1. ## 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 2. Originally Posted by dgolverk 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

3. 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:

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

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

$\displaystyle 200-20x=0$

$\displaystyle x=10$

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