Worked examples — Lagrange multipliers — one and two constraints

Before anything: three words we will use constantly.

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The scenario matrix

Every cell below is a distinct way this topic can bite. Each worked example names the cell(s) it covers.

#	Cell class	What makes it distinct	Example
C1	One constraint, both signs of $\lambda$	max and min from $\pm \lambda$	Ex 1
C2	Constraint level $c\ne 0$ + shadow price	verify $\lambda =d{f}^{\ast }/dc$, units	Ex 2
C3	Degenerate: $\nabla g=0$ (singular point)	method silently misses an extremum	Ex 3
C4	Boundary / product-of-vars, $xy=0$ branch	must check the branch you divided away	Ex 4
C5	Real-world word problem (economics)	translate words → $f,g$; interpret $\lambda$	Ex 5
C6	Two constraints, two real roots (max & min)	$\nabla f$ in span of two normals	Ex 6
C7	Two constraints, degenerate: parallel normals	when the method breaks / no solution	Ex 7
C8	Limiting behaviour: $\lambda \to 0$	constraint becomes inactive → unconstrained	Ex 8
C9	Exam twist: unbounded / no extremum exists	Lagrange returns a point that is a saddle on the curve	Ex 9

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Ex 1 — Both signs of $\lambda$ (cell C1)

Forecast: guess where on the circle $f$ is biggest. (Hint: $f$ weights $y$ more heavily.)

Figure s01 (below): the black unit circle carries four candidate points. The red ones $(0,\pm 1)$ are the maximum, the black ones $(\pm 1,0)$ the minimum; the red arrow shows $\nabla f$ at the top pointing straight out along the radius — parallel to $\nabla g$, exactly the Lagrange condition. Look at how $\nabla f$ has no sideways (tangential) component there.

1. Write the gradients. $\nabla f=(2x,4y)$, $\nabla g=(2x,2y)$. Why this step? Lagrange compares the two arrows; we need them explicitly.
2. Set $\nabla f=\lambda \nabla g$. $2x=\lambda \cdot 2x$ and $4y=\lambda \cdot 2y$. Why? At the extremum $\nabla f$ has no tangential component, so it is parallel to $\nabla g$.
3. Read each equation carefully. From the first: $2x(1-\lambda )=0$, so $x=0$ or $\lambda =1$. From the second: $2y(2-\lambda )=0$, so $y=0$ or $\lambda =2$. Why split like this? Dividing by $x$ or $y$ would silently drop the $x=0$ / $y=0$ solutions — the very poles of the circle. Never divide away a variable without checking it can be zero.
4. Enumerate branches.
   • $\lambda =1\Rightarrow y=0\Rightarrow x=\pm 1$: points $(\pm 1,0)$, $f=1$.
   • $\lambda =2\Rightarrow x=0\Rightarrow y=\pm 1$: points $(0,\pm 1)$, $f=2$.
5. Assign roles. $f=1$ (at $\lambda =1$) is the minimum; $f=2$ (at $\lambda =2$) is the maximum.

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Ex 2 — Nonzero constraint level & shadow price (cell C2)

Forecast: the closest point sits where the line's perpendicular from the origin lands. Guess it before computing.

Figure s02 (below): the black line is the constraint $x+2y=10$. The red arrow is the shortest path from the origin — it hits the line at right angles at the red point $(2,4)$. That perpendicularity is exactly what $\nabla f\parallel \nabla g$ forces here.

1. Gradients. $\nabla f=(2x,2y)$, $\nabla g=(1,2)$. Why? Distance-squared is smoother than distance and has the same minimizer — no square roots to differentiate.
2. $\nabla f=\lambda \nabla g$: $2x=\lambda$, $2y=2\lambda$. So $x=\lambda /2$, $y=\lambda$. Why keep $c$ symbolic? We want $f^{\ast }$ as a function of $c$ to test the shadow-price claim.
3. Impose the constraint $x+2y=c$: $\frac{\lambda }{2}+2\lambda =c\Rightarrow \frac{5\lambda }{2}=c\Rightarrow \lambda =\frac{2c}{5}$. Why substitute back into the constraint? Steps 1–2 only fixed the direction to the answer (the ratio $x:y$); the constraint is what pins the single actual location on the line.
4. Point & value. Substitute $\lambda =\frac{2c}{5}$ into $x=\lambda /2$ and $y=\lambda$: $x=\frac{c}{5},y=\frac{2c}{5}$, so $f^{\ast }=\frac{c^2}{25}+\frac{4c^2}{25}=\frac{5c^2}{25}=\frac{c^2}{5}$. Why compute $f^{\ast }$ as a formula in $c$? Because the whole point of this cell is to differentiate $f^{\ast }$ with respect to $c$ and compare it to $\lambda$ — a formula lets us do that.
5. Plug $c=10$: $x=2,y=4$, $f^{\ast }=\frac{100}{5}=20$; distance $=\sqrt{20}=2\sqrt{5}$. Why report distance too? The original question asked for the closest point, so we translate $f^{\ast }$ (a squared distance) back into an actual distance.

