A man cuts three pieces from a $91$ cm board. The second is $3$ cm longer than the shortest; the third is twice the shortest. The third piece must be at least $5$ cm longer than the second. What is the **maximum** length of the shortest piece?

Let $x$ be the shortest length. Then the second $= x + 3$, the third $= 2x$. Total length: $x + (x + 3) + 2x = 4x + 3 \leq 91 \Rightarrow x \leq 22$. Third at least 5 cm longer than second: $2x \geq (x + 3) + 5 \Rightarrow x \geq 8$. Combining: $8 \leq x \leq 22$. So the **maximum** shortest length is $\boxed{22}$ cm. Check at $x = 22$: pieces $22, 25, 44$. Total $= 91$ ✓. Third $- $ Second $= 19 \geq 5$ ✓.