Answer Happy

Case Study 2: Maximizing Profit MATH 1324 1 Background For many people, visits to cooperated dry are a regular part of life. In 2017, this industry generated by S5 Nation in revenue from about 29.500 location nationwide, the majority of which are individually owned and operated. One business owner decided to raise the price of a washing machine lond. They noticed that the reveto increased but wondered about bottone profile. The business owners serosted in learning about both the maximum poft Based on the price increase the business owner calculated the following revere. A), and cost.C), fico RX) -0.0057 +62 C) - 50x4520 where represents the number of washing machine loads. This leads to the following profit. P(x), function P(a)(x)-C) -(-0.005r+6)-50x520) -0.005550x520 For example, let x 20 washing loads. Then we have R(20) = -0.005(20)4+620) - 118 C(20) = 50(20) + 520 530, so this tells us that the revenue from operating 20 washing loads is $118 while the cost is $530. You can find the profit one of two ways, by using the profit function given or by subtracting the found cost from the found revenue, L. P/20) = -0.005(20)2 +5.50(20) - 520 = -412 P(20) = R(20) -C(20) 118 - 530-412. This means that operating 20 washing loads will result in a loss of $412,
1 Background For many people, visits to a coin-operated laundry are a regular part of life. In 2017. this industry generated nearly $5 billion in gross revenue from about 29.500 locations nationwide, the majority of which are individually owned and operated. One business owner decided to raise the price of a washing machine load. They noticed that the revenue increased but wondered about bottom-line profits. The business owner is in- terested in learning about both the maximum profit. Based on the price increase, the business owner calculated the following revenue, R(x), and cost, C(x), functions: R(x) = -0.005x + 6x C(x) = 50x + 520 where x represents the number of washing machine loads. This leads to the following profit, P(x), function: P(x) = R(x) -C(x) =(-0.00522 +6x)-(.50x + 520) = -0.0052 +5.50x - 520. For example, let x = 20 washing loads. Then we have R(20) = -0.005(20)2 +6(20) = 118 C(20) = .50(20) + 520 = 530. so this tells us that the revenue from operating 20 washing loads is $118 while the cost is $530. You can find the profit one of two ways, by using the profit function given or by subtracting the found cost from the found revenue, i.e.. P(20) = -0.005(20)2 +5.50(20) - 520 = -412 P(20) = R(20) -C(20) = 118 - 530 = -412. This means that operating 20 washing loads will result in a loss of $412.