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-   -   Optimization (Differential?) (https://www.askmehelpdesk.com/showthread.php?t=74828)

  • Mar 22, 2007, 11:56 AM
    soyuz02
    Optimization (Differential?)
    Hi,

    I am working on an assignment for University and I don't seem to be getting anywhere.

    The question states that a company charges 200$ for each order of 150 or fewer shares. The cost to the buyer on every share is reduced by 1$ for each order in excess of 150.

    For what size order is the revenue maximum?

    Ive come up with the following but it seemed to be too easy:

    Let the number of shares = s

    I have deduced that the company charges 200/150 so 4/3 per share.

    Total Revenue=200 + 4/3s - s

    The 200 is the min charge which the company seems to apply, 4/3s is what the company charges per share and -s subtracts 1 x number of shares from revenue.

    Any idea if this is correct?
    Where would I go from here?

    Normally I would have expected to come up with a quadratic equation which I could then have solved and differentiated to find the values.

    Thank you very much in advance!
  • Mar 22, 2007, 12:10 PM
    galactus
    You know Revenue is the number sold times the cost per item.





    Set to 0 and solve for x, we get x=25.

    Now, you can see the number to sell to maximize revenue?
  • Mar 22, 2007, 01:05 PM
    soyuz02
    You're a mircale!

    Id like to learn out of this however so maybe you can just quickly explain why or how you got the revenue equation in the first place, (200-x)(150+x).

    Thank you v much!
  • Mar 22, 2007, 02:02 PM
    galactus


    Revenue = cost per item times items sold.

    Let x = the number of units added. For every x items added to 150, the price drops by
    $1x dollars.

    Hence,

    Using what we found from differentiating, we find $175 should be charged per order and sell 175 of them.

    175*175=$30625 total max revenue.


    Does this help? I don't know how to explain it any better.


    Cheers,
    Cody
  • Mar 22, 2007, 03:02 PM
    soyuz02
    Thanks, you've helped me a great deal :)
  • Mar 22, 2007, 03:08 PM
    galactus
    You're very welcome. A large part of the ingrates who post never have the common courtesy to say thanks.

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