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Ex 3 — Degenerate: $\nabla g=0$ misses an extremum (cell C3)

Forecast: the curve looks like a leftward-pointing spike. Where is $x$ smallest? Will Lagrange find it?

Figure s03 (below): the black curve is the cusp $y^2=x^3$; the black arrow points in the "decrease $x$" direction. The red dot marks the cusp tip $(0,0)$ where the curve has a sharp corner and $\nabla g=0$ — there is no normal arrow to draw, which is exactly why Lagrange's equation cannot reach it.

1. Gradients. $\nabla f=(1,0)$, $\nabla g=(-3x^2,2y)$.
2. $\nabla f=\lambda \nabla g$: $1=-3\lambda x^2$ and $0=2\lambda y$. From the second: $\lambda =0$ or $y=0$. But $\lambda =0$ makes the first equation $1=0$ — impossible. So $y=0$. Why does this matter? $y=0$ forces $x^3=0$, i.e. the point $(0,0)$.
3. Check the point. At $(0,0)$: $\nabla g=(0,0)=0$. The multiplier equation $1=-3\lambda \cdot 0$ reads $1=0$ — no finite $\lambda$ works. Why? At the cusp the surface has no well-defined normal; $\nabla g=0$ is exactly the singular case the parent note warned about (Mistake C).
4. Resolve by hand. On the curve $x=y^{2/3}\ge 0$, so $x$ is minimized at $x=0$, the cusp $(0,0)$. The extremum is real — Lagrange's equation just couldn't produce it.

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Ex 4 — Product of variables, the $xy=0$ branch (cell C4)

Forecast: $f=xy$ is negative in quadrants II/IV, positive in I/III. So max and min should be symmetric $\pm$ values.

1. Gradients. $\nabla f=(y,x)$, $\nabla g=\left(\frac{x}{2},2y\right)$.
2. Set $\nabla f=\lambda \nabla g$: $y=\lambda \frac{x}{2}$, $x=\lambda 2y$. Why? At a constrained extremum $\nabla f$ has no tangential component along the ellipse, so it must be parallel to the ellipse's normal $\nabla g$ — that parallelism is precisely $\nabla f=\lambda \nabla g$.
3. Multiply them: $y\cdot x=\left(\lambda \frac{x}{2}\right)(2\lambda y)={\lambda }^{2}xy$. So $xy(1-{\lambda }^2)=0$. Why factor, not divide? Dividing by $xy$ hides the branch $xy=0$. Factoring keeps both branches.
4. Branch A: $xy=0$. Then $f=0$; points $(\pm 2,0)$ and $(0,\pm 1)$. These are saddle-value candidates (neither max nor min).
5. Branch B: ${\lambda }^2=1$, $\lambda =\pm 1$.
   • $\lambda =1$: $y=x/2$; constraint $\frac{x^2}{4}+\frac{x^2}{4}=1\Rightarrow x^2=2\Rightarrow x=\pm \sqrt{2}$, $y=\pm \frac{\sqrt{2}}{2}$. $f=xy=\sqrt{2}\cdot \frac{\sqrt{2}}{2}=1$.
   • $\lambda =-1$: $y=-x/2$; same magnitude, $f=-1$.

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Ex 5 — Real-world word problem: economics (cell C5)

Forecast: with capital 4× pricier, will the firm buy more labour or capital?

