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Revenue problem

  1. Apr 8, 2005 #1
    have final answer...cant figure out how to do it....i usually post my answers..but for this one i've been spending the past 2 days on it and not matching it with the correct answer in the textbook can someone please help me out as im homeschooling and have no teacher to ask...

    thanks in advance!!!

    The G-Food catering company finds that competitors cater lunch for a group of 100 people for $5 each. The manager of G-Food calculates that for each 25 cent discount per lunch, it is possible to sell an additional 10 lunches. If each lunch costs G-Food $2 to prepare, how many lunches should be prepared to maximize the profit?
  2. jcsd
  3. Apr 8, 2005 #2


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    The problem gives you a "base" of $5 each for 100 people- for the purposes of this problem I would interpret that as "if you charge $5 for each lunch, you will sell 100 lunches". Now, "for each 25 cent discount per lunch, it is possible to sell an additional 10 lunches.". Okay, let n, be the "number of '25 cent discounts'". Then the charge for each lunch is P= 5- 0.25n ($5 minus 25 cents for each discount). The number of lunches sold will be 100+ 10n (100 plus 10 for each discount). If each lunch costs $2 to prepare, the profit on each lunch is (5- 0.25n)- 2= 3- 0.25n.

    The TOTAL profit is, of course, the profit on each lunch times the number of lunches: (3- 0.25n)(100+ 10n). What should n be to make that a maximum? There are several ways to answer that- since that is a quadratic function, completing the square is probably easiest.

    And don't forget to ANSWER the question. After you know what n is, calculate the number of lunches that will be sold.
  4. Apr 8, 2005 #3
    well ye i know how to do that...but we did that last year...this question is from my calculus class...so i dont think its that easy!
  5. Apr 8, 2005 #4


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    Well, if this a calculus class, then it should be even easier- instead of completing the square, take the derivative and set it equal to 0!
  6. Apr 8, 2005 #5
    wat do u mean?
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