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Pls Help Calculus

  1. Dec 13, 2003 #1
    Pls Help!!! Calculus!!!

    A manufacturer can produce jackets at a cost of $50 per jacket. If he produces p jackets weekly and all the jackets are sold, the revenue is R(p)=4000[e^.01(p-100) + 1]. Weekly production must be at least 100 and can't exceed 250...... at wat price should the manufacturer sell the jackets to maximize profit... .and wat is the maximum weekly profit...
    PLEASE HELP!!!!!
  2. jcsd
  3. Dec 13, 2003 #2
    the e^(.01(p-100)) ....... the .01(p-100) is to the base e
  4. Dec 13, 2003 #3
    RE: Pls Help!!! Calculus!!!

    Solve [tex] \frac {dR} {dp} = 0 [/tex] for [tex]p[/tex]

    Then you can use the second derivative test to see if [tex]R(p)[/tex] is a maximum or minimum.

  5. Dec 13, 2003 #4
    read question carefully i need to find the max price of the jackets!!!..... i don't understand this question
  6. Dec 14, 2003 #5
    The manufacturer wants to maximize profit thus he wants to maximize revenue.

    What you need to do is find the number of jackets [tex]p[/tex] that maximizes revenue. The first and second derivatives of [tex]R(p)[/tex] help you find that value. Once you know the number of jackets that maximizes revenue then you can find the revenue that those jackets will generate. Then you can find the revenue that one jacket will generate which is the price. Once you know the price then you know the profit for each jacket.

    First you must find the derivative:


    Then solve:


    for [tex]p[/tex]. Substituting the [tex]p[/tex] thus found into the original [tex]R(p)[/tex] will give you the revenue.

    Then the price is:


    and the profit is:


  7. Dec 14, 2003 #6
    thanks doug... but i have a problem here.... i found dR/dp= e^.01(p+100)..... after i make that equal to 0.... but e^x can't be zero!!!..... so wat do i do???
  8. Dec 14, 2003 #7
    Then one of the endpoints of the interval takes the prize. Either


    or vice-versa.

  9. Dec 15, 2003 #8


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    Staff Emeritus
    Science Advisor

    HOLD ON!

    It's not the revenue you want to maximize but the profit!

    The revenue is given as [tex]4000(e^{.01(p-100)}+1)[/tex] (which is obviously increasing for all x) and the cost function ("cost of $50 per jacket") 50p. The profit is [tex]4000(e^{.01(p-100)}+1)- 50p[/tex].

    In order to find the maximum profit, you need to find
    [tex]\frac{dP}{dp}=\frac{d}{dp}4000(e^{.01(p-100)}+1)-50p[/tex] and set it equal to zero.
    That derivative is 400e^{.01(p-100)}- 50= 0 so
    400e^{.01(p-100)}= 50

    (Economists would say "marginal revenue equal marginal cost"- the revenue from selling "one more" jacket is equal to the cost of producing it.)

    Hmm. The "marginal revenue" is still an exponential with a positive exponent so increasing: the maximum profit is still produced by producing and selling the largest possible number: 250 where the profit is about $9500.
  10. Dec 15, 2003 #9
    Although we come up with the same answer your reasoning is more universal, shall we say
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