Question

. points WANEFMACZ 2.1.039. Quay set your teacher v The latest demand equation for your gaming website, www.mudbeast.net, is given by -500+1300 where is the number of users who log on per month and x is the log-on fee you charge. Your Internet provider bills you as follows: Site maintenance fee 50 per month Find the monthly cost as a function of the log-on fee x. C) - Find the monthly profit as a function of x. Determine the log-on fee you should charge to obtain the largest possible monthly profit. x = $ per log-on What is the largest possible monthly profit? Need Help? Redacht te Tube 2 points WANEFMAC7 2.1.502.XP. nyertes Ask Your Teacher The amount of carbon dioxide (in pounds per 15,000 cubic miles) released by a certain SUV depends on its fuel efficiency according to the formula W -2145x45,000 (12 3 3) where x is the fuel efficiency of an SUV in miles per gallon. According to the model, what is the fuel efficiency of the SUV with the least carbon dioxide pollution? x= mi per gallon Comment on the reliability of the model for fuel efficiencies that exceed your answer. Since the answer is close to the lower edge of the domain, fuel efficiencies that exceed the answer would produce a reliable prediction. Since the answer is close to the upper edge of the domain, fuel efficiencies that exceed the answer would not produce a reliable prediction. Need Help?

Answer 1

To determine the cost and profit equation, we must identify variable and fixed costs. Fixed cost is nothing but fixed amount paid to internet provider irrespective of log-on's Here, internet provider is charging $50 per month which is a fixed cost. $0.80 is charged per logon which is a variable cost. Cost function (Total cost] = Fixed cost + variable cost per log on x no. of users hence, Cost function [C] = 50 +0.80 since we need Cost in terms of x, we must plug in what we know about q in terms of x: So, the monthly cost as a function of x is C(x) = 50 +0.80(-500x + 1300) C(x) = 50 +0.80(-500x + 1300) C(x) = 50 - 400 x + 1040 C(x) = - 400 x + 1090

part 2: To find the profit function, we also need the revenue function Revenue = (price) (quantity), R(x) = ( x) * (9) The relationship between Revenue and profit is, Profit = Revenue - Total Cost P (2) =R () - C P(x) = x ( -500x +1300) - (-400x + 1090) P(x) = -500x 2 +1300x +400x - 1090 =-500x 2 + 1700x - 1090 Part 3: This function is quadratic, with negative leading coefficient. So, it opens downward and its maximum point is the vertex. To find the price x we should charge to maximize profits, we must find the x-coordinate of the vertex We can do this using the vertex formula Xmax =-1700 / 21-500) = 1.7 So the log-on fee should be set at $1.7 to maximize profits. X = $ 1.7 per log on

Part 4: To derive largest monthly profit, substitute the X value in profit equation. P(x) = -500x 2 + 1700x - 1090 P(1.7) = -500(1.7) 2+1700(1.7) - 1090 = $2,890 - 2,535 = $355 Thus, the largest monthly profit possible is $355.