This is a quadratics function question. D:

Question

A company charges 20$ for a subscription to something and gains 30,000 subscribers. It looses 1,000 subscribers for every 1$ they raise the subscription price.
If you could help me, that would be wonderful.

Answer 1

We do not see a question here. Could you complete the question?

Answer 2

A company charges 20$ for a subscription to something and gains 30,000 subscribers. It looses 1,000 subscribers for every 1$ they raise the subscription price.

Find the quadratic quation.
Find the rate at 20$. (So i suppose solve it for 20$)
Find maximum revenue. (I don't even know what its really asking or how to solve it.)

Answer 3

Let s(x)=number of subscribers at the rate of x.
We know that the marginal number of subscribers (derivative) is -1000 when x=20, or
d(s(x))/dx = -1000
Integrate to get
s(x)=-1000x+C
Since s(20)=30000, we get
30000=-1000*20+c
c=50000, or
s(x)=50000-1000x

The revenue R is the product of the number of customers and the price, x.
Therefore
R(x)=x*s(x)=50000x-1000x²

To find the maximum revenue, we set the marginal revenue (derivative of R) to zero:
R'(x)=50000-2000x=0
x=25
is the price that will maximum revenue at R(25)=50000*25-1000*25²=$625,000