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Assume $c(y)>0$ is differentiable along the ray, with sufficient regularity for the following variation, and fix the endpoints. Fermat principle makes the travel time stationary:For the variational integrand $f=\sqrt{1+y^{\prime 2}}/c$, the Euler-Lagrange equation isExpanding the derivative and multiplying by $c^2(1+y^{\prime 2}{)}^{3/2}$ givesand thereforeEquivalently, the Beltrami identity gives $1/[c(y)\sqrt{1+y^{\prime 2}}]=\text{constant}$. Mere continuity of $c$, as stated in the wording, does not by itself justify the appearance of $c^{\prime }$; the displayed classical ray equation requires this additional differentiability.

Introduce a constant Lagrange multiplier $\lambda$ for the fixed arc length. Up to an irrelevant constant term, the augmented functional isIt has no explicit $x$ dependence, so the Beltrami identity givesFor the hanging minimum choose $k>0$. Writing $y_0=-\lambda$, this becomes $1+y^{\prime 2}=(y-y_0{)}^2/k^2$. Its nonconstant smooth solution isOne may verify it without dividing by $y^{\prime }$ at the lowest point: the full Euler-Lagrange equation gives $(y+\lambda )y^{\prime \prime }=1+y^{\prime 2}$, which the expression satisfies everywhere. The branch with negative $k$ is the inverted stationary chain rather than the hanging minimum. The catenary minimizes the integral of height along a fixed-length curve, so the positive branch is also the global minimum, not merely a stationary candidate.

Define $h=f-y^{\prime }{f}_{y^{\prime }}$. For a density without explicit $x$ dependence, the chain rule givesThus the Euler-Lagrange equation implies the Beltrami identityWhere $y^{\prime }\ne 0$ this first integral is equivalent to the original equation. At a turning point it extends by continuity. On an entire constant segment, however, its derivative alone can vanish without imposing the full Euler-Lagrange equation, so such candidate paths must still be checked in the original equation.

The monotonic horizontal coordinate allows the path to be written as a graph. Its arc length element is $ds=\sqrt{1+y^{\prime 2}}dx$, so the variational density is $f(y,y^{\prime })=g(y)\sqrt{1+y^{\prime 2}}$. It has no explicit horizontal-coordinate dependence. The Beltrami identity becomesConsequently every smooth minimizing path satisfiesThis is a first integral of the weighted arc-length problem; fixed endpoint values remove the endpoint variation terms.

The original PDF places both endpoints at height $5/4$; the vertically stacked fraction can be misread in extracted text. The coefficient in the index is $4$, multiplying $\sqrt{y}$.

For the optical-length integrand $F=4\sqrt{y}\sqrt{1+y^{\prime 2}}$, the Beltrami identity applies because there is no explicit $x$ dependence. It giveswhere $h>0$ is constant. Therefore $y^{\prime 2}=y/h-1$. The full Euler-Lagrange equation is $2y{y}^{\prime \prime }=1+y^{\prime 2}$, and combining these gives $y^{\prime \prime }=1/(2h)$. This also excludes a constant curve, which would satisfy the squared first integral but not the full variational equation. Integration yieldsEqual endpoint heights at $x=-1,1$ require $x_0=0$. The height condition then becomes $h+1/(4h)=5/4$, or $4h^2-5h+1=0$. Its two positive roots are $h=1$ and $h=1/4$. Hence the two possible stationary rays areBoth stay in $y>0$, have the required endpoints and satisfy the variational equation. These are the two optical rays in a square-root refractive-index profile. Fermat stationarity supplies two rays; it does not assert that both globally minimize travel time.

The endpoint constraints are $y(-a)=y(a)=0$. The constants in the three costs contribute only $2aK$, $2a{K}^{\prime }$ and $2a{K}^{\prime \prime }$ respectively, so they do not affect the minimizing shapes. For companies B and C, the fixed arc length isThe original PDF has $(y^{\prime }{)}^2$ as company A's first variable term. The local TeX transcription has lost that derivative dot and incorrectly repeats $y^2$; the calculation below follows the PDF.

