classical isoperimetric problem

The points $a$ and $b$ on the $x$-axis have to be by an arc with a given length (http://planetmath.org/ArcLength) $l$ such that the area between the $x$-axis and the arc is as great as possible.

Denote the equation of the searched arc by  $y=y(x)$.  The task, which belongs to the isoperimetric problems (http://planetmath.org/IsoperimetricProblem), can be formulated as

$\text{to maximise}\mathit{\hspace{1em}}{\displaystyle {\int }_a^b}y𝑑x$(1)

under the constraint condition

${\displaystyle {\int }_a^b}\sqrt{1+y^{\prime \mathrm{\hspace{0.17em}2}}}𝑑x=l.$(2)

We have the integrands

$$f(x,y,y^{\prime })\equiv y,g(x,y,y^{\prime })\equiv \sqrt{1+y^{\prime \mathrm{\hspace{0.17em}2}}}.$$

The variation problem for the functional in (1) may be considered as a free variation problem (without conditions) for the functional ${\int }_a^b(f-\lambda g)𝑑x$ where $\lambda$ is a Lagrange multiplier.  For this end we need the Euler–Lagrange differential equation (http://planetmath.org/EulerLagrangeDifferentialEquation)

${\displaystyle \frac{\partial }{\partial y}}(f-\lambda g)-{\displaystyle \frac{d}{dx}}{\displaystyle \frac{\partial }{\partial y^{\prime }}}(f-\lambda g)=\mathrm{\hspace{0.33em}0}.$(3)

Since the expression $f-\lambda g$ does not depend explicitly on $x$, the differential equation (3) has, by the Beltrami identity, a first integral of the form

$$(f-\lambda g)-y^{\prime }\cdot (f_{y^{\prime }}^{\prime }-\lambda g_{y^{\prime }}^{\prime })=C_2,$$

which reads simply

$$y-\frac{\lambda }{\sqrt{1+y^{\prime \mathrm{\hspace{0.17em}2}}}}=C_2.$$

This differential equation may be written

$$y^{\prime }\equiv \frac{dy}{dx}=\frac{\sqrt{{\lambda }^2-{(y-C_2)}^2}}{y-C_2},$$

where one can separate the variables (http://planetmath.org/SeparationOfVariables) and integrate, obtaining the equation

$${(x-C_1)}^2+{(y-C_2)}^2={\lambda }^2$$

of a circle.  Here, the parametres $C_1,C_2,\lambda$ may be determined from the conditions

$$y(a)=y(b)=\mathrm{\hspace{0.33em}0},\text{arc length}=l.$$

Thus the extremal of this variational problem is a circular arc (http://planetmath.org/CircularSegment) connecting the given points.

Note that in every point of the arc, the angle of view of the line segment between the given points is constant.