Lagrange multipliers for constrained optimization: Worked examples

This page is a catalogue of cases. The parent note showed you the machinery; here we run it on every kind of problem the topic can throw at you — different signs of $λ$ , curved constraints, several constraints at once, degenerate cases where the method almost breaks, a word problem, and an exam-style twist.

Before we start, one tiny reminder of the only rule we use, spelled out in words:

Look at the figure: the grey curves are level curves of $f$ (points where $f$ has the same value, like contour lines on a map). The blue line is the constraint wall. At the star, a level curve just kisses the wall — they are tangent, so their perpendicular arrows line up. That tangency is the Lagrange condition.

The scenario matrix

Every Lagrange problem you will meet falls into one of these cells. The rest of the page fills each cell with a fully worked example.

#	Case class	What makes it special	Example
A	Straight-line constraint, minimize	$λ > 0$ , convex, unique answer	Ex 1
B	Symmetric objective, maximize	$λ > 0$ , symmetry shortcut	Ex 2
C	Curved (nonlinear) constraint	$\nabla g$ changes direction; multiple candidates	Ex 3
D	Negative multiplier	$λ < 0$ — relaxing the wall hurts	Ex 4
E	Zero / degenerate	objective already flat, or $\nabla g = 0$	Ex 5
F	Two constraints at once	vector $λ$ , sum of gradients	Ex 6
G	Word problem (real world)	translate words → $f, g$ ; read $λ$ as price	Ex 7
H	Exam twist	unknown constant $c$ ; solve for a target value	Ex 8

Ex 1 — Cell A: line constraint, minimize

Forecast: by symmetry between $x$ and $y$ , guess the answer sits at equal coordinates. Where?

Build the Lagrangian. Why this step? It fuses objective and wall into one function whose flat points are the constrained optima. $L = x^{2} + y^{2} - λ (x + y - 1)$
Differentiate in $x, y$ . Why? These say " $\nabla f$ is parallel to $\nabla g$ ." $2 x - λ = 0, 2 y - λ = 0 \Rightarrow x = y = \frac{λ}{2}$
Differentiate in $λ$ . Why? It returns the wall so we don't float off it. $x + y = 1$
Solve. Why? Substitute $x = y = λ /2$ : $λ = 1$ , so $x = y = \frac{1}{2}$ .

Verify: point $(\frac{1}{2}, \frac{1}{2})$ lies on the line ( $\frac{1}{2} + \frac{1}{2} = 1$ ✓). Distance $^{2} = \frac{1}{4} + \frac{1}{4} = \frac{1}{2}$ . Since $f$ is a bowl (convex) and the wall a straight line, this single stationary point is the global minimum. $λ = 1 > 0$ : pushing the line out ( $c : 1 \to 1.01$ ) raises the min distance $^{2}$ by about $0.01$ .

The red arrow $\nabla f$ (pointing away from origin) and blue arrow $\nabla g$ point the same way at the star — that is why $λ > 0$ .

Ex 2 — Cell B: symmetric objective, maximize

Forecast: everyone "knows" the square wins — but why does the algebra force it?

Lagrangian. $L = x y - λ (2 x + 2 y - 20)$ .
Partials in $x, y$ . Why? Parallel-gradient condition. $y - 2 λ = 0, x - 2 λ = 0 \Rightarrow x = y = 2 λ$
Symmetry read-off. Why this step? The equations are identical after swapping $x \leftrightarrow y$ , so the optimum must have $x = y$ — the symmetry of the problem is inherited by the answer.
$λ$ -partial + solve. $2 x + 2 y = 20 \Rightarrow x + y = 10 \Rightarrow 2 λ + 2 λ = 10 \Rightarrow λ = 2.5$ , so $x = y = 5$ .

Verify: area $= 5 \times 5 = 25$ . Nudge perimeter to $21$ : new side $5.25$ , area $27.5625$ , gain $2.5625 \approx λ = 2.5$ ✓. Units: $λ$ has units of area per unit perimeter (m²/m = m), matching "shadow price."

Ex 3 — Cell C: curved constraint, multiple candidates

Forecast: a curved wall means $\nabla g$ points radially and keeps turning. So expect two answers — a max and a min on opposite sides.

