Using the mechanical principle that the centre of mass places itself as low as possible, determine the equation of the curve formed by a flexible homogeneous wire or a thin chain with length $l$ when supported at its ends in the points  $P_{1}=(x_{1},\,y_{1})$  and  $P_{2}=(x_{2},\,y_{2})$.

We have an isoperimetric problem

under the constraint

where both the path integrals are taken along some curve $c$.  Using a Lagrange multiplier $\lambda$, the task changes to a free problem

(cf. example of calculus of variations).

The Euler–Lagrange differential equation, the necessary condition for (3) to give an extremal $c$, reduces to the Beltrami identity

where $a$ is a constant of integration.  After solving this equation for the derivative $y^{{\prime}}$ and separation of variables, we get

which may become clearer by notating  $y\!-\!\lambda:=u$;  then by integrating

we choose the new constant of integration $b$ such that  $x=b$  when  $u=a$:

We can write two equivalent results

i.e.

Adding these allows to eliminate the square roots and to obtain

or

This is the sought form of the equation of the chain curve.  The constants $\lambda,\,a,\,b$ can then be determined for putting the curve to pass through the given points $P_{1}$ and $P_{2}$.