615. From a round log of radius $R$, it is required to cut out a rectangular beam of maximum strength. It is known that the strength of the beam is directly proportional to the product of its width and the square of its height. What should be the dimensions of the beam so that its strength is maximized?
Solution. Let the width of the beam be denoted by $x$, its height by $y$, and its strength (in bending) by $J$, so we have $J=k x y^{2}$, where $k>0$ is a proportionality coefficient.

Thus, we need to find the maximum of the function $J=k x y^{2}$. To express the strength of the beam in terms of one of the unknowns, note that $x^{2}+y^{2}=4 R^{2}$, from which $y^{2}=4 R^{2}-x^{2}$ and $J=k x\left(4 R^{2}-x^{2}\right)$.

We find the derivative and set it to zero:

$$
J^{\prime}=4 k R^{2}-3 k x^{2}=0 \quad \text { or } \quad k\left(4 R^{2}-3 x^{2}\right)=0
$$

Since $k \neq 0$, we have $4 R^{2}-3 x^{2}=0$, from which $x= \pm 2 R / \sqrt{3}$.

By the context of the problem, only the positive root $x=2 R / \sqrt{3}$ is of interest. Since $J^{\prime \prime}=-6 k x<0$ at $x=2 R / \sqrt{3}$, we conclude that the strength of the beam will be maximal when $x=2 R / \sqrt{3}$.