Worked examples — Isoperimetric problems — constraints (Lagrange multipliers in variational sense)

Before we start, one plain-words reminder of the pieces, because we use them on every line.

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The scenario matrix

Here is every class of situation an isoperimetric problem can put in front of you. Each row is a distinct "shape" of difficulty; the last column names the example on this page that lands on it.

#	Case class	What makes it different	Covered by
C1	$f$ has no explicit $x$ → use Beltrami	first integral shortcut	Ex 1, Ex 2
C2	$f$ does contain $x$ → Beltrami fails, use full E–L	must integrate $\frac{d}{dx}{F}_{y^{\prime }}$	Ex 3
C3	Constraint too small / degenerate ($\ell \le$ chord)	no valid curve; boundary case	Ex 4
C4	Limiting value: $\ell \to$ minimum → curve flattens to line	$\lambda \to \infty$ / arc → chord	Ex 4
C5	Sign of $\lambda$ (max vs min flips its sign)	which way the curve bends	Ex 2, Ex 5
C6	Both endpoints equal / symmetric interval	$x_0$ pinned by symmetry	Ex 2, Ex 5
C7	Real-world word problem (physics units)	translate words → $f,g$	Ex 6
C8	Exam twist: constraint is $\int y^2$, not arc-length	$g$ depends on $y$, not $y^{\prime }$	Ex 7

Now we work them.

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Ex 1 — Baseline sanity: minimise energy at fixed length (case C1)

Forecast: guess the shape and where its lowest point sits before reading on.

Step 1 — Augment. $F=f-\lambda g=(y-\lambda )\sqrt{1+y^{\prime 2}}$. Why this step? Adding $-\lambda g$ lets us search all curves freely; $\lambda$ later absorbs the length rule.

Step 2 — Beltrami, because $F$ has no explicit $x$. Why this tool and not full E–L? When $x$ never appears inside $F$, the Beltrami Identity $F-y^{\prime }{F}_{y^{\prime }}=C_1$ is a first integral — it hands us a first-order ODE for free instead of a second-order one. Choosing it saves an entire integration. Here $C_1$ is our first integration constant. $\frac{y-\lambda }{\sqrt{1+y^{\prime 2}}}=C_1.$

Step 3 — Separate and integrate. Rearrange the Beltrami relation to isolate the slope: $\sqrt{1+y^{\prime 2}}=\frac{y-\lambda }{C_1}\Rightarrow y^{\prime 2}=(\frac{y-\lambda }{C_1}{)}^2-1.$ Why introduce a new letter $t$? The right side is $(\text{something}{)}^2-1$, and the identity ${\cosh }^{2}t-1={\sinh }^{2}t$ is built to turn exactly that shape into a perfect square. So we let the substitution $\frac{y-\lambda }{C_1}=\cosh t$ — this is a change of variable that makes $\sqrt{(\cdots {)}^2-1}=\sinh t$ clean, with no square root left. Here $t$ is just a running label for position; it will end up being $t=\frac{x-x_0}{C_1}$. With that substitution $y^{\prime }=\sinh t$ and $dy=C_1\sinh tdt$, and since $y^{\prime }=dy/dx$ we get $dx=C_{1}dt$, so integrating gives $x-x_0=C_{1}t$. Undoing the substitution ($t=(x-x_0)/C_1$, $y-\lambda =C_1\cosh t$): $y(x)=\lambda +C_1\cosh \left(\frac{x-x_0}{C_1}\right).$ Which constants are these? $C_1$ came from Beltrami (Step 2); the second integration produced $x_0$, so $x_0$ plays the role of $C_2$ — the horizontal shift. Together with $\lambda$ that is the promised trio $C_1,C_2(=x_0),\lambda$.

Step 4 — Use symmetry. The endpoints are equal and the interval $[-1,1]$ is symmetric, so the lowest point of the $\cosh$ sits at the middle: $==x_0=0==$. Why this step? $\cosh$ is even about its vertex; equal boundary heights force the vertex onto the axis of symmetry. In the figure below this is the pink dashed line: the curve is a mirror image across it.

Look at the figure: the yellow curve is the $\cosh$; the two blue dots are the pinned endpoints at equal height; the pink dot is the vertex sitting exactly on the pink symmetry axis $x_0=0$ — the lowest point of the chain.

