1. **Problem statement:** Kelly needs to rent tables for exactly 240 people using 6-person rectangular tables and 10-person round tables. The constraints are: - Maximum 34 tables total - Maximum 15 rectangular tables available Costs: - Rectangular table: 28 each - Round table: 51 each We want to minimize the total cost. 2. **Define variables:** Let $x$ = number of rectangular tables Let $y$ = number of round tables 3. **Formulate equations:** People seating: $$6x + 10y = 240$$ Table count: $$x + y \leq 34$$ Availability: $$x \leq 15$$ Non-negativity: $$x \geq 0, y \geq 0$$ 4. **Express $y$ from seating equation:** $$y = \frac{240 - 6x}{10}$$ 5. **Check constraints on $y$:** Since $y$ must be integer and $\geq 0$, $240 - 6x$ must be divisible by 10 and $y \geq 0$. 6. **Check table count constraint:** $$x + y \leq 34$$ Substitute $y$: $$x + \frac{240 - 6x}{10} \leq 34$$ Multiply both sides by 10: $$10x + 240 - 6x \leq 340$$ $$4x + 240 \leq 340$$ $$4x \leq 100$$ $$x \leq 25$$ Since $x \leq 15$ from availability, the stricter constraint is $x \leq 15$. 7. **Check integer values of $x$ from 0 to 15 where $y$ is integer and non-negative:** We want $240 - 6x$ divisible by 10. Check $x$ values: - $x=0$: $y=24$ (valid, $x+y=24 \leq 34$) - $x=5$: $y= (240 - 30)/10=21$ (valid, $5+21=26 \leq 34$) - $x=10$: $y= (240 - 60)/10=18$ (valid, $10+18=28 \leq 34$) - $x=15$: $y= (240 - 90)/10=15$ (valid, $15+15=30 \leq 34$) 8. **Calculate cost for each valid pair:** $$C = 28x + 51y$$ - $x=0, y=24$: $C=0 + 51 \times 24=1224$ - $x=5, y=21$: $C=28 \times 5 + 51 \times 21=140 + 1071=1211$ - $x=10, y=18$: $C=280 + 918=1198$ - $x=15, y=15$: $C=420 + 765=1185$ 9. **Minimum cost is $1185$ with $15$ rectangular and $15$ round tables.** --- 1. **Problem statement:** Meat market wants to mix ground beef and pork to minimize cost. Constraints: - Beef is 76% lean, costs 0.90 per lb - Pork is 68% lean, costs 0.66 per lb - Meat loaf must be at least 70% lean - At least 50 lb pork - No more than 200 lb beef - Pounds are integers 2. **Define variables:** Let $b$ = pounds of beef Let $p$ = pounds of pork 3. **Lean percentage constraint:** Total lean meat: $$0.76b + 0.68p$$ Total weight: $$b + p$$ Lean percentage constraint: $$\frac{0.76b + 0.68p}{b + p} \geq 0.70$$ Multiply both sides by $b+p$: $$0.76b + 0.68p \geq 0.70b + 0.70p$$ Simplify: $$0.76b - 0.70b \geq 0.70p - 0.68p$$ $$0.06b \geq 0.02p$$ Divide both sides by 0.02: $$3b \geq p$$ 4. **Other constraints:** $$p \geq 50$$ $$b \leq 200$$ $$b, p \geq 0$$ 5. **Cost function to minimize:** $$C = 0.90b + 0.66p$$ 6. **Use $p \leq 3b$ from lean constraint and $p \geq 50$:** Since $p \leq 3b$ and $p \geq 50$, we have: $$50 \leq p \leq 3b$$ To minimize cost, use smallest $p$ possible (since pork is cheaper), so set $p=50$. Then from $p \leq 3b$: $$50 \leq 3b \Rightarrow b \geq \frac{50}{3} \approx 16.67$$ 7. **Check cost at $p=50$ and $b=17$ (integer):** $$C = 0.90 \times 17 + 0.66 \times 50 = 15.3 + 33 = 48.3$$ 8. **Check if increasing $b$ reduces cost:** Cost increases with $b$ because beef is more expensive. 9. **Check if increasing $p$ reduces cost:** Increasing $p$ increases cost since $p$ is positive cost. 10. **Check $b=17$, $p=50$ satisfies all constraints:** - $b \leq 200$ true - $p \geq 50$ true - $p \leq 3b$ true since $50 \leq 51$ 11. **Final answer:** Beef = 17 lb, Pork = 50 lb, minimum cost = $48.3$