= Solution
Use the maximizing form of the <Pontryagin maximum principle>. For a cost-minimization problem there are a nonnegative multiplier $\lambda_0$ and a costate $\psi$, not both identically zero, with <Hamiltonian> $\mathscr H=\psi b-\lambda_0c$. The optimal control maximizes $\mathscr H$, the costate obeys $\dot\psi=-\mathscr H_x$, and free terminal time at the fixed terminal state zero gives $\mathscr H(\tau)=\lambda_0C'(\tau)$. There is no fixed terminal costate value since the terminal state is fixed. In the normal case scale $\lambda_0=1$.
For the energy problem $b=-u$, $c=up(t)$, $C=-R$, so $\psi$ is constant and the normal <Hamiltonian> is $u(p_*-p(t))$, where $p_*=-\psi$. Maximization gives full power for $p(t)<p_*$ and zero power for $p(t)>p_*$; either value, or a fractional value, is allowed on a price tie. On the final active arc, the free-time condition gives
$$
\boxed{p_*+R'(\tau_*)=p(\tau_*).}
$$
This is the <threshold energy scheduling with a completion reward> rule. In this example the normal case is justified: an abnormal constant costate would require either no processing or processing at all times, and free-time transversality would then force that nonzero costate to vanish.
For $p_*=1/4$, processing starts at $t=1/2$. Energy $1/2$ is completed at $\tau=1$, before the price rises again. Its cost and reward are
$$
\boxed{\int_{1/2}^{1}(t-1)^2dt=1/24,\qquad R(1)=1/2,\qquad J=-11/24.}
$$
Any threshold carrying completion into day two has $\tau\geq2$ and reward at most $1/3$. Since energy cost is nonnegative, $J\geq-1/3>-11/24$, so it is suboptimal.
For a global check, even any completion time $\tau\geq5/4$ has reward at most $4/9<11/24$ and cannot beat the displayed candidate. Also $\tau\geq1/2$ since power is bounded by one. At a fixed $\tau\in[1/2,5/4]$, the cheapest half-unit of processing consists of the last half-unit interval $[\tau-1/2,\tau]$: every earlier point has a price at least as large as every selected point. Exchanging any earlier powered interval with a later unpowered cheaper one proves this rearrangement claim, also for fractional controls.
Write $h=3/2-\tau\in[1/4,1]$. This minimal cost minus reward is
$$
J(h)=h^2/2-h/4+1/24-\frac1{5/2-h},
$$
with
$$
J'(h)=h-1/4-\frac1{(5/2-h)^2},\qquad
J''(h)=1-\frac2{(5/2-h)^3}>0.
$$
It has its unique minimum at $h=1/2$, where $J'=0$. The selected interval is then $[1/2,1]$ and its threshold is $h^2$. Consequently
$$
\boxed{p_*=1/4\text{ is globally optimal}.}
$$
The terminal condition also checks: $1/4+R'(1)=1/4-1/4=p(1)=0$.
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