Verify: With $x_0=0$, $y(-1)=y(1)=\lambda +C_1\cosh (1/C_1)$ automatically — both endpoints match with one equation, confirming symmetry is consistent. Units: $y,x,\ell$ all lengths, dimensionally clean. ✓

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Ex 2 — Dido, concrete numbers: fixed perimeter → circular arc (cases C1, C5, C6)

Forecast: the chord is length $2$; the string is length $\pi$. Guess the shape and its area.

Step 1 — Augment. $f=y,g=\sqrt{1+y^{\prime 2}}$, so $F=y-\lambda \sqrt{1+y^{\prime 2}}$.

Step 2 — Beltrami (no explicit $x$): $F-y^{\prime }{F}_{y^{\prime }}=C$ gives $y-\frac{\lambda }{\sqrt{1+y^{\prime 2}}}=C$.

Step 3 — Rearrange to a circle, one algebra step at a time. Start from Beltrami and isolate the square-root term: $\frac{\lambda }{\sqrt{1+y^{\prime 2}}}=y-C.$ Let $u=y-C$ (a shifted height). Square and invert: $1+y^{\prime 2}=\frac{{\lambda }^2}{u^2}\Rightarrow y^{\prime 2}=\frac{{\lambda }^2-u^2}{u^2}\Rightarrow y^{\prime }=\frac{du}{dx}=\frac{\sqrt{{\lambda }^2-u^2}}{u}.$ Why separate now? We have $du/dx$ as a function of $u$ alone, so we can split the variables: $\frac{udu}{\sqrt{{\lambda }^2-u^2}}=dx.$ The left side integrates to $-\sqrt{{\lambda }^2-u^2}$ (its derivative is exactly $u/\sqrt{{\lambda }^2-u^2}$), so $-\sqrt{{\lambda }^2-u^2}=x-x_0\Rightarrow (x-x_0{)}^2+u^2={\lambda }^2.$ Putting back $u=y-C$: $(x-x_0{)}^2+(y-C{)}^2={\lambda }^2.$ This is a circle of radius $\lambda$ centred at $(x_0,C)$ — that is where the circle comes from. Symmetry ($C6$) puts the centre on the $y$-axis, so $x_0=0$.

Step 4 — Fit the endpoints. $y(\pm 1)=0$ means $1+C^2={\lambda }^2$. Why this step? Both boundary points lie on the circle; that is one equation linking $C$ and the radius $\lambda$.

Step 5 — Fit the length. A circle of radius $\lambda$ subtending half-angle $\theta$ (where $\sin \theta =1/\lambda$) over the chord has arc length $2\lambda \theta =\pi$. Try $\lambda =1$: then $\sin \theta =1$, $\theta =\pi /2$, arc $=2\cdot 1\cdot \pi /2=\pi$. ✓ So $\lambda =1$, $C=0$: a semicircle of radius $1$. Why $\lambda >0$ (case C5)? We are maximising area, so the curve bows outward (upward) — a positive radius. Minimising would flip the sign and bow it inward.

Step 6 — Area. Half a unit disc: $J=\frac{1}{2}\pi (1{)}^2=\frac{\pi }{2}$.

Look at the figure: the yellow arc is the semicircle of length $\pi$; the blue segment is the chord of length $2$; the shaded pink region is the enclosed area $\pi /2$; the pink dot at the origin is the centre, and the radius arrow shows $\lambda =1$ — the multiplier literally measured as a radius.

Verify: arc length of a unit semicircle $=\pi (1)=\pi =\ell$ ✓. Area $=\pi /2\approx 1.5708$ ✓. The multiplier $\lambda =1$ is the radius, matching the parent's punchline. ✓

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Ex 3 — When $x$ appears: Beltrami fails, use full E–L (case C2)

Forecast: will the answer be a $\cosh$, a line, or a polynomial?

Step 1 — Augment. $g=y$, so $F=y^{\prime 2}+xy-\lambda y=y^{\prime 2}+(x-\lambda )y$.

Step 2 — Notice Beltrami is illegal here. Why? $F$ contains $x$ explicitly (the $xy$ term). Beltrami requires $\partial F/\partial x=0$. Using it would be a mistake; we must use the full Euler–Lagrange Equation.

