The maximum value of the objective function $\mathrm{z}=4 x+6 y$ subject to $3 x+2 y \leq 12, x+y \geq 4, x$, $y \geq 0$ is

$$\frac{\mathrm{d}}{\mathrm{~d} x}\left(\cos ^{-1}\left(\frac{x-\frac{1}{x}}{x+\frac{1}{x}}\right)\right)=$$

If $\bar{a}, \bar{b}, \bar{c}$ are non-coplanar vectors and $\overline{\mathrm{p}}=\frac{\overline{\mathrm{b}} \times \overline{\mathrm{c}}}{[\overline{\mathrm{a}} \overline{\mathrm{b}} \overline{\mathrm{c}}]}, \overline{\mathrm{q}}=\frac{\overline{\mathrm{c}} \times \overline{\mathrm{a}}}{[\overline{\mathrm{a}} \overline{\mathrm{b}} \overline{\mathrm{c}}]}, \overline{\mathrm{r}}=\frac{\overline{\mathrm{a}} \times \overline{\mathrm{b}}}{[\overline{\mathrm{a}} \overline{\mathrm{b}} \overline{\mathrm{c}}]}$, then $2 \overline{\mathrm{a}} \cdot \overline{\mathrm{p}}+\overline{\mathrm{b}} \cdot \overline{\mathrm{q}}+\overline{\mathrm{c}} \cdot \overline{\mathrm{r}}=$

The approximate value of $\tan ^{-1}(0.999)$ is (use $\pi=3.1415$ )