Lagrangian. $L = x + y - λ (x^{2} + y^{2} - 2)$ .
Partials. Why? Same rule, but now $\nabla g = (2 x, 2 y)$ depends on position. $1 - 2 λ x = 0, 1 - 2 λ y = 0 \Rightarrow x = y = \frac{1}{2 λ}$
Impose the circle. Why? Curved constraint, so substitute. $2 (\frac{1}{2 λ})^{2} = 2 \Rightarrow \frac{1}{2 λ ^{2}} = 2 \Rightarrow λ^{2} = \frac{1}{4} \Rightarrow λ = \pm \frac{1}{2}$
Two branches. Why two? Each sign of $λ$ is a genuine candidate.
- $λ = + \frac{1}{2}$ : $x = y = 1$ , $f = 2$ → maximum.
- $λ = - \frac{1}{2}$ : $x = y = - 1$ , $f = - 2$ → minimum.

Verify: both points satisfy $1^{2} + 1^{2} = 2$ ✓. At the max, $\nabla f = (1, 1)$ points outward like $\nabla g = (2, 2)$ (same direction, $λ > 0$ ); at the min the radial arrow points inward, opposite $\nabla f$ ( $λ < 0$ ). A curved wall must be checked at all candidates — this is the cell the line problems never showed.

Ex 4 — Cell D: negative multiplier

This is the minimizing half of Ex 3. We already found the candidate $λ = - \frac{1}{2}$ , $x = y = - 1$ , $f = - 2$ .

Read the sign. Why this step? With $L = f - λ (g - c)$ , the optimal value changes by $\approx λ Δ c$ . Here $λ = - \frac{1}{2} < 0$ .
Interpret. Increase $c$ from $2$ to $2.2$ (bigger circle). Predicted change in min: $λ Δ c = - \frac{1}{2} (0.2) = - 0.1$ , i.e. the minimum drops to about $- 2.1$ .

Verify (exact): on radius $r = c$ the minimum of $x + y$ is $- 2 c$ . At $c = 2$ : $- 2$ . At $c = 2.2$ : $- 4.4 \approx - 2.0976$ . Change $\approx - 0.0976 \approx - 0.1$ ✓. Lesson of the cell: a negative $λ$ is normal and meaningful — "relaxing the constraint" made the minimum smaller, not larger.

Ex 5 — Cell E: zero / degenerate inputs

Forecast: E1 has nothing to optimize; E2 has a constraint that is a single point.

E1 steps.

Lagrangian. $L = 3 - λ (x + y - 1)$ .
Partials. $\partial_{x} L = - λ = 0$ , $\partial_{y} L = - λ = 0 \Rightarrow λ = 0$ . Why $λ = 0$ ? Because $\nabla f = (0, 0)$ : the objective doesn't change anywhere, so nudging the wall changes nothing — zero shadow price.
Constraint. $x + y = 1$ : every point on the line is an optimum.

Verify (E1): $f = 3$ for all feasible points; there is no unique minimizer, and $λ = 0$ correctly reports "the wall's location is worthless to me." ✓

E2 steps.

Inspect $g$ . $x^{2} + y^{2} = 0$ forces $x = y = 0$ — the constraint is a single point, not a curve.
Check $\nabla g$ . $\nabla g = (2 x, 2 y) = (0, 0)$ at that point. Why does this break the method? Lagrange assumes $\nabla g \neq = 0$ so a well-defined perpendicular direction exists. Here there is no wall to be tangent to.
Resolve by hand. The only feasible point is $(0, 0)$ , so it is trivially both the max and min: $f = 0$ .

Verify (E2): feasible set $= {(0, 0)}$ , $f (0, 0) = 0$ ✓. Cell lesson: always confirm $\nabla g \neq = 0$ (a constraint qualification); when it fails, fall back to direct reasoning. This subtlety is exactly where KT Conditions add care for inequalities.

Ex 6 — Cell F: two constraints at once

Forecast: two walls → two multipliers → $\nabla f = λ_{1} \nabla g_{1} + λ_{2} \nabla g_{2}$ .