Step 3 — Apply E–L. $F_y=x-\lambda$, $F_{y^{\prime }}=2y^{\prime }$, so $F_y-\frac{d}{dx}{F}_{y^{\prime }}=(x-\lambda )-2y^{\prime \prime }=0\Rightarrow y^{\prime \prime }=\frac{1}{2}(x-\lambda ).$

Step 4 — Integrate twice. $y=\frac{x^3}{12}-\frac{\lambda x^2}{4}+C_{1}x+C_2.$

Step 5 — Boundary conditions. $y(0)=0\Rightarrow C_2=0$. $y(1)=0\Rightarrow \frac{1}{12}-\frac{\lambda }{4}+C_1=0\Rightarrow C_1=\frac{\lambda }{4}-\frac{1}{12}$.

Step 6 — Constraint fixes $\lambda$. The strategy: we now have $y$ written with only $\lambda$ unknown (since $C_1,C_2$ are already expressed through $\lambda$), so the single remaining equation — the constraint — must determine $\lambda$. Compute ${\int }_0^{1}ydx=\ell$: ${\int }_0^1(\frac{x^3}{12}-\frac{\lambda x^2}{4}+C_{1}x)dx=\frac{1}{48}-\frac{\lambda }{12}+\frac{C_1}{2}=\ell .$ Why group this way? We substitute the Step-5 value $C_1=\frac{\lambda }{4}-\frac{1}{12}$ so that every term is either a pure number or a multiple of $\lambda$ — then collecting the $\lambda$-terms on one side isolates it in one move: $\underset{\text{pure numbers}}{\underbrace{\frac{1}{48}-\frac{1}{24}}}+\underset{\text{coeff. of }\lambda }{\underbrace{(-\frac{1}{12}+\frac{1}{8})}}\lambda =-\frac{1}{48}+\frac{\lambda }{24}=\ell .$ Hence $\frac{\lambda }{24}=\ell +\frac{1}{48}$, giving $\lambda =24\ell +\frac{1}{2}$.

Verify: pick $\ell =0$. Then $\lambda =\frac{1}{2}$, $C_1=\frac{1}{8}-\frac{1}{12}=\frac{1}{24}$, and $y=\frac{x^3}{12}-\frac{x^2}{8}+\frac{x}{24}$. Check $y(1)=\frac{1}{12}-\frac{1}{8}+\frac{1}{24}=0$ ✓ and ${\int }_0^{1}ydx=0$ ✓ (done in VERIFY). ✓

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Ex 4 — Degenerate & limiting budget (cases C3, C4)

Forecast: guess the shape as $\ell \downarrow 2$, and whether $\ell =1.5$ is even solvable.

Step 1 — Lower bound of $\ell$ is the straight chord. Why? The shortest curve joining $(-1,0)$ to $(1,0)$ is the straight segment, length $2$. No admissible curve can be shorter.

Step 2 — Case C3, $\ell =1.5$. Since $1.5<2$, no curve exists. The problem is infeasible — the constraint surface $K[y]=\ell$ is empty for these endpoints. Why this matters: always check $\ell \ge$ chord before hunting for a multiplier. A "solution" you compute here would be spurious.

Step 3 — Limiting case C4, $\ell \to 2^{+}$. The arc must become nearly straight, so the radius $\lambda \to \infty$ and the arc collapses onto the chord. Area $J\to 0$. Why? A huge-radius circular arc over a fixed chord is almost flat; a flat curve encloses almost no area.

Step 4 — Quantify the small-bulge area. For a circular segment over a chord of length $L=2$ with maximum height (sagitta) $h$, the enclosed area is exactly $A={\int }_{-1}^1\left(\text{arc height}\right)dx.$ For a shallow parabola-like bulge $y\approx h(1-x^2)$ (the leading shape of a low arc over $[-1,1]$), this integral is $A=h{\int }_{-1}^1(1-x^2)dx=h\cdot \frac{4}{3}=\frac{2}{3}Lh$ with $L=2$. So $A\approx \frac{2}{3}Lh\to 0$ as $h\to 0$. Why this approximation? Near the flat limit the exact circular arc is indistinguishable from this parabola to leading order, and the parabola integrates in one line — enough to see $A\to 0$ linearly in the sagitta.

