Lagrange multipliers — one and two constraints: Question bank

Vocabulary reminder before we start (every symbol is earned here so you never meet an undefined one):

$f$ — the thing we optimize (the "height"). Its gradient $\nabla f$ is the arrow pointing in the direction $f$ increases fastest. See Gradient and directional derivative.
$g = 0$ , $h = 0$ — the constraints: surfaces (or the curve where they cross) you are forced to stay on. $\nabla g$ , $\nabla h$ are arrows sticking straight out of those surfaces — the normals. See Tangent planes and normal vectors.
$λ$ (lambda), $μ$ (mu) — the multipliers: numbers that say how much of each normal arrow you mix to reproduce $\nabla f$ .
$T$ — a tangent direction: a direction you're actually allowed to walk while staying feasible.

Everything below is the parent method poked from every angle.

Spot the error

"To minimize

f

x + y = 4

, I set

\nabla f = 0

and solved."

Wrong tool:

\nabla f = 0

finds unconstrained critical points and ignores the fence. You need

\nabla f = λ \nabla g

plus the constraint

x + y - 4 = 0

"I wrote

\nabla f = λ \nabla g

and got

λ

; done."

Incomplete: you never used

g = 0

, so the point isn't located. Dropping the constraint equation is the classic under-determination error.

"On the intersection of two surfaces I used only

\nabla f = λ \nabla g

You forgot the second surface. The feasible set is a curve, so

\nabla f

must be balanced by both normals:

\nabla f = λ \nabla g + μ \nabla h

\nabla g

and

\nabla f

point the same way, so

λ > 0

always."

Not so. Parallel includes antiparallel;

\nabla f = λ \nabla g

allows

λ < 0

when the two arrows point opposite ways.

"The tangency of level curves means

f

and

g

have equal values there."

No. Tangency means their level curves touch (parallel normals), not that

f = g

. It's a geometric alignment of directions, not of function values. See Level sets and contours.

"Since Lagrange gave one point, that's the answer."

The method can yield several stationary points (min, max, saddle on the curve); one root of the algebra is rarely the whole story. Always enumerate and compare.

"At a singular point where

\nabla g = 0

, the equation

\nabla f = λ \nabla g

still pins things down."

It collapses to

\nabla f = 0

regardless of

λ

, so the method gives no information; such points must be checked separately — Lagrange can silently miss them.

"Two constraints in

R^{3}

give 3 equations."

Miscount. It's 3 gradient equations + 2 constraint equations = 5 equations in the 5 unknowns

(x, y, z, λ, μ)

Why questions

Why must

\nabla f

be normal to the constraint surface at an extremum?

If it had any tangential component, you could walk along the surface in that direction and still change

f

— so it wouldn't be an extremum. Only a purely normal

\nabla f

can't be improved.

Why does

\nabla f

live in the plane spanned by

\nabla g, \nabla h

for two constraints?

The one allowed tangent

T

is perpendicular to both normals;

\nabla f \cdot T = 0

forces

\nabla f

to have no

T

-part, i.e. to lie in the plane those two normals define.

Why is

λ

called a "shadow price"?

Because

λ = d f^{*} / d c

where the constraint level is

c

: it's the marginal improvement in the optimum per unit of loosening the constraint — literally its economic price. See Dual problem and shadow prices.

Why do we differentiate

f (r (t))

along curves inside the surface?

Because on the constraint we can only move along such curves; a constrained extremum is an ordinary extremum of

f

restricted to every feasible path, giving

\frac{d}{d t} f = 0

Why does the chain rule turn

\frac{d}{d t} f = 0

into

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

The chain rule says the rate of change of

f

along

r (t)

is the dot product of the gradient with the velocity

r^{'}

; setting it to zero is exactly the extremum condition. See Gradient and directional derivative.

Why doesn't the sign or magnitude of

λ

affect where the optimum is?

λ

only balances the lengths of the two gradient arrows; the location is fixed by requiring parallel directions plus satisfying

g = 0

Why does Lagrange generalize to KKT conditions for inequalities?

An inequality is either slack (multiplier

= 0

, constraint idle) or tight (behaves like an equality with a sign-restricted multiplier); KKT bundles both cases into the same gradient-balance idea.

Edge cases

What if

\nabla g = 0

at a candidate point?

The surface has no well-defined normal there (a singular point);

\nabla f = λ \nabla g

becomes

\nabla f = 0

for any

λ

, so you must inspect that point by hand — Lagrange can miss extrema there.

What if

\nabla g

and

\nabla h

are parallel (dependent) at a point?

They no longer span a plane, so the intersection isn't a clean curve; the two-constraint setup degenerates and

λ, μ

aren't uniquely determined — a constraint-qualification failure.

What if the constraint set is a closed region with a boundary, not just

g = 0

Lagrange only searches the smooth surface

g = 0

; extrema can sit on a boundary edge not described by

g = 0

, so those edges need a separate check.

What happens to the method as the constraint is removed (no

g

You're back to

\nabla f = 0

— the unconstrained case, which is exactly the

λ = 0

limit where the constraint exerts no pull.

Can a constrained problem have no Lagrange solution yet still have a max?

Yes — if the max occurs at a singular point or an unincluded boundary, the gradient-balance equations have no root there, so it's missed unless separately examined.

Why does the paraboloid-plus-plane problem give two roots for

z

The intersection curve dips down then rises, so

z

has both a minimum and a maximum on it — two stationary points, matching the two roots of

2 x^{2} + 2 x - 3 = 0

What if

\nabla f = 0

but the point still lies on

g = 0

Then

\nabla f = λ \nabla g

holds with

λ = 0

: it's an unconstrained critical point that happens to be feasible, a perfectly valid (constraint-inactive) Lagrange solution.

What if

f

is constant on the whole constraint surface?

Every feasible point is trivially an extremum;

\nabla f

is either zero or already normal everywhere, so

λ

is unconstrained/undetermined — the problem is degenerate.

Recall Fast self-check

$\nabla f$ has a tangential component at a feasible point — extremum or not? ::: Not an extremum: you can still raise $f$ by stepping along that tangential component. Two parallel gradients but $g \neq = 0$ — valid Lagrange point? ::: No: parallel gradients are necessary but you must also be on the surface, $g = 0$ .

Connections

Lagrange multipliers — one and two constraints — the parent method these traps probe.
Gradient and directional derivative — why a tangential $\nabla f$ means "not yet optimal".
Tangent planes and normal vectors — the normals $\nabla g, \nabla h$ these traps hinge on.
Unconstrained optimization — critical points — the $λ = 0$ limit.
KKT conditions — the inequality generalization several "why" items point to.
Level sets and contours — tangency of level curves at the optimum.
Dual problem and shadow prices — the economic reading of $λ$ .

Question bank — Lagrange multipliers — one and two constraints

True or false — justify

Spot the error

Why questions

Edge cases

Connections