A company manufactures and sells x e-book readers per month The monthly cost and
ID: 3248837 • Letter: A
Question
A company manufactures and sells x e-book readers per month The monthly cost and or price-demand equations are. respectively. C (x) = 350x + 50,000 p = 500 - 0.025 x A. Find the maximum revenue. B. 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? C. 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? D. Include a graph of the function you are maximizing and the derivative. E. Interpret the graph and the derivative.Explanation / Answer
Given
e-book readers = x
Monthly cost C(x) =350x + 50,000
Price demand p = 500- 0.025x
a) revenue R(x) =price demand * quantity
R(x) = (500-0.025x)*x
R(x)=500x – 0.025x2
Now to maximise the revenue R'(x) and set it is equal to zero
R'(x) =500 -0.05x
Since R'(x) =0 hence 500-0.5x =0
[NOTE:(if R(x) =x2 , then R'(x) =2x, and if R(x) =x, then R'(x) =1)]
x = 500/0.05
=10,000
Hence the maximum revenue=10,000
b).profit =revenue - cost
P(x) = 500x – 0.025x2 -350x-50000
P(x) =150x -0.025x2 -50000
To maximise the profit P'(x) and set it is equal to zero
P'(x) =150-0.05x
Since P'(x) =0, hence 150-0.05x =0
x=150/0.05 =3000
Maximum monthly profit =3000
Charge per each reader = 500-0.025x
=500-0.025 (3000)
Charge per each reader =500-75 = 425
c) Government decide to tax the company $20 for each reader, hence the profit equation is
P(x) =150x -0.025x2 -50000 -20x
=130x-0.025x2 - 50000
To maximise the profit P'(x) and set it is equal to zero
P(x) =130x-0.025x2 - 50000
P'(x) =130-0.05x
Since P'(x) =0, hence , 130-0.05x =0
=130/0.05 =2600
Maximum monthly profit =2600
Charge per each reader = 500-0.025x
=500-0.025 (2600)
Charge per each reader =500-65 = 435
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