Company A. The relevant functional is $J_A=\int (y^{\prime 2}+y^2-2xy)dx$. The Euler-Lagrange equation givesIts general solution is $y=x+c_1\cosh x+c_2\sinh x$. Adding and subtracting the endpoint equations gives $c_1=0$ and $c_2=-a/\sinh a$. ThusThis curve is odd, negative on the left half and positive on the right half; company A has no lower-half-plane restriction. It is the unique global minimum. Indeed, for any endpoint-zero variation $\eta$, integration by parts and the differential equation cancel all linear terms, leavingunless $\eta =0$.

Company B. Minimize $\int ydx$ under the length constraint. Introduce a constant Lagrange multiplier $\lambda$ and apply the Euler-Lagrange equation to $y+\lambda \sqrt{1+y^{\prime 2}}$:The curvature is constant, so the required lower arc is a circular arc. Writing its radius as $R\ge a$ and imposing the endpoints givesSince $\sqrt{1+y_B^{\prime 2}}=R/\sqrt{R^2-x^2}$, its length determines the radius:Equivalently, introduce $0<\theta \le \pi /2$ with $a=R\sin \theta$. Then $L/(2a)=\theta /\sin \theta$, which increases strictly from $1$ to $\pi /2$. The stipulated range of $L$ therefore selects one minor arc. At $L=\pi a$ it becomes the lower semicircle, with vertical endpoint tangents; it is still a single-valued finite-length graph.

Global minimality follows from the convexity of the augmented slope cost with $\lambda =R$. For a competitor $y=y_B+\eta$,Consequently the augmented cost difference is at least $\int (\eta +x{\eta }^{\prime })dx=[x\eta {]}_{-a}^a=0$. Competitors have the same length, so the multiplier contribution cancels and the original cost is minimized. This proves that a circular arc maximizes area below a fixed chord: minimizing negative signed height maximizes the area between the fence and the chord.

Company C. The variable cost is $J_C=\int yds=\int y\sqrt{1+y^{\prime 2}}dx$. The augmented integrand is $(y+\lambda )\sqrt{1+y^{\prime 2}}$. Since it has no explicit $x$ dependence, the Beltrami identity givesThe hanging branch with $b>0$ integrates to $y+\lambda =b\cosh [(x-x_0)/b]$. Equal endpoint heights force $x_0=0$, and the endpoint value fixes $\lambda =b\cosh (a/b)$. ThusIts length isWriting $q=a/b$, the equation $L/(2a)=\sinh q/q$ has one positive solution for every $L>2a$: its derivative has numerator $q\cosh q-\sinh q>0$. The curve is a catenary, is even, and lies strictly below the chord except at the endpoints.

A direct argument proves this is a global minimum, rather than merely an extremum of an augmented functional. Parametrize any competitor by arc length $s\in [-L/2,L/2]$, from its left endpoint to its right endpoint. Then $x^{\prime 2}+y^{\prime 2}=1$, $\int x^{\prime }ds=2a$, and integration by parts givesThe Cauchy-Schwarz inequality gives the last step. Equality requires $x^{\prime }=b/\sqrt{b^2+s^2}$ and $y^{\prime }=s/\sqrt{b^2+s^2}$, whose integrals are precisely the displayed catenary with the stated length. Thus it attains a lower bound valid for every competitor. This proves that a catenary minimizes the integral of height along a fixed-length curve.

The sketch uses $a=1$ and $L=2.7$ for the two fixed-length cases. The vertical scale in the company A panel is enlarged to show its small odd deflection. No comparison between the three companies' absolute prices is required or possible from the shapes alone, since their additive cost constants are independent.

