SOLUTION: A telephone company in a town has 500 subscribers on its list and collects fixed charges of Rs 300/- per subscriber per year. The company proposes to increase the annual subscripti

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Question 842454: A telephone company in a town has 500 subscribers on its list and collects fixed charges of Rs 300/- per subscriber per year. The company proposes to increase the annual subscription and it is believed that for every increase of
Re 1/- one subscriber will discontinue the service. Find what increase will bring maximum profit?
Answer by Theo(13342) About Me  (Show Source):
You can put this solution on YOUR website!
i believe your equation is going to be:
y = (300+x) * (500-x)
300+x represents the cost per subscriber.
500-x represents the number of subscribers.
every time the cost per subscriber goes up 1, the number of subscribers goes down 1.
if you multiply these factors out, you will get:
y = 150,000 - 300x + 500x - x^2 which you can simplify to:
y = 150,000 + 200x - x^2
place this equation in standard form to get:
y = -x^2 + 200x + 150,000
the maximum point in this equation will be at x = -b/2a
a = -1
b = 200
c = 150,000
x = -b/2a will be at x = -200/-2 which would be at x = 100.
when x = 100, y will be equal to (300+100) * (500-100) = 400 * 400 = 160,000.
when x = 100, profit will be at 160,000.
you can graph this equation to see what it looks like.
it will be a parabola that points up and opens down because the coefficient of the x^2 term is negative.
here's the graph:

I added 2 horizontal lines.

the first horizontal line is at 150,000 which is the profit when x = 0.
when x = 0, the cost per subscriber is equal to 300 and the number of subscribers is equal to 500.
the second horizontal line at 160,000 which is the profit when x = 100.
when x = 100, the cost per subscriber is equal to 400 and the number of subscribers is also equal to 400.
you can see this easier from the equation in the form of:
y = (300+x) * (500-x)
the 300+x represents the cost per subscriber.
the 500-x represents the number of subscribers.
for find minimum point of the equation, conversion to standard form made that easy because the formula for finding the maximum point in the equation is derived from the standard form of the equation.