Answer :
Answer:
Part A: the a quadratic function that reflects the company's revenue.
R = (800+40x)(80-2x) = 64,000 + 1,600 x - 80 x²
Part B: The price should the company charge to maximize its revenue = $60
Step-by-step explanation:
company that sells 800 phones each week when it charges $80 per phone. It sells 40 more phones per week for each $2 decrease in price
Part A: Find the a quadratic function that reflects the company's revenue.
Let the number of weeks = x, and the revenue R(x)
So, the number of sold phones = 800 + 40x
And the cost of the one phone = 80 - 2x
∴ R = (800+40x)(80-2x)
∴ R = 64,000 + 1,600 x - 80 x²
Part B: What price should the company charge to maximize its revenue?
The equation of the revenue represent a parabola
R = 64,000 + 1,600 x - 80 x²
The maximum point of the parabola will be at the vertex
see the attached figure
As shown, the maximum will be at the point (10, 72000)
Which mean, after 10 weeks
The number of sold phones = 800 + 40*10 = 1,200 phones
The price of the phone = 80 - 2 * 10 = 80 - 20 = $60
So, the price should the company charge to maximize its revenue = $60

In this exercise we have to use the knowledge of quadratic function to calculate the value of the company in this way we can say uqe;
A) [tex]R= 64,000 + 1,600 X - 80 X^2[/tex]
B)[tex]V= \$60[/tex]
A) First, we find the a quadratic function that reflects the company's revenue:
[tex]R = (800+40X)(80-2X)\\ R = 64,000 + 1,600X - 80 X^2[/tex]
B)The equation of the revenue represent a parabola:
[tex]R = 64,000 + 1,600X - 80 X^2[/tex]
As shown, the maximum will be at the point 10, which mean, after 10 weeks the number of sold phones
[tex]S = 800 + 40*10 = 1,200 \\P=1,200/200=60[/tex]
See more about quadratic function at brainly.com/question/4119784