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Profit Equation (need to maximze profit)

  1. Apr 28, 2004 #1
    An automobile manufacturer finds that 80,000 cars can be sold if each is priced at $12,000. However, the number sold increases by 24 for every $1 decrease in the price. The manufacturer has fixed costs of $45,000,000; in addition, it costs $5,500 to produce each car. How should the cars be priced to maximize profits?

    I can do this by trial and error but I would like to know how to set up and do this problem the right way. Thanks.
     
  2. jcsd
  3. Apr 28, 2004 #2
    Construate the function of profit(price of a car), what means find out how does the profit depend on the price. Then find the maximum of that function using derivatives.
     
  4. Apr 28, 2004 #3

    HallsofIvy

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    "An automobile manufacturer finds that 80,000 cars can be sold if each is priced at $12,000. However, the number sold increases by 24 for every $1 decrease in the price."

    Think of this as y= mx+b where if x= 12000, y= 8000. Also "number sold increases --by 24 for every $1 decrease in the price is the same as saying that the slope of the line is m= -24. y= -24x+ b so 8000= -24(12000)+ b. b= 8000+ 24(12000)= 296000.
    That is: number sold= 296000- 24(price) or N= 296000-24p.

    Of course, the total money brought in is just the number sold times the price:
    Np or 296000p- 24p2.

    From that, you have to deduct the 45000000 fixed costs and 5500 for each car: a total cost of 450000000+ 5500N= 450000000-5500(296000-24p).

    The profit will be: income - cost or
    profit= 296000p- 24p2-450000000-5500(296000-24p).

    What value of p will make that a maximum?

    (I can think of two ways of doing it: set the derivative equal to 0 or, since this is a quadratic, complete the square to find the vertex of the parabola. One of those should be familiar to you.)
     
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