 Solution

Integrate once and use $h=e^{-\alpha x}$ to obtain $p^{\prime }=6e^{2\alpha x}+C{e}^{3\alpha x}$. The two zero endpoint pressures determineConsequentlyThe constant $M={\int }_0^1{h}^{-2}dx/{\int }_0^1{h}^{-3}dx$ is a weighted average of $h$ with positive weight $h^{-3}$, so $e^{-\alpha }<M<1$. Since $e^{-\alpha x}$ decreases strictly, $p^{\prime }$ changes sign exactly once, from positive to negative. Thus the unique maximum is atFor small $\alpha$, $\log M=-\alpha /2-5{\alpha }^2/24+O({\alpha }^4)$, so $\boxed{x_m=\frac{1}{2}+\frac{5}{24}\alpha +O({\alpha }^3)\sim \frac{1}{2}}$. The exponential-gap slider bearing provides a particularly simple example where a converging gap produces positive load-bearing pressure.