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Calculus please help asap-due tomorrow

  1. May 9, 2005 #1
    I am stuck on a calculus problem.. I have most of the answer but I suddenly got confused, and cant 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.???
     
  2. jcsd
  3. May 9, 2005 #2
    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.
     
    Last edited: May 9, 2005
  4. May 9, 2005 #3
    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?
     
  5. May 9, 2005 #4
    It is right in the equation, since x=1, the number is 100+10x = 110.
     
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