Please check back the original question before reposting.
http://www.jiskha.com/display.cgi?id=1298510239
A company charges 16$ for a subscription. It has 30,000 subscribers. It looses 1,000 subscribers for every 1$ they raise the price.
1.Find the quadratic quation.
(Can you tell me how to pull the numbers out of the question and put it into the equation?)
Find the rate at 16$. (So i suppose solve it for 16$?)
Find max revenue. (I don't even know what its really asking or how to solve it.)
5 answers
I already looked over what you had posted. Unfortunatly, i didn't quite understand what you had said.
You have titled the post "algebra 2". Does that mean that you have done calculus, or should the solution be without calculus?
We usually solve maximization problems by linear programming or calculus. However,
this problem can also be solved by tabulation without using calculus. If that is the case, let me know.
We usually solve maximization problems by linear programming or calculus. However,
this problem can also be solved by tabulation without using calculus. If that is the case, let me know.
In Kentucky, where i am from, we do algebra and algebra two before pre-cal and calculus. I didn't realize that the problem could be solved in multiple ways and i apologise.
But the solution should be done without calculus.
But the solution should be done without calculus.
This question is different from the previous where the original price was at $20 / subscription, but the solution method remains the same.
This question is telling us:
1. The company charges $16 per subscription, at this rate, it has recruited 30000 customers, or a revenue of 30000*16=$480000.
2. If the company raises the price of the subscription, it will do so at the cost of losing 1000 customers for every dollar it raises.
This means that the number of customers N is a linear function of the price x, and the slope m is -1000.
Let N(x)=mx+b
But we know that N(16)=30000, so
30000=-1000(16)+b
or
b=30000+16000=46000
So now the function N(x) is completely defined:
N(x)=-1000x+46000
If we define the revenue R(x) as the product of the number of customers multiplied by the price, then
R(x)=x*N(x)=-1000x²+46000x
So R(16)=-1000(16)²+46000*16=$480000 as before.
To calculate the maximum, we will have to do this by tabulation:
For a given x, calculate R(x), and present it in table form:
R(16)=480000 (as before)
try
R(20)=520000
R(25)=525000
R(30)=480000
So we know that the maximum revenue occurs at 20<x<30.
Give a few more tries, and you should get the maximum.
In this case, the optimal price x0 happens to be a whole number of dollars. But you still have to prove that the price gives the maximum revenue by calculating revenues around the optimal price, like R(x0-0.1), and R(x+0.1). They should both be smaller than R(x0).
Post what you've got for confirmation if you wish.
This question is telling us:
1. The company charges $16 per subscription, at this rate, it has recruited 30000 customers, or a revenue of 30000*16=$480000.
2. If the company raises the price of the subscription, it will do so at the cost of losing 1000 customers for every dollar it raises.
This means that the number of customers N is a linear function of the price x, and the slope m is -1000.
Let N(x)=mx+b
But we know that N(16)=30000, so
30000=-1000(16)+b
or
b=30000+16000=46000
So now the function N(x) is completely defined:
N(x)=-1000x+46000
If we define the revenue R(x) as the product of the number of customers multiplied by the price, then
R(x)=x*N(x)=-1000x²+46000x
So R(16)=-1000(16)²+46000*16=$480000 as before.
To calculate the maximum, we will have to do this by tabulation:
For a given x, calculate R(x), and present it in table form:
R(16)=480000 (as before)
try
R(20)=520000
R(25)=525000
R(30)=480000
So we know that the maximum revenue occurs at 20<x<30.
Give a few more tries, and you should get the maximum.
In this case, the optimal price x0 happens to be a whole number of dollars. But you still have to prove that the price gives the maximum revenue by calculating revenues around the optimal price, like R(x0-0.1), and R(x+0.1). They should both be smaller than R(x0).
Post what you've got for confirmation if you wish.