SOLUTION: A local Dunkin Donuts sells their original blend of Dunkin DOnuts ground coffee for $8.99 per pound. On average, the local Dunkin Donuts sells about 48 one-pound bags a day. After

Question 675658: A local Dunkin Donuts sells their original blend of Dunkin DOnuts ground coffee for $8.99 per pound. On average, the local Dunkin Donuts sells about 48 one-pound bags a day. After doing some market research, they found that for each $0.50 increase in the price that they will lose one costumer per day. What price should they set the coffee at in order to maximize their profit? What price above $8.99 will cause them to lose money? Sketch a graph to support all of your results.
Answer by jsmallt9(3758) (Show Source): You can put this solution on YOUR website!
First of all, this problem is badly worded. Profits (and losses) cannot be determined without some way of determining the costs of producing and selling the coffee. The problem tells us nothing about costs. All we know is how to determine revenue (the amount of money received from the sale of the coffee). So instead of "maximize profit" (which is impossible to find) we are going to maximize revenue. And instead of finding the price which causes them to "lose money" (which is also impossible to find) we are going to find the price above $8.99 that causes them to have less revenue than the revenue they make selling 48 pounds at $8.99.

The revenue Dunkin Donuts gets from selling coffee is simply the number of pounds they sell times the price for each pound. So we need to find expressions for the number of pounds and the price of each pound. These expressions will be based on how many $0.50 increases have been made:
Let n = the number of $0.50 increases in the price that have been made. Since 1 customer is lost for each of these increases, the number of pounds sold will be the original number minus n:
number of pounds sold = 48-n
And the price of each pound will be $8.99 plus n times $0.50:
price per pound = 8.99 + n*0.50 = 8.99 + 0.50n

Now we can write an equation for revenue:

Simplifying...

This is a quadratic equation. So we'll put the terms in the standard order:

The graph of a quadratic equation is a parabola. In this case, with the negative coefficient of the squared term, the parabola opens downward. Once we understand this we should recognize that the maximum revenue will occur at the vertex of this downward opening parabola. To find this maximum we will take advantage of the fact that the vertex occurs when x (or, in this case, n) equals -b/2a. So the value for the n of the vertex will be:

which simplifies to:

This is the value of n the gives the maximum revenue. But before we calculate the price, we must decide if n can be a non-integer like 15.01. The problem does not suggest that other price increases are possible. So we must round this value to an integer. I think we can safely round this n to 15. (you can double check this by trying 16, too and see which value results in the higher revenue.)
price per pound = 8.99 + 0.50n = 8.99 + 0.50(15)
I'll leave it up to you to work this out.

To find the price where the revenue starts declining we can take advantage of the symmetry of parabolas. When n = 0 the revenue is $431.52. When n = 15.01 we get the maximum revenue (at the vertex). Because of the symmetry, it will take another 15.01 change to n (IOW, n = 30.02) to bring the revenue back down to the original revenue. Any n greater than 30.02 will cause the revenue to fall below the initial revenue of $431.52. Again we should round n to an integer. We must round to the next integer, 31. (Not 30. an n of 30 would have a revenue slightly higher than $431.52. (Try it and see.) We must round up to ensure we go past the "break-even" n of 30.02.) So the price which brings the revenue down below $431.52 is:
price per pound = 8.99 + 0.50n = 8.99 + 0.50(31)
I'll leave it up to you to work this out.

Here's what the important part of the graph looks like:

NOTE: The scaling on the two axes are different! So the parabola is really much steeper than you see here. (I did this do you could see all the important parts of the graph without having to make it an enormous graph.)
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