How can calculus help maximize profit for Goodfood catering company?

[m0286]

I am stuck on a calculus problem.. I have most of the answer but I suddenly got confused, and can't figure it out any further, the question is:

The goodfood catering company finds that competitors cater lunch for a group of 100 people for $5 each. The manager of Goodfood calculates the for each 25 cent discount per lunch, its possible to sell an additional 10 lunches. If each lunch costs goodfood $2 to prepare, how many lunches should be prepared to maximize profit.
This is what I got so far:

let P represent profit, let x represent # of discounted of lunches
P=(3-0.25x)(100+10x)
=-2.5x^2+5x+300
for the derivative i got x=1.
When i substituted that into the above equationi got:
=-2.5(1)^2+5(1)+300
=302.5 HERES WHERE I AM LOST!
Is this 302.5, the amount of profit they make or is this the number of lunches they should make to make greatest profit.?

 

[robert Ihnot]

302.5 is the amount of profit. The price of the lunch is $2.75, which represents X=1. When I wrote the problem out I used:

($5.00-.25x)(100+10x)-(100+10x)(2.00), which is Revenue minus Expenses = Profit.

Note that the profit would have been $300 had we not reduced the price. This happens to be the same value we would get if we dropped the price by $.50: ($2.50)(120) = $300. And it is downhill from there.

 

[m0286]

Thanks for that help, but would you be able to help me with how I would find how many lunches should be prepared to reach maximum profit?

 

[robert Ihnot]

It is right in the equation, since x=1, the number is 100+10x = 110.