29. A triangular box is to be cut from an equilateral triangle of length $30 \mathrm{~cm}$. Find the largest possible volume of the box (in $\mathrm{cm}^{3}$ ).
29. Answer: 500 .
Let the length and the height of the bax be $a$ and $h$, respectively. Note that $a+2 \sqrt{3} h=30$. Then the volume of the box is
$$
\frac{\sqrt{3} a^{2}}{4} h=\frac{1}{2}\left(\frac{a}{2} \cdot \frac{a}{2} \cdot 2 \sqrt{3} h\right) \leq \frac{1}{2}\left(\frac{a+2 \sqrt{3} h}{3}\right)^{3}=\frac{1}{2} \times 10^{3}=500 .
$$

The equality holds if $a / 2=2 \sqrt{3} h$, i.e., $a=20$ and $h=5 / \sqrt{3}$.