If $f$ has no explicit $x$ dependence, the chain rule givesConsequentlyon an Euler-Lagrange equation solution. Thus the Beltrami identity supplies the first integralThe opposite sign is an equally valid convention. The conserved value can be zero for particular integrands or paths; absence of explicit $x$ alone does not guarantee a nonzero numerical value. The useful statement is that this expression is constant along an extremal.

The generalized momentum of this Lagrangian isThe Euler-Lagrange equation therefore givesBecause $\mathcal{L}$ has no explicit time dependence, the energy form of the Beltrami identity gives a conserved quantity $E=\dot{x}\partial \mathcal{L}/\partial \dot{x}-\mathcal{L}$. HereFor a direct check of conservation of energy, differentiate:This is the rest-plus-kinetic energy of a relativistic particle, in units with light speed one, plus its potential energy.

With fixed endpoint values, a smooth extremal satisfies the Euler-Lagrange equation $F_y-(d/dx)F_{y^{\prime }}=0$. If $F$ has no explicit $x$ dependence, differentiation givesThus $F-y^{\prime }{F}_{y^{\prime }}$ is constant, the Beltrami identity.

For this integrand, $F_y=2y$ and $F_{y^{\prime }}=1+2y^{\prime }$, so the equation is $y^{\prime \prime }=y$. ConsequentlyThe two constants are fixed uniquely by the prescribed values at distinct endpoints. Equivalently, writing these values as $y_1,y_2$,Its first integral is $y^2-y^{\prime 2}$, which is constant. Using only that first integral without the Euler-Lagrange equation could admit spurious constant functions, so the differential equation was used directly.

For any nonzero endpoint-vanishing perturbation $\eta$, the linear terms in $J[y+\eta ]-J[y]$ vanish by the Euler-Lagrange equation and integration by parts; also $\int {\eta }^{\prime }=0$. The remainder isHence the extremal is the unique strict global minimum, and cannot be a maximum.

For a variation $y+\epsilon \eta$ with $\eta (0)=\eta (1)=0$, stationarity givesThe boundary term vanishes. The fundamental lemma of the calculus of variations therefore gives the Euler-Lagrange equation $d{F}_{y^{\prime }}/dx-F_y=0$.

Differentiating the proposed first integral givesThus it is constant along an extremal when $F_x=0$, proving the Beltrami identity.

For the positive branch relevant to the endpoint data, $F=\sqrt{1+y^{\prime 2}}/y$. The identity becomesThe Euler-Lagrange equation itself gives $y{y}^{\prime \prime }=-(1+y^{\prime 2})$, so $(y^2{)}^{\prime \prime }=-2$. Integrating, or solving the first integral, yields a circle $y^2+(x-c{)}^2=R^2$. The equal endpoint values force $c=1/2$ and $R^2=5/4$. Hence the unique positive smooth extremal isIt stays positive on the entire interval and satisfies the second-order equation, including its turning point. Merely differentiating the first integral after dividing by $y^{\prime }$ would need extra care there. Geometrically this is an arc of a hyperbolic geodesic in the upper half-plane model.

Discard the constant factor $2\pi$ when forming the Euler-Lagrange equation. For $F=y\sqrt{1+y^{\prime 2}}$ it becomesThe Beltrami identity, since $F$ has no explicit $x$ dependence, gives $F-y^{\prime }{F}_{y^{\prime }}=y/\sqrt{1+y^{\prime 2}}=C>0$. Combining with the differential equation yields $y^{\prime \prime }=y/C^2$. Its exponential solution has coefficients whose product is $C^2/4$, by $y^{\prime 2}=y^2/C^2-1$. Positivity therefore puts it in the formIts constants must satisfy $y_1=C\cosh ((x_1-K)/C)$ and $y_2=C\cosh ((x_2-K)/C)$. Depending on the endpoint data, these equations may have no solution or more than one. A constant function can satisfy a formal first-integral equation but fails the full Euler-Lagrange equation; it is not an extra extremal.

