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Stephxox
09-28-2008, 03:09 PM
I`m stuck on a math question! Help please!

here it is:

A large dealership has been selling new cars at $6000 over the factory price. Salse have been averaging 80 cars per month. Because of inflation, the $6000 markup is going to be increased. The marketing manager has determined that, for ever $100 increase, there will be 1 less car sold each month. What should the markup be in order to maximize the revenue.

This is what I have:

MaxR= (100-1)(80+x)

Would this be the correct equation. I`m pretty sure I know how to do the question, I`m just wondering if, in fact, that is even remotly close to the right equation I should use.

Mrspi
09-28-2008, 03:23 PM
I`m stuck on a math question! Help please!

here it is:

A large dealership has been selling new cars at $6000 over the factory price. Salse have been averaging 80 cars per month. Because of inflation, the $6000 markup is going to be increased. The marketing manager has determined that, for ever $100 increase, there will be 1 less car sold each month. What should the markup be in order to maximize the revenue.

This is what I have:

MaxR= (100-1)(80+x)

Would this be the correct equation. I`m pretty sure I know how to do the question, I`m just wondering if, in fact, that is even remotly close to the right equation I should use.

It's difficult to assess the "correctness" of your equation, because you did not identify what x represents. I'm pretty certain, though, that your equation is incorrect. 80 is the current number of cars sold each month. If you increase the markup, that number is supposed to go DOWN, so I sure wouldn't expect to see 80 plus anything in a correct equation.

Let x = number of $100 increases

For each of these increases, the number of cars sold monthly will go down 1. So, if you have x increases, the number of cars sold will be 80 - 1*x, or 80 - x.

The markup is currently $6000. If you make x increases of $100 each in the markup, the new markup will be 6000 + 100x.

revenue = markup * number of sales

revenue = (6000 + 100x)(80 - x)

Find the maximum value of the expression on the right side....

Stephxox
09-28-2008, 03:53 PM
Thankyou soo much.
I got the correct answer!
:D :D :D