The sales are a linear function of the mark-up , so:

for some and , we are told that a $10 increas

in mark up will reduce sales by 1 car, so , then as

current sales are 80 for a markup of $600, we have:

so , and:

Now profit is:

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:

,

so we seek solutions of:

,

or:

which has a solution , which corresponds to a maximum

of as is negative there.

RonL