Asked by c:
This is a quadratics function question. D:
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.
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.
Answers
Answered by
MathMate
We do not see a question here. Could you complete the question?
Answered by
c:
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.)
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.)
Answered by
MathMate
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
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
There are no AI answers yet. The ability to request AI answers is coming soon!
Submit Your Answer
We prioritize human answers over AI answers.
If you are human, and you can answer this question, please submit your answer.