4. To protect valuable equipment from bad weather, a tent in the shape of a pyramid is required, with a rectangle as its base, and one of the lateral edges should be perpendicular to the base. Find the maximum possible volume of the tent, given that no edge of the pyramid should be longer than 2 meters.

$8 /(9 \sqrt{3})$.
Solution. (1) Let the base be a rectangle with sides $x$ and $y$, and height $z$. Then the conditions of the problem are equivalent to finding $x, y, z$ such that $\frac{x y z}{3} \Rightarrow \max$ while simultaneously satisfying the conditions $x>0, y>0, z>0, x^{2}+y^{2}+z^{2} \leqslant 4$. (2) Let the tent of maximum volume have height $z$. We will prove that $x=y$. For a given $z$, the condition $\frac{x y z}{3} \Rightarrow$ max is equivalent to $\left\{\begin{array}{l}x>0, y>0, x y \\ x^{2}+y^{2}+z^{2} \leqslant 4\end{array} \Rightarrow \max .\left\{\begin{array}{l}x>0, y>0, x y \Rightarrow \max , \\ y^{2}+y^{2} \leqslant 4-z^{2} .\end{array}\right.\right.$ According to the Cauchy inequality, $x=y=\sqrt{4-z^{2}}$.

(3) Thus, the base is a square with side $x$. Then the conditions of the problem are equivalent to finding $x, z$ such that $\frac{x^{2} z}{3} \Rightarrow$ max while simultaneously satisfying the conditions $0<x \leqslant 2$; $0<z \leqslant 2, 2 x^{2}+z^{2} \leqslant 4, x=\sqrt{4-z^{2}}$. Consider the function $f(z)=\frac{\left(4-z^{2}\right) \cdot z}{3}$ and the condition $f(z) \Rightarrow \max \Leftrightarrow$ $\frac{\left(4-z^{2}\right) \cdot z}{3} \Rightarrow \max \Leftrightarrow\left(4-z^{2}\right) \cdot z \Rightarrow \max \Leftrightarrow 4 z-z^{3} \Rightarrow \max$. Let $g(z)=4 z-z^{3}: g^{\prime}(z)=4-3 z^{2}$, $4-3 z^{2}=0$ when $z=2 / \sqrt{3}$. The sign analysis of the derivative shows that this is a point of maximum. (4) Thus, $x^{2}+y^{2}+z^{2}=4, 3 x^{2}=4, x=y=z=2 / \sqrt{3}, V=x^{3} / 3=8 /(9 \sqrt{3})$.