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 $s$ are a linear function of the mark-up $p$, so:

$
s=a\times p+c
$

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:

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

so $c=140$, and:

$s=-0.1\,p+140$

Now profit $f$ is:

$
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:

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

so we seek solutions of:

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

or:

$
-0.2\,p+140=0
$

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

RonL

3. 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:

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

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

$200-20x=0$

$x=10$

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