= Solution
Impose $y=x$ before varying. Writing $\delta=\delta_1+\delta_2$, the restricted integrand is
$$
f_{\mathrm{line}}=\alpha(1+\beta x)+\alpha(\gamma_1+\gamma_2)(1+\beta x)\dot x+\frac\delta2\dot x^2.
$$
Its middle term is the <total derivative> of $\alpha(\gamma_1+\gamma_2)(x+\beta x^2/2)$, so it can be dropped for fixed endpoints by part (b). The remaining <Euler-Lagrange equation> is $\delta\ddot x=\alpha\beta$. Integrating twice and imposing $x(0)=0$, $x(T)=S$ gives
$$
\boxed{x(t)=y(t)=\frac STt-\frac{\alpha\beta}{2(\delta_1+\delta_2)}t(T-t),\qquad0\leq t\leq T.}
$$
This is the unique <global minimizer> among paths on the specified line. To prove it, add any admissible <variation> $\eta$ with $\eta(0)=\eta(T)=0$. The change in fuel, after dropping the fixed <boundary term>, is
$$
F[x+\eta]-F[x]=\int_0^T[\alpha\beta\eta+\delta\dot x\dot\eta]dt+\frac\delta2\int_0^T\dot\eta^2dt=\frac\delta2\int_0^T\dot\eta^2dt\geq0.
$$
The first <integral> vanishes by <integration by parts> and the equation for $x$. Equality forces $\eta=0$. No monotonicity or restriction to the segment between endpoints was imposed in the question; the minimizing coordinate can initially decrease for sufficiently large $T$.
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