Lagrange multipliers — one and two constraints

4.4.15 · Maths › Multivariable Calculus

YE KAAM KYUN KARTA HAI? (Scratch se derivation)

KIYA CHAHIYE HUMEIN: $f (x)$ ko maximize/minimize karna jab $x$ surface $g (x) = 0$ pe rehne ke liye forced ho.

Step 1 — Constraint ke saath move karne ke baare mein socho. Maano $r (t)$ koi bhi smooth curve hai jo constraint surface ke andar rehti hai, toh $g (r (t)) = 0$ sabhi $t$ ke liye. Constrained extremum point $r (t_{0}) = x^{*}$ pe, value $f (r (t))$ ka ek ordinary extremum hona chahiye, toh

$\frac{d}{d t} f (r (t))_{t_{0}} = 0.$

Ye step kyun? Constraint surface pe hum free nahi hain — hum sirf un curves pe travel kar sakte hain jo surface mein rehti hain. Constrained max simply restricted function ka ordinary max hai.

Step 2 — Chain rule apply karo.

$\nabla f (x^{*}) \cdot r^{'} (t_{0}) = 0.$

Ye surface ki har tangent direction $r^{'} (t_{0})$ ke liye hold karta hai. Toh $\nabla f$ sabhi tangent directions ke orthogonal hai — yaani $\nabla f$ constraint surface ke normal hai.

Step 3 — Lekin $\nabla g$ bhi surface ke normal hai. $g (r (t)) = 0$ differentiate karo: $\nabla g \cdot r^{'} = 0$ , toh $\nabla g$ bhi normal hai. Ek hi surface ke normal do vectors (one-dim normal space) parallel hone chahiye:

Do constraints

Ab point ko dono $g (x) = 0$ aur $h (x) = 0$ satisfy karne chahiye. Geometrically feasible set do surfaces ka intersection curve hai.

Do multipliers kyun? Us intersection curve ke saath, allowed tangent direction $T$ dono $\nabla g$ aur $\nabla h$ ke perpendicular honi chahiye. Extremum pe $\nabla f \cdot T = 0$ , toh $\nabla f$ ka $T$ ke along koi component nahi — matlab $\nabla f$ , $\nabla g$ aur $\nabla h$ se spanned plane mein lie karta hai:

Worked Example 1 — One constraint (ek classic)

$f (x, y) = x y$ maximize karo subject to $g = x^{2} + y^{2} - 1 = 0$ (unit circle pe point).

$\nabla f = (y, x)$ , $\nabla g = (2 x, 2 y)$ . $\nabla f = λ \nabla g$ set karo:

$y = 2 λ x$ (kyun: pehla component)
$x = 2 λ y$ (kyun: doosra component)

Multiply karo: $x y = 4 λ^{2} x y$ . Agar $x y \neq = 0$ , toh $4 λ^{2} = 1$ , $λ = \pm \frac{1}{2}$ . $y = 2 λ x$ se $λ = \frac{1}{2}$ ke saath: $y = x$ . Constraint mein plug karo: $2 x^{2} = 1$ , $x = \pm \frac{1}{2}$ .

Max value $f = x y = \frac{1}{2}$ at $(\frac{1}{2}, \frac{1}{2})$ . Kyun accept karein: check karo $λ = - \frac{1}{2}$ se $y = - x$ aata hai, $f = - \frac{1}{2}$ (min hai).

Worked Example 2 — Distance / shadow-price meaning

Line $x + y = 4$ pe origin se closest point. $f = x^{2} + y^{2}$ minimize karo, $g = x + y - 4 = 0$ .

$\nabla f = (2 x, 2 y) = λ (1, 1)$ ⟹ $2 x = 2 y = λ$ ⟹ $x = y$ . Constraint: $2 x = 4$ , $x = y = 2$ . $f^{*} = 8$ . $λ$ meaning check karo: $g = x + y - c$ ke saath, $f^{*} = c^{2} /2$ , toh $d f^{*} / d c = c = 4$ , aur indeed $λ = 2 x = 4$ . ✔ Multiplier sensitivity ke barabar hai.

Worked Example 3 — Two constraints

$f = x + y + z$ maximize karo circle pe jo plane $g = x + y + z - 1 = 0$ ... ruko, ye degenerate hai. Use karo: $f = z$ minimize karo subject to $g = x^{2} + y^{2} - z = 0$ (paraboloid) aur $h = x + y + z - 3 = 0$ (plane).

