I am trying to minimise the functional
$$V[z]= \int_{-a}^{a} z \sqrt{1+(z’)^2}dx.$$
Note $f(z, z’, x)= z \sqrt{1+(z’)^2} $ hasn’t got any dependence on $x$, so
$$f-z’ \frac{\partial f}{\partial z’}=0$$
at the extremum.
$$\frac{\partial f}{\partial z’}=z \frac{1}{2} (1+(z’)^2)^{-1/2} 2z’$$
So the equation becomes
$$z \left( \sqrt{1+(z’)^2} - (z’)^2/\sqrt{1+(z’)^2} \right)=0.$$
$$1+(z’)^2-(z’)^2=0.$$
$$1=0.$$
I’ve obviously got to be doing something wrong here, but I cannot figure out what.
I’m expecting a solution of the form $z-z_0=B \cosh (x/B)$ since the functional describes the potential energy of a uniform rope in a gravitational field.
Your reduction of the E-L equation is incorrect. The difference should be constant, but that constant need not be zero.
$$ f - z'\partial_{z'} f = c $$
Substitute the given values
$$ z\sqrt{1+z'^2} - zz'^2/\sqrt{1+z'^2} = c$$
Rearrange to get
$$ z = c\sqrt{1+z'^2} $$
This looks like it will give you the $\cosh$ you're looking for.