Multi-constraint Lagrangian. Why a sum? Each wall pins one direction; $\nabla f$ must be a blend of both wall-arrows. $L = x^{2} + y^{2} + z^{2} - λ_{1} (x + y + z - 1) - λ_{2} (x - y)$
Partials. $2 x - λ_{1} - λ_{2} = 0, 2 y - λ_{1} + λ_{2} = 0, 2 z - λ_{1} = 0$
Use symmetry $x = y$ . From $g_{2}$ , $x = y$ . Subtracting the first two partials: $2 x - 2 y = 2 λ_{2} \Rightarrow λ_{2} = 0$ . Why? The two planes are already symmetric in $x, y$ , so the second wall costs nothing here.
Solve. Then $x = y = λ_{1} /2$ , $z = λ_{1} /2$ . Plug into $g_{1}$ : $3 (λ_{1} /2) = 1 \Rightarrow λ_{1} = \frac{2}{3}$ , so $x = y = z = \frac{1}{3}$ .

Verify: $x + y + z = 1$ ✓, $x - y = 0$ ✓, $f = 3 \cdot \frac{1}{9} = \frac{1}{3}$ . Any other point on the line has larger $f$ (convex bowl) ✓.

Ex 7 — Cell G: real-world word problem

Forecast: cost-minimizing under an area target — expect a square again, and $λ$ to read as "metres of fence saved per extra m² required."

Translate words → math. Objective = perimeter, wall = area. Why? Getting $f, g$ right is 90% of a word problem.
Lagrangian. $L = 2 x + 2 y - λ (x y - 100)$ .
Partials. $2 - λ y = 0$ , $2 - λ x = 0 \Rightarrow x = y = \frac{2}{λ}$ .
Constraint + solve. $x y = 100 \Rightarrow (2/ λ)^{2} = 100 \Rightarrow 2/ λ = 10 \Rightarrow x = y = 10$ , and $λ = \frac{2}{10} = 0.2$ .

Verify: area $10 \times 10 = 100$ ✓, perimeter $= 40$ m. Units of $λ$ : metres per m² = m $^{- 1}$ ; predicted extra fence for one more m² of area $= λ \cdot 1 = 0.2$ m. Exact: perimeter for area $A$ is $4 A$ , derivative $2/ A = 2/10 = 0.2$ ✓.

Ex 8 — Cell H: exam twist (solve for the constant)

Forecast: we solve the family in terms of $c$ , then invert.

Solve generally. Why? Keep $c$ as a symbol so we can tune it. From Ex 1's pattern: $x = y = c /2$ , $λ = c$ .
Minimum value. $f_{m i n} = 2 (c /2)^{2} = \frac{c ^{2}}{2}$ .
Set equal to target. $\frac{c ^{2}}{2} = 8 \Rightarrow c^{2} = 16 \Rightarrow c = \pm 4$ .
Interpret both roots. Why two? The line $x + y = 4$ and $x + y = - 4$ are equidistant from the origin (symmetric across it), so both give min distance $^{2} = 8$ .

Verify: for $c = 4$ : $x = y = 2$ , $f = 4 + 4 = 8$ ✓; $λ = 4$ , and $\frac{d}{d c} (\frac{c ^{2}}{2}) = c = 4 = λ$ ✓ (shadow price matches). For $c = - 4$ : $x = y = - 2$ , $f = 8$ ✓.

Recall Quick self-test

Which sign does $λ$ take when relaxing the wall improves a minimization? ::: Negative — a lower minimum means $λ < 0$ (see Ex 4). Why must a curved constraint be checked at every stationary candidate? ::: Because $\nabla g$ keeps rotating, giving several tangency points — some maxima, some minima (Ex 3). What breaks the whole method, and what do you do? ::: When $\nabla g = 0$ (constraint qualification fails); solve the feasible set directly instead (Ex 5). In two constraints, what replaces $\nabla f = λ \nabla g$ ? ::: $\nabla f = λ_{1} \nabla g_{1} + λ_{2} \nabla g_{2}$ (Ex 6).

Where this goes next: equality-only Lagrange becomes the KKT system once you add inequality walls — see KT Conditions, and its starring role in Support Vector Machines, Regularization, Duality Theory, Convex Optimization, and the softer cousin Penalty Methods.