0 ⇒$ constrained maximum, $B<0 ⇒$ minimum (for one constraint, two variables). Many introductory courses just compare $f$ values if domain bounded — mention both.

6. Exam Points

• Form $L = f + λ g$ and solve $L_x=L_y=L_λ=0$ — 5-7 marks, almost certain.
• Find maxima/minima of $x y$ given $x+y=a$, or $x²+y²$ given $g$ — 7 marks, very frequent.
• Two-constraint case — 5 marks, frequent.
• State geometrically $∇f ∥ ∇g$ — 3 marks.
• Second-order / bordered Hessian mention — 3 marks, occasional.