1. **Stating the problem:** We have two types of dinghies: Fiberglass and Wooden-hulled. Fiberglass profit = 13, craftsmen needed = 2, apprentices needed = 3. Wooden profit = 20, craftsmen needed = 3, apprentices needed = 5. Total craftsmen available = 12, total apprentices available = 19. We want to find the ratio of the two types to maximize profit. 2. **Define variables:** Let $x$ = number of fiberglass dinghies. Let $y$ = number of wooden dinghies. 3. **Constraints:** Craftsmen: $2x + 3y \leq 12$ Apprentices: $3x + 5y \leq 19$ 4. **Profit function to maximize:** $$P = 13x + 20y$$ 5. **Find feasible points by solving constraints:** From craftsmen constraint: $$2x + 3y \leq 12$$ From apprentices constraint: $$3x + 5y \leq 19$$ 6. **Find intercepts for constraints:** For craftsmen: If $x=0$, $3y=12 \Rightarrow y=4$ If $y=0$, $2x=12 \Rightarrow x=6$ For apprentices: If $x=0$, $5y=19 \Rightarrow y=\frac{19}{5} = 3.8$ If $y=0$, $3x=19 \Rightarrow x=\frac{19}{3} \approx 6.33$ 7. **Find intersection of constraints:** Solve system: $$\begin{cases} 2x + 3y = 12 \\ 3x + 5y = 19 \end{cases}$$ Multiply first by 3 and second by 2: $$\begin{cases} 6x + 9y = 36 \\ 6x + 10y = 38 \end{cases}$$ Subtract first from second: $$6x + 10y - (6x + 9y) = 38 - 36$$ $$y = 2$$ Substitute $y=2$ into first equation: $$2x + 3(2) = 12$$ $$2x + 6 = 12$$ $$2x = 6$$ $$x = 3$$ 8. **Evaluate profit at vertices:** - At $(0,0)$: $P=0$ - At $(0,3.8)$: $P=13(0)+20(3.8)=76$ - At $(6,0)$: $P=13(6)+20(0)=78$ - At $(3,2)$: $P=13(3)+20(2)=39+40=79$ 9. **Conclusion:** Maximum profit is 79 at $x=3$, $y=2$. Ratio of fiberglass to wooden dinghies is $3:2$.