1. Translate. Constraint $g=x+4y-c=0$ with budget level $c=100$ (spend the whole budget). Maximize $f=\sqrt{x}\sqrt{y}$. Why full budget? Output rises in both inputs, so the optimum spends everything. Why keep $c$ symbolic? We will need $f^{\ast }$ as a function of $c$ to interpret $\lambda$.
2. Gradients. $\nabla f=\left(\frac{1}{2}{x}^{-0.5}{y}^{0.5},\frac{1}{2}{x}^{0.5}{y}^{-0.5}\right)$, $\nabla g=(1,4)$.
3. Equations. $\frac{1}{2}\sqrt{\frac{y}{x}}=\lambda$ and $\frac{1}{2}\sqrt{\frac{x}{y}}=4\lambda$.
4. Divide the two (both sides positive, so safe): $\frac{\sqrt{y/x}}{\sqrt{x/y}}=\frac{1}{4}\Rightarrow \frac{y}{x}=\frac{1}{4}\Rightarrow x=4y$. Why divide here? It cancels $\lambda$ instantly, giving the ratio of inputs — the economic heart of the answer.
5. Budget in terms of $c$: $x+4y=4y+4y=8y=c\Rightarrow y=\frac{c}{8},x=\frac{c}{2}$. Why solve for general $c$? So the optimal-value formula $f^{\ast }(c)$ falls out directly in the next step.
6. Optimal value as a function of $c$. $f^{\ast }(c)=\sqrt{x}\sqrt{y}=\sqrt{\frac{c}{2}}\sqrt{\frac{c}{8}}=\sqrt{\frac{c^2}{16}}=\frac{c}{4}$. Why? This is the general solution before plugging numbers; differentiating it gives the shadow price with no guesswork.
7. Plug $c=100$. $x=50,y=12.5$, output $f^{\ast }=\frac{100}{4}=25$. Multiplier $\lambda =\frac{1}{2}\sqrt{y/x}=\frac{1}{2}\sqrt{12.5/50}=\frac{1}{2}\cdot \frac{1}{2}=0.25$.

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Ex 6 — Two constraints, two real roots (cell C6)

Forecast: the intersection is a tilted circle. $z$ should hit a highest and a lowest point on it.

Figure s04 (below): a side view of the tilted intersection circle (black). The red dot is the top of the circle, $(-1,-1,2)$, where $z$ is largest; the black dot at the bottom, $(1,1,-2)$, is where $z$ is smallest. Two extremes on one closed curve — that's why we expect two real roots.

1. Gradients. $\nabla f=(0,0,1)$, $\nabla g=(2x,2y,2z)$, $\nabla h=(1,1,1)$.
2. $\nabla f=\lambda \nabla g+\mu \nabla h$: $0=2\lambda x+\mu ,0=2\lambda y+\mu ,1=2\lambda z+\mu$. Why a combination? On the intersection curve, $\nabla f$ must lie in the span (plane) of the two normals — no single normal suffices.
3. Subtract first two: $2\lambda (x-y)=0\Rightarrow \lambda =0$ or $x=y$. If $\lambda =0$ then eqn 3 gives $1=\mu$ but eqns 1,2 give $\mu =0$ — contradiction. So $x=y$. Why subtract the two equations? Both equal $0$, so their difference eliminates the unknown $\mu$ in one stroke and hands us a clean relation between $x$ and $y$ — the fastest route to collapse three variables toward one.
4. Use constraints to find the points. $h$: $2x+z=0\Rightarrow z=-2x$. $g$: $2x^2+z^2=6\Rightarrow 2x^2+4x^2=6\Rightarrow 6x^2=6\Rightarrow x=\pm 1$. Why now bring in $g,h$? Step 3 fixed a direction on the curve; the two constraint equations pin the actual points where that direction meets the circle.
   • $x=1$: $y=1,z=-2$: point $(1,1,-2)$, $z=-2$ (minimum).
   • $x=-1$: $y=-1,z=2$: point $(-1,-1,2)$, $z=2$ (maximum).
5. Recover the multipliers. Take $x=-1,y=-1,z=2$. From eqn 1: $\mu =-2\lambda x=2\lambda$. Substitute into eqn 3: $1=2\lambda z+\mu =2\lambda (2)+2\lambda =6\lambda \Rightarrow \lambda =\frac{1}{6}$, then $\mu =2\lambda =\frac{1}{3}$. Why bother with $\lambda ,\mu$? They confirm $\nabla f$ genuinely sits in the span of the two normals (a valid solution exists), and $\lambda ,\mu$ are the two shadow prices — sensitivities of $z^{\ast }$ to relaxing the sphere and the plane respectively.
6. Why two roots? The tilted circle has a top and a bottom in the $z$-direction; the two signs $x=\pm 1$ are exactly those two points — one max, one min.

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Ex 7 — Two constraints, degenerate parallel normals (cell C7)

Forecast: look hard at the two planes before computing.

1. Read the constraints. $h=2x+2y+2z-5=2(x+y+z)-5$. But $g=0$ says $x+y+z=1$, so $h=2(1)-5=-3\ne 0$. Why check first? The two planes are parallel ($\nabla h=2\nabla g$) and don't coincide — so the feasible set is empty.
2. Confirm via gradients. $\nabla g=(1,1,1)$, $\nabla h=(2,2,2)=2\nabla g$. They are linearly dependent (one is a stretch of the other). Why fatal? Lagrange's two-constraint formula needs $\nabla g,\nabla h$ independent so their span is a genuine plane. If they are parallel, the "plane" collapses to a single line and the method's assumption fails (parent Mistake C, two-constraint version).
3. Conclusion. No feasible point exists ⇒ no extremum. Lagrange correctly refuses to produce a solution.

