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 $y$ in terms of $x$ from seating equation:** $$y = \frac{240 - 6x}{10}$$ 5. **Check constraints on $y$:** Since $y$ must be integer and non-negative, $240 - 6x$ 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 limit is $x \leq 15$. 7. **Find integer $x$ values from 0 to 15 such that $y = \frac{240 - 6x}{10}$ is integer and non-negative:** Check $x$ values: - $x=0$: $y=24$ (valid, $x+y=24 \leq 34$) - $x=5$: $y=\frac{240-30}{10}=21$ (valid, $5+21=26 \leq 34$) - $x=10$: $y=\frac{240-60}{10}=18$ (valid, $10+18=28 \leq 34$) - $x=15$: $y=\frac{240-90}{10}=15$ (valid, $15+15=30 \leq 34$) 8. **Calculate cost for each valid pair:** $$\text{Cost} = 28x + 51y$$ - $x=0, y=24$: $28*0 + 51*24 = 1224$ - $x=5, y=21$: $28*5 + 51*21 = 140 + 1071 = 1211$ - $x=10, y=18$: $28*10 + 51*18 = 280 + 918 = 1198$ - $x=15, y=15$: $28*15 + 51*15 = 420 + 765 = 1185$ 9. **Minimum cost is $1185$ with $15$ rectangular and $15$ round tables.** --- 1. **Problem statement:** A meat market wants to mix ground beef and ground pork to make meat loaf at minimum cost. Constraints: - Ground beef: 76% lean, cost 0.90 per lb, max 200 lb - Ground pork: 68% lean, cost 0.66 per lb, at least 50 lb - Meat loaf must be at least 70% lean - Pounds must be integers 2. **Define variables:** Let $b$ = pounds of ground beef Let $p$ = pounds of ground 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 $p \geq 50$ and $p \leq 3b$, then: $$50 \leq p \leq 3b$$ So: $$3b \geq 50 \Rightarrow b \geq \frac{50}{3} \approx 16.67$$ 7. **Try to minimize cost by choosing $b$ and $p$ at boundary:** Since $p$ is cheaper per pound, minimize $b$ and maximize $p$ within constraints. Try $p = 3b$ (max $p$ for given $b$): $$C = 0.90b + 0.66(3b) = 0.90b + 1.98b = 2.88b$$ Minimize $C$ by minimizing $b$ subject to $b \geq 16.67$ and $b \leq 200$. Choose $b = 17$ (integer), then $p = 3*17 = 51$ (integer and $\geq 50$). 8. **Check lean percentage:** $$\frac{0.76*17 + 0.68*51}{17 + 51} = \frac{12.92 + 34.68}{68} = \frac{47.6}{68} \approx 0.7$$ Meets lean requirement. 9. **Calculate cost:** $$C = 0.90*17 + 0.66*51 = 15.3 + 33.66 = 48.96$$ 10. **Check if increasing $b$ reduces cost:** Try $b=18$, $p=54$: $$C=0.90*18 + 0.66*54=16.2 + 35.64=51.84 > 48.96$$ So $b=17$, $p=51$ is minimum cost. **Final answers:** - Rectangular tables: 15 - Round tables: 15 - Minimum cost for tables: 1185 - Ground beef: 17 pounds - Ground pork: 51 pounds - Minimum cost for meat: 48.96