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Derivative and marginal profit

  1. Nov 10, 2009 #1
    1. The problem statement, all variables and given/known data

    A company's Marginal profit is P'(x) = -0.0105x^2 + 0.14x + 25 where x represents hundreds of items and profit is thousands of dollars
    a) Estimate the change in profit it the production is raised from 80 to 81 hundred units. Should the manager increase production?
    b)How many hundreds of items should the company be making in order to maximize profit?

    2. Relevant equations

    ∆y≈dy=f '(c)dx

    3. The attempt at a solution

    ∆P = dP = P'(c) dx
    C=8000, dx=100
    dP= [-0.01059(8000)^2 + 0.14(8000) + 25](100)
    So solving this equation I got dP = -67085500
    And since the profit is expressed in thousands of dollars, I divided my answer by 1000. So I got $ -67085.5 for the answer to question (a). So I wrote profit [increase] by $ 67085.5
    Am I doing right? Because my prof explained a sample probelm similar to this, and the answer was positive, so she wrote :decreases by xxx. And I thought if the answer is negative, I should write opposite from her. So I concluded it is increasing. But I am not sure.

    And about problem (b), will I be able to get maximum if I set P'(x)=0 and have a test?
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Nov 10, 2009 #2

    Mark44

    Staff: Mentor

    One thing I see right away is that you are using different units than are given in the problem. x is in units of hundreds, and the marginal profit is in terms of thousands of dollars. Do your calculations in terms of those units, and then, later, do your conversions. For example, C = 80 and dx = 1. See what you get with those numbers.
     
  4. Nov 10, 2009 #3
    Oops! Thank you so much. the answer was -31, using 80 as c and dx as 1. So profit has $ 31000 increased?
     
  5. Nov 10, 2009 #4

    Mark44

    Staff: Mentor

    The profit will decrease by about $31,000.
     
  6. Nov 10, 2009 #5
    Thank you so much. :)
     
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