For fixed endpoints and a regular integral constraint, Lagrange multipliers replace the constrained stationarity problem by stationarity of $I-\lambda J$, with constant $\lambda$. Writing $H=F-\lambda G$, the Euler-Lagrange equation is $H_y-(d/dx)H_{y^{\prime }}=0$. Regularity here means the first variation of $J$ is not identically zero; abnormal constraint extrema otherwise need separate consideration.

If $H$ has no explicit $x$ dependence, thenThis is the Beltrami identity, giving the required constant of integration.

For the chain, $ds=\sqrt{1+y^{\prime 2}}dx$, so its fixed arc length is exactly $J[y]=L$, and its gravitational energy is $\rho g{\int }_{-a}^{a}y\sqrt{1+y^{\prime 2}}dx$. The positive constant $\rho g$ can be suppressed in minimization. UseThe equation, with a consistent sign convention for the multiplier, isThe first integral becomes $(y-\lambda )/\sqrt{1+y^{\prime 2}}=c$. It implies $y^{\prime \prime }=(1+y^{\prime 2})/(y-\lambda )$; solving givesThe equal endpoint heights force $x_0=0$, by evenness and strict increase of $\cosh$ on the positive half-line. Thus the extrema have the asserted catenary form. The length and height conditions giveFor the hanging minimum choose $c>0$, the upward-convex branch lying below its endpoints. For $L>2a$ this positive $c$ is unique: writing $t=a/c$, the ratio $L/(2a)=\sinh (t)/t$ increases continuously from one to infinity. The negative-$c$ branch describes an upward arch, not the hanging minimum. If $L=2a$, only the horizontal straight chain is admissible and is the limiting degenerate case; $L<2a$ is impossible.

For a regular isoperimetric constraint, introduce a constant Lagrange multiplier $\lambda$ and set $L=F-\lambda G$. With fixed endpoint values, the Euler-Lagrange equation isIf $F$ and $G$ have no explicit $x$ dependence, the Beltrami identity gives the first integralThe regularity assumption excludes the exceptional case in which the constraint itself has zero first variation for every admissible perturbation.

Parametrize the surface of revolution by $r(z,\theta )=(f(z)\cos \theta ,f(z)\sin \theta ,z)$. The magnitude of $r_z\times r_{\theta }$ is $f\sqrt{1+f^{\prime 2}}$. Integrating the resulting surface area element over $\theta$, and integrating the circular cross-sectional areas for the volume, givesHere $f\ge 0$ because it is a radius. For a regular connected closed surface, take $f>0$ in the interior; the closing endpoints can have unbounded $f^{\prime }$ even when the surface itself is smooth.

After dividing $A-\lambda V$ by $\pi$, its integrand is $L=2f\sqrt{1+f^{\prime 2}}-\lambda f^2$. The Euler-Lagrange equation and Beltrami identity becomeAs $f$ tends to zero at an endpoint, $0\le 2f/\sqrt{1+f^{\prime 2}}\le 2f$ also tends to zero, so $C=0$. On the positive interior this implies $\lambda >0$ and, writing $R=2/\lambda$,The Euler-Lagrange equation also yields $f^{\prime \prime }=-R^2/f^3$. Thereforeincluding at the maximum where $f^{\prime }=0$. Integrating and using the first integral givesThus the regular extremal is a sphere. The two zeros fix its center and radius:Equivalently, the prescribed volume fixes $R=(3V/(4\pi ){)}^{1/3}$. If $a,b$ are fixed in advance, their separation and the prescribed volume must be compatible with these equations; there is no positive regular stationary profile of the stated type for an incompatible volume.

For minimality, the three-dimensional isoperimetric inequality states that every sufficiently regular closed surface enclosing volume $V$ haswith equality precisely for a sphere. Our extremal has $A=4\pi R^2$ and $V=4\pi R^3/3$, so it attains equality and is a global minimum of area at the fixed volume. This also makes clear the connected-surface assumption behind uniqueness: allowing several disconnected components introduces stationary collections of spheres, rather than the single positive generating profile used here.

