= Solution
Take one-dimensional <steady state> <thermal conduction>, constant <thermal conductivity> $k>0$, and positive <volumetric heat generation rate>. Write $q_0=q'''_0$ for brevity, and let $x$ increase away from the illuminated face. <Conservation of energy> and <Fourier's law> give
$$
\frac{d}{dx}\left(-k\frac{dT}{dx}\right)=q_0e^{-\mu x},
\qquad T(0)=T(a)=T_0.
$$
Thus $T''=-(q_0/k)e^{-\mu x}<0$: the profile is strictly concave and any stationary point is its unique maximum. Two <integrations> and the equal-surface <boundary conditions> yield the <exponential slab heating> solution
$$
\boxed{T(x)-T_0=\frac{q_0}{k\mu^2}
\left[1-e^{-\mu x}-(1-e^{-\mu a})\frac{x}{a}\right].}
$$
Its derivative is
$$
T'(x)=\frac{q_0}{k\mu}
\left[e^{-\mu x}-\frac{1-e^{-\mu a}}{\mu a}\right].
$$
Define $\beta=\mu a$ and $C=(1-e^{-\beta})/\beta$. Setting $T'(x_m)=0$ gives \b[the distance from the illuminated surface]:
$$
\boxed{x_m=\frac1\mu\ln\frac{\mu a}{1-e^{-\mu a}}=-\frac{\ln C}{\mu}.}
$$
For $\beta>0$, $e^{-\beta}<C<1$, so $0<x_m<a$. More specifically $C=e^{-\beta/2}\sinh(\beta/2)/(\beta/2)>e^{-\beta/2}$, hence $x_m<a/2$: the peak is nearer the more strongly heated face.
The apparent singularity at $\mu=0$ is removable. Expansion gives
$$
\frac{x_m}{a}=\frac12-\frac{\beta}{24}+O(\beta^3),
\qquad
T(x)-T_0\longrightarrow\frac{q_0}{2k}x(a-x).
$$
The uniform-heating limit therefore has its maximum at $a/2$, as symmetry requires.
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