Question

Please help!

Answer 1

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