May 7, 2010

Profit Maximization



A manufacturer can sell X items (X >= 0) at a price of Rs. (330 - X)
each. The cost of producing X items is ((X^2) + 10X + 12).
Determine the no. of items sold so that the manufacturer can make
the maximum profit..

3 comments:

  1. João MarquesMay 7, 2010 at 7:44 PM

    my answer is 80.
    we have the formula of the income money for x items:
    x*(330-x)
    and we have the formula for the cost:
    x^2+10*x+12

    we want to maximize the profit
    max (income - cost)

    max(330*x-x^2-x^2-10*x-12)
    = max(-2*x^2+320*x-12)

    parabola.... max point is x=-b/2*a
    =-320/-4 = 80

    ReplyDelete
  2. PratapMay 23, 2010 at 10:39 PM

    I second what Marques answer is.

    80

    Profit = (330-x)x - (x^2 +10x+12)
    = 320x - 2x^2-12
    = 2x(160-x) -12

    So we need to find max (2x(160-x))
    which is achieved at 80

    ReplyDelete
  3. UnknownMay 31, 2010 at 2:07 AM

    yes that is true, the maximum point is when the derivative of -2x^2+320x-12 equals 0 (horizontal slope)

    that is -4x+320=0 ==> x=-320/-4 = 80

    ReplyDelete