1. The problem asks to find the value of $x$ for which the volume of the cuboid is maximum. 2. The volume function $V(x)$ is given or derived as a quadratic expression. To find the maximum volume, we differentiate $V(x)$ with respect to $x$ and set the derivative equal to zero. 3. The derivative is given as $$\frac{dV}{dx} = 16 - 2x - 6x^2.$$ Setting this equal to zero gives the quadratic equation: $$16 - 2x - 6x^2 = 0.$$ 4. Rearranging: $$-6x^2 - 2x + 16 = 0.$$ Multiply both sides by $-1$ to simplify: $$6x^2 + 2x - 16 = 0.$$ 5. Use the quadratic formula: $$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$ where $a=6$, $b=2$, and $c=-16$. 6. Calculate the discriminant: $$\Delta = 2^2 - 4 \times 6 \times (-16) = 4 + 384 = 388.$$ 7. Calculate $x$: $$x = \frac{-2 \pm \sqrt{388}}{12} = \frac{-2 \pm 19.6977}{12}.$$ 8. Two solutions: $$x_1 = \frac{-2 + 19.6977}{12} = 1.4748,$$ $$x_2 = \frac{-2 - 19.6977}{12} = -1.8081.$$ 9. Since $x$ must be positive for the cuboid dimensions, we take: $$x = 1.47$$ (to 3 significant figures). 10. This is the value of $x$ that maximizes the volume. **Final answer:** $$\boxed{1.47}$$