Please check ans: how many calendars to max profit

[ochocki]

The Utah Ski Club sells calendars to raise money. The profit P, in cents, from selling x calendars is given by the equation P(x) = 360x - x^2

a. Find how many calendars must be sold to maximize profit.

b. Find the maximum profit.

To do this, I used the vertex formula x = (-b) / (2a)

Since a = -1 and b = 360, when plugged into the formula, I get x = 180. I believe this is the answer to part (a).

Now to get part (b). I was a little unsure. I tried plugging in 180 to x to see my y value which gave me 97200. If this is the answer to b. I am having a hard time trying to relate the two numbers. Can anyone please explain this to me?

 

[stapel]

You have:

. . . . .P(x) = -x² + 360x

...so a = -1, b = 360, and c = 0. You have the correct formula for the x-value of the vertex, h. This is the value that maximizes the profit P.

. . .h = (-b) / (2a) = (-360) / (-2) = 180

So they need to sell 180 units. The profit for 180 units is P(180). I get a different value from yours; I have a feeling you lost a "minus" sign somewhere. I'm not sure what you mean by "relating" the two numbers, since the exercise clearly defines the relationship: "the profit P gained by selling x calendars, measured in cents".

Eliz.

 

[ochocki]

Do you get the value 32400 for P(180)? I think you were right about the negative, I don't have the answer in my book, can someone please confirm for me.

 

[stapel]

Do you get the value 32400 for P(180)?

Yes. So the maximizing number of calendars is x = 180, and the maximal profit is $324.00.

Eliz.