Put $x=k^2$ and $T=\lambda/\sigma$. Differentiating $Ra$ gives the exact stationarity equation
$$
(\pi^2+x)^2(2x-\pi^2)=\pi^2T^2.
$$
When $T\gg1$, the optimum has $x\gg\pi^2$, so $2x^3\sim\pi^2T^2$. Therefore
$$
\boxed{k\sim\left(\frac{\pi^2}{2}\right)^{1/6}
\left(\frac\lambda\sigma\right)^{1/3}}.
$$


Solution

Past exam of the mathematics course of the University of Cambridge / 2021 / iii / Paper 331 / 1 / f / Solution

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2021/iii/paper-331/1/f/solution