The autonomous Lagrangian has the Beltrami identityHere $L$ is homogeneous of degree two in the velocities, so $\dot{u}{L}_{\dot{u}}+\dot{v}{L}_{\dot{v}}=2L$. Thus $L$ itself is constant, say $C\ge 0$.

For $C>0$, put $\ell =\sqrt{L}$. Its Euler-Lagrange expression along the curve isThe first term vanishes by stationarity for $L$, and the second by constancy of $L$. Thus the same path is stationary for its length functional $\int \sqrt{L}dt$.

On a regular segment described as a graph $v=v(u)$, reparametrize by $u$. Since $\dot{v}=v_u\dot{u}$,Choose the orientation with increasing $u$; reversing it does not change stationarity. Invariance of length under regular reparametrization therefore givesThis is the principle of length extremals from a constant quadratic energy. The stated nonnegativity also permits the degenerate case $C=0$. Then the length is zero, its absolute minimum since the integrand is nonnegative, so it is an extremal in the minimizing sense. The differentiated square-root Euler-Lagrange equation above is only used where $L>0$; it would not be legitimate to divide by $\sqrt{L}$ on a null curve. For a genuine surface metric and a regular nonconstant path, $C>0$ automatically.

With fixed endpoint values, the Euler-Lagrange equation is $F_y-\frac{d}{dx}{F}_{y^{\prime }}=0$. If $F$ has no explicit $x$ dependence, differentiation givesThus the Beltrami identity is $F-y^{\prime }{F}_{y^{\prime }}=\text{constant}$.

For this integrand, $F_y=2y$ and $F_{y^{\prime }}=1+2y^{\prime }$, so $y^{\prime \prime }=y$. The extremals arewhere the two fixed endpoint values determine $A,B$. The Beltrami constant is $y^2-y^{\prime 2}$, consistent with this solution.

For a nonzero smooth perturbation $\eta$ vanishing at the endpoints, expand the quadratic functional. The first variation vanishes at an extremal, whileThe linear term $\int y^{\prime }$ depends only on the fixed endpoints. Consequently the extremal is a strict minimum and cannot be a maximum.

Independent endpoint-fixed variations of the two coordinates give the Euler-Lagrange equationsDifferentiating the sphere parametrization gives the Riemannian metric $d{s}^2=a^2(d{\theta }^2+{\sin }^2\theta d{\varphi }^2)$. Its energy functional is the stated $I$. Its stationary paths are geodesics with affine, constant-speed parametrization; conversely a geodesic can be so parametrized. This explains the use of energy rather than arc length.

For $F=a^2({\dot{\theta }}^2+{\sin }^2\theta {\dot{\varphi }}^2)/2$, the equations areThe second gives the conserved generalized momentum ${\sin }^2\theta \dot{\varphi }=h$. Since $F$ has no explicit time dependence, its Beltrami identity gives constant $\dot{\theta }{F}_{\dot{\theta }}+\dot{\varphi }{F}_{\dot{\varphi }}-F=F$. Dividing by $a^2/2$ gives the second first integral,Thus $E$ is the squared speed divided by $a^2$, while $h$ records the rotational symmetry about the polar axis. The coordinate equations apply away from the polar coordinate singularities.

For $F(y,y^{\prime })=y\sqrt{1+y^{\prime 2}}$, the Euler-Lagrange equation isSince $F$ has no explicit $x$ dependence, the Beltrami identity givesSolving this first integral, or substituting into the second-order equation, yieldsThe constants satisfy $y_1=C\cosh ((x_1-K)/C)$ and $y_2=C\cosh ((x_2-K)/C)$. There need not be a solution for arbitrary endpoint data, and when there is one these equations can have more than one pair of constants. These are the possible stationary catenoids, so the axisymmetric second variation of a catenoid still matters when deciding whether they minimize surface area.

