1. **State the problem:** We want to maximize the profit from products A and B given constraints on sales, availability, and raw material usage. 2. **Define variables:** Let $x$ = units of product A produced per day. Let $y$ = units of product B produced per day. 3. **Constraints:** - At least 80% of total sales are from A and B combined, so $x + y \geq 0.8(x + y)$ is always true, so this is redundant. - Maximum daily availability is 240 lb of raw material. - Usage rates: 2 lb per unit of A, 4 lb per unit of B. - So, raw material constraint: $$2x + 4y \leq 240$$ - Maximum units for A is 60, so $$x \leq 60$$ - Non-negativity: $$x \geq 0, y \geq 0$$ 4. **Profit function to maximize:** $$P = 20x + 50y$$ 5. **Summarize constraints:** $$\begin{cases} 2x + 4y \leq 240 \\ x \leq 60 \\ x \geq 0 \\ y \geq 0 \end{cases}$$ 6. **Find corner points of feasible region:** - When $x=0$, $2(0) + 4y \leq 240 \Rightarrow y \leq 60$ - When $y=0$, $2x + 0 \leq 240 \Rightarrow x \leq 120$, but $x \leq 60$ limits $x$ to 60. - Intersection of $2x + 4y = 240$ and $x=60$: $$2(60) + 4y = 240 \Rightarrow 120 + 4y = 240 \Rightarrow 4y = 120 \Rightarrow y = 30$$ 7. **Evaluate profit at corner points:** - At $(0,0)$: $P=0$ - At $(0,60)$: $P=20(0)+50(60)=3000$ - At $(60,0)$: $P=20(60)+50(0)=1200$ - At $(60,30)$: $P=20(60)+50(30)=1200+1500=2700$ 8. **Conclusion:** Maximum profit is $3000$ at $x=0$, $y=60$. Final answer: Produce 0 units of A and 60 units of B for maximum profit of 3000.