Maximum revenue and profit A company manufactures and sells

Maximum revenue and profit. A company manufactures and sells x e-book readers per month. The monthly cost and price-demand equations are, respectively, C(x) = 350x + 50,000 p = 500 - 0.025x 0 x 20,000 Find the maximum revenue. How many readers should the company manufacture each month to maximize its profit? What is the maximum monthly profit? How much should the company charge for each reader? If the government decides to tax the company $20 for each reader it produces, how many readers should the company manufacture each month to maximize its profit? What is the maximum monthly profit? How much should the company charge for each reader?

Solution

This solution has checked & certain answers a) Revenue = (500 - 0.025 x)x = 500x - 0.025 X^2 0<= x <= 20000 Now differentiation of expression = 0 gives us 500 - 0.05X = 0.. Hence, x = 10,000 Answer: Max revenue = 500x - 0.025 X^2 = $ 2,500,000 b) Profit = (500 - 0.025 x)x - 350x - 50,000 = 150x - 0.025 X^2 - 50,000 0<= x <= 20000 Now differentiation of expression = 0 gives us 150 - 0.05X = 0.. Hence, x = 3000 Answer: 3000 readers should be manufactured for maximum profit maximum monthly profit = $ 175,000 Price per reader = 500-.025*3000 = $425 c) Profit = (500 - 0.025 x)x - 350x -20x - 50,000; = 130x - 0.025 X^2 - 50,000; 0<= x <= 20000; Now differentiation of expression = 0 gives us 130 - 0.05X = 0.. Hence, x = 2600 Answer: 2600 readers should be manufactured for maximum profit maximum monthly profit = $ 119,000 Price per reader = 500-.025*2600 = $435 b)