40. Suppose $C D$ is the longest edge of the tetrahedron $A B C D, A B=a, C K$ and $D L$ are the altitudes of the triangles $A B C$ and $A B D$ respectively, and $D M$ is the altitude of the tetrahedron $A B C D$. Then $C K^{2} \leq 1-a^{2} / 4$, since $C K$ is a leg of the right triangle whose other leg has length not less than $a / 2$ and whose hypotemuse has length not greater than 1 (AKC or $B K C$ ). In the similar way we can show that $D L^{2} \leq 1-a^{2} / 4$. Since $D M \leq D L$, then $D M^{2} \leq 1-\alpha^{2} / 4$. It follows that
$$\begin{aligned} V & =\frac{1}{3}\left(\frac{a}{2} C K\right) D M \leq \frac{1}{6} a\left(1-\frac{a^{2}}{4}\right)=\frac{1}{24} a(2-a)(2+a) \\ & =\frac{1}{24}\left[1-(a-1)^{2}\right](2+a) \leq \frac{1}{24} \cdot 1 \cdot 3=\frac{1}{8} \end{aligned}$$