I need a hint on this exercise in Calculus of Variation.
Let $X=\{u\in C^{1}(-1,1)\cap C[-1,1]:u(\pm 1)=0\}$ and consider the length and the area functionals $F,G:X\rightarrow\mathbb{R}$: \begin{equation*} F(u)=\int_{-1}^{1}\sqrt{1+u'(x)^{2}}dx \end{equation*} \begin{equation*} G(u)=\int_{-1}^{1}u(x)dx \end{equation*} For a fixed constant $v>0$ consider the minimum problem: \begin{equation*} \min\{F(u):u\in X \ \text{such that} \ G(u)=v\} \end{equation*} Assuming its existence we compute its solution.
Given $\phi$, $\psi\in C^{\infty}_{c}(-1,1)$ define $H:\mathbb{R}^{2}\rightarrow\mathbb{R}$, $H(\epsilon,\tau)=G(u+\epsilon\phi+\tau\psi)$. Show that there exists $\psi$ and a function $\epsilon\mapsto\tau(\epsilon)$ such that $H(\epsilon,\tau(\epsilon))=v$ for all $\epsilon$.