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Math Help - Optimization

  1. #1
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    Post Optimization

    I have a question that might be on my test tomorrow but I'm not sure how to do it.
    Can you guys help me?

    A large car dealership has been selling new cars at $600 over the factory proce. Sales have been averaging 80 cars per month. Due to inflation the $600 mark-up is going to be increased. The marketing manager has determined that for $10 increase there will be one less car each month. What should be the new markup in order to maximize profit?


    Thank you very much
    Last edited by dgolverk; December 13th 2006 at 02:30 PM.
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  2. #2
    Grand Panjandrum
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    Quote Originally Posted by dgolverk View Post
    I have a question that might be on my test tomorrow but I'm not sure how to do it.
    Can you guys help me?

    A large car dealership has been selling new cars at $600 over the factory proce. Sales have been averaging 80 cars per month. Due to inflation the $600 mark-up is going to be increased. The marketing manager has determined that for $10 increase there will be one less car each month. What should be the new markup in order to maximize profit?


    Thank you very much
    The sales s are a linear function of the mark-up p, so:

    <br />
s=a\times p+c<br />

    for some a and c, we are told that a $10 increas
    in mark up will reduce sales by 1 car, so a=1/10, then as
    current sales are 80 for a markup of $600, we have:

    <br />
80=(-1/10)\times 600+c<br />

    so c=140, and:

    s=-0.1\,p+140

    Now profit f is:

    <br />
f=s\times p=(-0.1\,p+140)p=-0.1p^2+140p<br />

    To find the mark-up that maximises profit we need to take the derivative
    of profit set this to zero and solve for the mark-up. Now:

    \frac{df}{dp}=-0.2p+140,

    so we seek solutions of:

    \frac{df}{dp}=0,

    or:

    <br />
-0.2\,p+140=0<br />

    which has a solution p=700, which corresponds to a maximum
    of f as d^2f/dp^2 is negative there.

    RonL
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  3. #3
    Eater of Worlds
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    You could also think of it this way.

    \underbrace{(600+10x)}_{\text{amount over mark up}}\cdot\overbrace{(80-x)}^{\text{number sold}}

    Expand and differentiate:

    -10x^{2}+200x+48000

    \frac{d}{dx}[-10x^{2}+200x+48000]=200-20x

    200-20x=0

    x=10

    (600+10(10))(80-10)=(700)(70)
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