Question a-smal

5.1 Linear Inequalities in One or Two Variables - Systems of inequalities and feasible region
0:00

A small business owner budgets $2,200 to purchase candles. The owner must purchase a minimum of 200 candles to maintain the discounted pricing. If the owner pays $4.90 per candle to purchase small candles and $11.60 per candle to purchase large candles, what is the maximum number of large candles the owner can purchase to stay within the budget and maintain the discounted pricing?

Enter your answer:

A small business owner budgets $ 2,200 to purchase candles. The owner must purchase a minimum

Hard-difficulty · SAT Math · Linear Inequalities in One or Two Variables — Systems of inequalities and feasible region. Read the question above, select your answer, and check the full explanation below to understand exactly why the correct choice works.

Answer explanation

The correct answer is 182 . Let s represent the number of small candles the owner can purchase, and let l represent the number of large candles the owner can purchase. It’s given that the owner pays $4.90 per candle to purchase small candles and $11.60 per candle to purchase large candles. Therefore, the owner pays 4.90s dollars for s small candles and 11.60l dollars for l large candles, which means the owner pays a total of 4.90s+11.60l dollars to purchase candles. It’s given that the owner budgets $2,200 to purchase candles. Therefore, 4.90s+11.60ld2,200. It’s also given that the owner must purchase a minimum of 200 candles. Therefore, s+le200. The inequalities 4.90s+11.60ld2,200 and s+le200 can be combined into one compound inequality by rewriting the second inequality so that its left-hand side is equivalent to the left-hand side of the first inequality. Subtracting l from both sides of the inequality s+le200 yields se200-l. Multiplying both sides of this inequality by 4.90 yields 4.90se4.90200-l, or 4.90se980-4.90l. Adding 11.60l to both sides of this inequality yields 4.90s+11.60le980-4.90l+11.60l, or 4.90s+11.60le980+6.70l. This inequality can be combined with the inequality 4.90s+11.60ld2,200, which yields the compound inequality 980+6.70ld4.90s+11.60ld2,200. It follows that 980+6.70ld2,200. Subtracting 980 from both sides of this inequality yields 6.70ld2,200. Dividing both sides of this inequality by 6.70 yields approximately ld182.09. Since the number of large candles the owner purchases must be a whole number, the maximum number of large candles the owner can purchase is the largest whole number less than 182.09 , which is 182 .