Step 5 — The two ends of the range. At $\ell =2$ exactly, $\theta \to 0$, $\lambda \to \infty$, $J\to 0$. At $\ell =\pi$ (Ex 2) we found $J=\pi /2$, the semicircle. So area rises monotonically from $0$ as $\ell$ climbs from $2$ toward $\pi$.

Verify: at $\ell =\pi$, $J=\pi /2\approx 1.571$; the semicircle uses the largest $\ell$ compatible with a single-valued $y(x)$ arc (half-angle $\pi /2$). Any $\ell$ between $2$ and $\pi$ gives an area strictly between $0$ and $\pi /2$. Feasibility test $\ell \ge 2$ confirmed. ✓

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Ex 5 — Sign of $\lambda$: maximise vs minimise (cases C5, C6)

Forecast: which way does the arc bow now?

Step 1 — Same augmented $F$, opposite optimisation. The E–L / Beltrami algebra is identical: we still get a circle $(x{)}^2+(y-C{)}^2={\lambda }^2$.

Step 2 — The sign flips. Why? Minimising signed area wants $y$ as negative as possible, so the arc bows downward. The centre is now above the chord; the radius $\lambda$ effectively carries the opposite sign relative to the maximisation case.

Step 3 — Solution. The minimiser is the lower semicircle: $y(x)=-\sqrt{1-x^2}$, area $J=-\pi /2$.

Verify: arc length $=\pi$ ✓ (same circle, mirrored). Signed area $=-\pi /2\approx -1.571$ ✓. Sign of $\lambda$ is exactly the sign of the bulge — matching the parent's "$\lambda$ = radius" reading (up to orientation). ✓

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Ex 6 — Real-world word problem with units (case C7)

Forecast: guess whether $10\text{ m}$ of rope over a $6\text{ m}$ gap sags by more or less than $1\text{ m}$.

Step 1 — Translate words → $f,g$, and recall the shape. A hanging rope minimises potential energy $J=\int y\sqrt{1+y^{\prime 2}}dx$ subject to length $K=\int \sqrt{1+y^{\prime 2}}dx=\ell$. This is exactly the setup of Ex 1, whose solution we derived there: $y(x)=\lambda +C\cosh \left(\frac{x-x_0}{C}\right),\text{(from Ex 1, Step 3)}$ with $C$ the Beltrami constant. We reuse that result here rather than re-deriving it.

Step 2 — Symmetry pins $x_0$. Pins at $x=\pm 3$ are symmetric, so $x_0=0$ (case C6, as in Ex 1).

Step 3 — Length formula for a catenary. Differentiate the recalled shape: $y^{\prime }=\sinh \left(\frac{x}{C}\right)$ (since $x_0=0$). Then the arc-length integrand is $\sqrt{1+y^{\prime 2}}=\sqrt{1+{\sinh }^2\frac{x}{C}}=\cosh \frac{x}{C},$ using the identity $1+{\sinh }^2={\cosh }^2$. So $\ell ={\int }_{-3}^3\cosh \frac{x}{C}dx=2C\sinh \frac{3}{C}.$ Why this step? Writing $\sqrt{1+y^{\prime 2}}$ back in terms of the known derivative $y^{\prime }=\sinh (x/C)$ collapses the messy root into a plain $\cosh$, whose integral is $C\sinh$ — turning arc length into a clean closed form.

Step 4 — Plug $C=2$. $\ell =2\cdot 2\cdot \sinh (1.5)=4\sinh (1.5)\approx 4(2.1293)=8.517\text{ m}$. This is not $10\text{ m}$ — so $C=2$ is inconsistent with a $10\text{ m}$ rope. The correct $C$ solves $2C\sinh (3/C)=10$, giving $C\approx 1.62\text{ m}$. Why check this? Never trust a handed-down constant; the length constraint is what fixes $C$.

Step 5 — Sag. With the true $C\approx 1.62$: sag $=y(3)-y(0)=C(\cosh \frac{3}{C}-1)=1.62(\cosh (1.852)-1)\approx 1.62(3.199-1)\approx 3.56\text{ m}$.

