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February 10th, 2008, 02:39 PM   #1
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Word Problem

Need some help with this word problem. You don't have to give me the answer, just show me the steps and show your work, mathematically.

1) The cost of a ticket to a hockey arena searting 800 people is $3. At this price every ticket is sold. A survey indicates that if the price is increased, attendance will fall by 100 for every dollar of increase. What ticket price results in the greatest revenue. What is the greatest revenue.
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February 10th, 2008, 03:19 PM   #2
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Try writing a function what represents the number of peoples than will come to the arena if the price is x. Then, using this function, you can find the function that represents the revenue for a certain price x, and studying that you can get your answer. Still, if you want to see the answer, highlight the end of my post.

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So, if the number of people estimated to come to the arena falls by 100 for every new 1$ rising of the ticket cost then you should have a linear function describing it. The function should have the following values:

f(3)=800
f(4)=700
..

But these two will suffice. Now, since it has to be linear, it has to look like this:

f(x)=ax+b

and from our data, we get:

a*3+b=800

a*4+b=700

now, subtracting the two equations, we get:

4*a+b-3*a-b=700-800

a=-100

Plugging that into one of them (I'll use the first one) gives us:

(-100)*3+b=800

b=1100

So the equation looks like:

f(x)=1100-100x=100*(11-x)

Now, this isn't a hard function to come up with. You could use the fact that at the beginning it should equal 800, and as the difference x-3 grows (where x is the price), the function gets smaller at the same rate times 100. This way you could write:

f(x)=800-(x-3)*100

but that's the same thing. So, now we have this, we can write the revenue function. It equals the number of people that bought the ticket times the ticket's cost. That is:

R(x)=x*f(x)=100*x*(11-x)

Now we have to find this functions extrema. Since I don't know how much you know about functions, I'll write two ways to solve this. So, the first way: the first step is to realize our R function is an parabola. The second thing is to remember the fact that a parabola has it's extrema in a point that's exactly in the middle of it's roots (that is, the x component). The third step is to see how obvious it is that the roots are x=0 and x=11. So, the extrema happens at x_ex=(11-0)/2=5.5. And the value of the function at that point is f(x_ex)=3025. So, we've found out that if we set the ticket cost at 5.5$, the amount of money we'll get is 3025.

Now, the second way: find the derivative of R(x).

That is:

R'(x)=100(11-x)+100*x*(-1)=0

1100-100x-100x=0

200x=1100

x=5.5

And that's it! Hope I've helped!
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February 10th, 2008, 07:37 PM   #3
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Increasing the ticket price by $n loses $300n due to 100n unsold tickets, but raises an additional $(800 - 100n)n.

Hence the increase in revenue is $(500n - 100n²), i.e., $(625 - (25 - 10n)²), which has a greatest value of $625 when n = 25/10 = 2.5.

The corresponding new ticket price and revenue are easy to calculate.

This answer makes various assumptions, which ought to be stated. The most important one is that all tickets sold must have the same price.
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