$\nabla f = (0, 0, 1)$ , $\nabla g = (2 x, 2 y, - 1)$ , $\nabla h = (1, 1, 1)$ . $\nabla f = λ \nabla g + μ \nabla h$ :

$0 = 2 λ x + μ$
$0 = 2 λ y + μ$
$1 = - λ + μ$

Pehle do se: $2 λ x = 2 λ y$ ⟹ $x = y$ (agar $λ \neq = 0$ ). Constraints: $z = 2 x^{2}$ aur $2 x + z = 3$ ⟹ $2 x + 2 x^{2} = 3$ ⟹ $2 x^{2} + 2 x - 3 = 0$ ⟹ $x = \frac{- 1 \pm 7}{2}$ . Do roots kyun: curve neeche jaata hai phir upar; ek min hai, ek max hai $z$ ka curve pe.

Common mistake Common errors ko steel-man karna

Mistake A: "Bas $\nabla f = 0$ set kar do." Sahi lagta hai kyunki aise hi hum unconstrained extrema dhundhte hain. Fix: constraint pe, $\nabla f$ usually ZERO nahi hota — bas surface ke bahar point karna chahiye, $\nabla g$ se balance hota hua.

Mistake B: Constraint equation ko bhool jaana. $\nabla f = λ \nabla g$ directions deta hai; location pin karne ke liye abhi bhi $g = 0$ chahiye. Tumhe saare $n + 1$ equations chahiye.

Mistake C: Un points ko ignore karna jahan $\nabla g = 0$ (singular points) ya boundaries pe — Lagrange wahan extrema miss kar sakta hai. Sahi lagta hai algebra par trust karna. Fix: method ki assumption check karo (gradients nonzero & independent).

Mistake D (two constraints): Ye assume karna ki $\nabla f = λ \nabla g$ ya $= μ \nabla h$ alag alag. Fix: ye dono ka combination hona chahiye.

Recall Feynman: 12-saal ke bacche ko samjhao

Socho tum ek bounded hilly field mein chal rahe ho aur tumhe fence path pe rehte hue highest spot chahiye. Chalte raho; jab tak ground tumhari walking direction mein upar jaata hai, chalte raho. Tum high point pe ruk jaate ho jab slope sirf fence ke across sideways jaata hai, tumhare path ke along nahi. "Path ke sideways" exactly yahi hai " $\nabla f$ fence ke across point karta hai" = fence ke normal $\nabla g$ ke parallel. Do fences cross karte hain toh tum sirf wahan khade ho sakte ho jahan wo cross karein, aur do pulls balance karte ho — wahi do multipliers hain.

Flashcards

What condition does Lagrange's method impose at a constrained extremum (one constraint)?

\nabla f = λ \nabla g

AND

g = 0

Geometric meaning of

\nabla f = λ \nabla g

\nabla f

constraint surface ke normal hai (

\nabla g

ke parallel); koi tangential direction

f

ko increase nahi karti.

Why is

\nabla g

normal to the surface

g = 0

g (r (t)) = 0

differentiate karne se

\nabla g \cdot r^{'} = 0

milta hai har tangent

r^{'}

ke liye.

Interpretation of the multiplier

λ

Constraint level relax karne ke saath optimal value ka rate of change:

λ = d f^{*} / d c

(shadow price).

Condition with TWO constraints?

\nabla f = λ \nabla g + μ \nabla h

, with

g = 0

and

h = 0

Why two multipliers for two constraints?

\nabla f

ka intersection curve ke tangent ke along koi component nahi hona chahiye, toh wo

span {\nabla g, \nabla h}

mein lie karta hai.

How many equations/unknowns for

f : R^{3}

with two constraints?

5 equations (3 gradient + 2 constraints), 5 unknowns

(x, y, z, λ, μ)

A case where Lagrange can MISS an extremum?

Jahan

\nabla g = 0

(singular point) ho ya boundary pe jo

g = 0

se capture nahi hoti.

Max of

x y

x^{2} + y^{2} = 1

1/2

(\frac{1}{2}, \frac{1}{2})

Connections

Gradient and directional derivative — kyun $\nabla f$ steepest ascent deta hai.
Tangent planes and normal vectors — $\nabla g$ surface normal ke roop mein.
Unconstrained optimization — critical points — $λ = 0$ / no-constraint limit.
KKT conditions — inequality-constraint generalization.
Level sets and contours — optimum pe level curves ki tangency.
Dual problem and shadow prices — $λ$ ka economic meaning.