Verify: $2(1.62)\sinh (3/1.62)=3.24\sinh (1.852)\approx 3.24(3.086)\approx 10.0\text{ m}$ ✓ (length recovered). Units all metres ✓. The lesson: the false $C=2$ gave the wrong length — the constraint corrects it. ✓

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Ex 7 — Exam twist: constraint depends on $y$, not $y^{\prime }$ (case C8)

Forecast: what familiar functions have ${\int }_0^{\pi }(\cdot {)}^2$ easy to normalise?

Step 1 — Augment. $f=y^{\prime 2}$, $g=y^2$, so $F=y^{\prime 2}-\lambda y^2$.

Step 2 — Full E–L ($x$ absent, but let's be safe): $F_y=-2\lambda y$, $F_{y^{\prime }}=2y^{\prime }$, so $-2\lambda y-\frac{d}{dx}(2y^{\prime })=0\Rightarrow y^{\prime \prime }+\lambda y=0.$ Why this step? The multiplier turned a constrained variational problem into the eigenvalue equation $y^{\prime \prime }=-\lambda y$ — the constraint's "price" $\lambda$ is the eigenvalue.

Step 3 — Split on the sign of $\lambda$ (the edge case). The character of the solution depends entirely on the sign of $\lambda$:

• If $\lambda <0$, write $\lambda =-k^2$: then $y^{\prime \prime }=k^{2}y$ has solutions $y=A{e}^{kx}+B{e}^{-kx}$ (hyperbolic). The only way such a combination vanishes at both $x=0$ and $x=\pi$ is $A=B=0$, i.e. $y\equiv 0$ — which breaks the constraint $\int y^2=1$. So $\lambda <0$ is impossible here.
• If $\lambda =0$, then $y^{\prime \prime }=0$, $y=ax+b$; both endpoints zero forces $y\equiv 0$ again — impossible.
• If $\lambda >0$, write $\lambda =k^2$: then $y^{\prime \prime }+k^{2}y=0$ gives oscillating solutions $y=A\sin (kx)+B\cos (kx)$, which can vanish at both ends. This is the only viable sign.

Why $\lambda$ must be positive: only oscillatory (trigonometric) solutions can hit zero at two separate points while being non-trivial; that requires $\lambda >0$.

Step 4 — Boundary conditions. With $y(0)=0$ we need $B=0$; with $y(\pi )=0$ we need $\sin (k\pi )=0$, so $k=n$ integer. Thus $y=A\sin (nx)$ with $\lambda =n^2$, $n=1,2,3,\dots$

Step 5 — Normalise using the constraint. ${\int }_0^{\pi }{A}^2{\sin }^2(nx)dx=A^2\cdot \frac{\pi }{2}=1\Rightarrow A=\sqrt{2/\pi }$.

Step 6 — Pick the minimiser. The smallest $J={\int }_0^{\pi }{y}^{\prime 2}$ comes from the smallest $\lambda =1$: $y=\sqrt{2/\pi }\sin x$, and then $J=\lambda \int y^2=\lambda =1$. Why $\lambda$ = value of $J$? Multiply $y^{\prime \prime }+\lambda y=0$ by $y$ and integrate by parts: $\int y^{\prime 2}=\lambda \int y^2=\lambda$. The multiplier equals the optimum — the shadow price made literal.

Verify: $A^2\pi /2=(2/\pi )(\pi /2)=1$ ✓. Minimum $J=1$ at $n=1$ ✓. Higher modes $n=2,3$ give larger $J=4,9$ — stationary but not minimal ✓ (checked in VERIFY). ✓

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Recall Quick self-test

When is Beltrami legal? Only when the augmented F = f - lambda g has no explicit x; otherwise use the full Euler–Lagrange equation. How many unknowns and how many conditions in a standard isoperimetric solve? Three unknowns C1, C2, lambda; three conditions — two boundary values plus the integral constraint. What does lambda mean numerically? It is the marginal value dJ*/d-ell: the amount the optimum J improves per unit of extra budget (the radius in Dido, the eigenvalue in Ex 7). What happens if ell is below the chord length? No admissible curve exists — the problem is infeasible (case C3).

Back to the parent topic. Related: Catenary Curve, Isoperimetric Inequality, Brachistochrone Problem, Lagrange Multipliers (finite-dimensional), Calculus of Variations — Fundamental Lemma.