For a normal constrained extremal, introduce a constant Lagrange multiplier and use $L=F-\lambda G$. The Euler-Lagrange equation isWhen both integrands have no explicit $x$ dependence, the Beltrami identity gives $L-y^{\prime }{L}_{y^{\prime }}=\text{constant}$, as follows by differentiating this expression and using the Euler-Lagrange equation.

Parametrize the surface by $(f(z)\cos \theta ,f(z)\sin \theta ,z)$. The cross product of its two coordinate tangents has magnitude $f\sqrt{1+(f^{\prime }{)}^2}$, so integration in $\theta$ gives $A=2\pi {\int }_a^{b}f\sqrt{1+(f^{\prime }{)}^2}dz$. Integrating the cross-sectional disk areas gives $V=\pi {\int }_a^b{f}^{2}dz$.

Dividing the augmented functional $A-\lambda V$ by $\pi$, useThe Euler-Lagrange equation isThe Beltrami identity becomes $2f/\sqrt{1+(f^{\prime }{)}^2}-\lambda f^2=C$. At an endpoint $f\to 0$, the absolute value of the first term is at most $2f$, so $C=0$, even if the profile has an infinite endpoint slope. For a nondegenerate regular closed surface take $f>0$ on $(a,b)$. Then $\lambda >0$, and, with $R=2/\lambda$,Combining this identity with the Euler-Lagrange equation yields $f^{\prime \prime }=-R^2/f^3$. Consequently $(f^2{)}^{\prime \prime }=2((f^{\prime }{)}^2+f{f}^{\prime \prime })=-2$. The two zero endpoint values uniquely determineThe positive radius profile is therefore a semicircle and generates a sphere, centred at height $(a+b)/2$, with radius $R=(b-a)/2$. In particularThe graph is smooth in the interior; its vertical slopes at the two poles are compatible with a smooth spherical surface. Its volume is $4\pi R^3/3$, the volume associated with this stationary candidate.

For minimality, the three-dimensional isoperimetric inequality says that every closed surface enclosing volume $V$ has $A^3\ge 36\pi V^2$, with equality for a sphere. The displayed extremal attains equality, proving that it minimizes area at its fixed enclosed volume. The uniqueness above concerns the regular connected positive profile with the two stated endpoint heights.

Use the chain rule along a sufficiently smooth extremal of the functional. For $E=y^{\prime }{F}_{y^{\prime }}-F$,The bracket vanishes by the Euler-Lagrange equation. When $F$ has no explicit $x$ dependence, $F_x=0$, giving the Beltrami identity:

For the specified functional,Since $y>0$, the Beltrami identity gives $y\sqrt{1+y^{\prime 2}}=R$ for some $R>0$. This suggests a circular arc, but it is worth deriving its equation without dividing by $y^{\prime }$: such division would lose the point where the arc is horizontal.

Substitute the displayed partial derivative and $F_y=-\sqrt{1+y^{\prime 2}}/y^2$ in the full Euler-Lagrange equation. Combining terms over the common denominator givesThus $(y{y}^{\prime }{)}^{\prime }=-1$, so $y{y}^{\prime }=h-x$ and a second integration givesThis establishes the semicircular extremals of the hyperbolic length functional and excludes constant $y$, which would satisfy the first-order Beltrami identity for a suitable $R$ but not the full Euler-Lagrange equation.

The two prescribed endpoints giveSubtracting yields $4h=4(d^2-c^2)$. ThereforeIn particular the centre is $\boxed{(d^2-c^2,0)}$. Admissibility requires $c,d>0$, as the prescribed heights must lie in $y>0$. Under that condition the radicand is positive throughout $[-1,1]$: it is a concave quadratic polynomial whose two endpoint values are $4c^2$ and $4d^2$. The positive root consequently gives one smooth admissible arc, uniquely determined by the endpoints.