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Ex 8 — Limiting behaviour $\lambda \to 0$: constraint goes inactive (cell C8)

Forecast: the unconstrained minimum of $f$ is $(3,4)$, where $x+y=7$. What does the constraint do at $c=7$?

1. Gradients. $\nabla f=(2(x-3),2(y-4))$, $\nabla g=(1,1)$.
2. Equations. $2(x-3)=\lambda$, $2(y-4)=\lambda$. So $x-3=y-4\Rightarrow y=x+1$.
3. Constraint. $x+(x+1)=c\Rightarrow x=\frac{c-1}{2},y=\frac{c+1}{2}$.
4. Multiplier. $\lambda =2(x-3)=2\left(\frac{c-1}{2}-3\right)=c-7$. Why watch this? $\lambda$ measures how hard the constraint "pushes". As $c\to 7$, $\lambda \to 0$.
5. Interpret. At $c=7$ the constrained minimizer is $(3,4)$ — the unconstrained minimizer, which already lies on the line. The constraint costs nothing, so its shadow price is zero. This is the $\lambda =0$ limit linking to Unconstrained optimization — critical points.

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Ex 9 — Exam twist: Lagrange point that is not a maximum (cell C9)

Forecast: sketch the hyperbola. Does $x$ have a maximum on it at all?

Figure s05 (below): the black hyperbola has two branches. The red dot at $(1,0)$ is the vertex Lagrange lands on; the black annotation shows $x$ climbing without bound as you slide up the right branch — so that red dot is the branch's minimum of $x$, not a maximum.

1. Gradients. $\nabla f=(1,0)$, $\nabla g=(2x,-2y)$. Why? We still form both arrows even for a "twist" problem — the trap is in interpreting the result, not in the setup.
2. Equations. $1=2\lambda x$ and $0=-2\lambda y$. The second gives $\lambda =0$ (rejected, since it kills equation 1) or $y=0$. So $y=0$. Why reject $\lambda =0$? If $\lambda =0$, equation 1 reads $1=0$, impossible — so the only live branch is $y=0$.
3. Impose the constraint. With $y=0$: $x^2-1=0\Rightarrow x=\pm 1$. Candidates $(\pm 1,0)$, with $\lambda =\frac{1}{2x}=\pm \frac{1}{2}$. Why compute $\lambda$ here? It confirms both candidates are legitimate stationary points (finite non-zero $\lambda$); the problem is not that Lagrange failed, but what the points mean.
4. Classify honestly. Parametrize the right branch as $x=\sqrt{1+y^2}$. As $|y|\to \infty$, $x\to \infty$ — $x$ grows without bound. So on that branch $x$ has no maximum; the point $(1,0)$ is where $x$ is smallest (the vertex), i.e. a minimum, not a maximum. Why parametrize? It converts the implicit curve into an explicit function of one variable so we can watch $x$'s behaviour directly and settle max-vs-min instead of guessing.
5. Conclusion. The problem as posed has no maximum of $x$ on the hyperbola ($\sup x=+\infty$). Lagrange truthfully returned $(1,0)$ as a stationary point, but it is the branch minimum; $(-1,0)$ is the analogous vertex on the left branch. Moral: always ask whether an extremum can even exist (is the objective bounded on the feasible set?) before labelling a Lagrange point.

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Recall Self-test

On the unit circle, why does $f=x^2+2y^2$ give two different $\lambda$ values? ::: Because the max (at $(0,\pm 1)$, $\lambda =2$) and min (at $(\pm 1,0)$, $\lambda =1$) are different constrained extrema, each with its own parallel-scaling factor. What single check would have instantly killed Ex 7? ::: Test if $\nabla g$ and $\nabla h$ are linearly independent — here $\nabla h=2\nabla g$, so the feasible set can be empty and the method is inapplicable.

Connections

• Lagrange multipliers — one and two constraints — the parent theory these examples exercise.
• Gradient and directional derivative — every $\nabla f$ arrow above.
• Tangent planes and normal vectors — $\nabla g,\nabla h$ as normals (Ex 6, 7).
• Unconstrained optimization — critical points — the $\lambda \to 0$ limit (Ex 8).
• KKT conditions — what to do when constraints are inequalities / boundaries appear (Ex 9).
• Level sets and contours — tangency picture behind Ex 1, 4.
• Dual problem and shadow prices — the $\lambda$ interpretation in Ex 2, 5.