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Thread: 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; Dec 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 $\displaystyle s$ are a linear function of the mark-up $\displaystyle p$, so:

    $\displaystyle
    s=a\times p+c
    $

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

    $\displaystyle
    80=(-1/10)\times 600+c
    $

    so $\displaystyle c=140$, and:

    $\displaystyle s=-0.1\,p+140$

    Now profit $\displaystyle f$ is:

    $\displaystyle
    f=s\times p=(-0.1\,p+140)p=-0.1p^2+140p
    $

    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:

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

    so we seek solutions of:

    $\displaystyle \frac{df}{dp}=0$,

    or:

    $\displaystyle
    -0.2\,p+140=0
    $

    which has a solution $\displaystyle p=700$, which corresponds to a maximum
    of $\displaystyle f$ as $\displaystyle d^2f/dp^2$ 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.

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

    Expand and differentiate:

    $\displaystyle -10x^{2}+200x+48000$

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

    $\displaystyle 200-20x=0$

    $\displaystyle x=10$

    $\displaystyle (600+10(10))(80-10)=(700)(70)$
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