 Solution

Take one-dimensional steady state thermal conduction, constant thermal conductivity $k>0$, and positive volumetric heat generation rate. Write $q_0=q_0^{\prime \prime \prime }$ for brevity, and let $x$ increase away from the illuminated face. Conservation of energy and Fourier's law giveThus $T^{\prime \prime }=-(q_0/k)e^{-\mu x}<0$: 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 solutionIts derivative isDefine $\beta =\mu a$ and $C=(1-e^{-\beta })/\beta$. Setting $T^{\prime }(x_m)=0$ gives the distance from the illuminated surface: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 givesThe uniform-heating limit therefore has its maximum at $a/2$, as symmetry requires.