# Thread: Maximizing revenue equation cannot understand the method

1. ## Maximizing revenue equation cannot understand the method

Here is how my text book puts it:

Maximizing Revenue
a. $R = f (p) = p . q = p(2400-20p) = 2400p - 20p^2$
b. Manipulate the demand function algebraically by subtracting $q$ from both sides, adding $20p$ to both sides and dividing both sides by $20$

The result should be $p = 120-0.05q$

I can't figure out how to get to that result. Any help would be greatly appreciated.

2. ## Re: Maximizing revenue equation cannot understand the method

Originally Posted by chicapsy
Here is how my text book puts it:

Maximizing Revenue
a. $R = f (p) = p . q = p(2400-20p) = 2400p - 20p^2$
b. Manipulate the demand function algebraically by subtracting $q$ from both sides, adding $20p$ to both sides and dividing both sides by $20$

The result should be $p = 120-0.05q$

I can't figure out how to get to that result. Any help would be greatly appreciated.
What are you trying to do?

You have a demand function:

$q=2400-20p$

where $q$ denotes the quantity that can be sold (the demand) at unit price $p$. Then the revenue is:

$R=p.q=2400p-20p^2$

Now there are several methods of finding the price $p$ that maximises the revenue $R$. If you know calculus you can differentiate the revenue with respect to price, set the derivative to zero and solve for the price which will be the price that maximises revenue... Alternatively you can complete the square (which if you do not have calculus you will have been taught either in your current course of your algebra or pre-calculus course).

Either way you will find the price which maximises revenue is 60 quatloos from which you can find the demand and revenue at the maximum.

CB

3. ## Re: Maximizing revenue equation cannot understand the method

Originally Posted by CaptainBlack
What are you trying to do?

You have a demand function:

$q=2400-20p$

where $q$ denotes the quantity that can be sold (the demand) at unit price $p$. Then the revenue is:

$R=p.q=2400p-20p^2$

Now there are several methods of finding the price $p$ that maximises the revenue $R$. If you know calculus you can differentiate the revenue with respect to price, set the derivative to zero and solve for the price which will be the price that maximises revenue... Alternatively you can complete the square (which if you do not have calculus you will have been taught either in your current course of your algebra or pre-calculus course).

Either way you will find the price which maximises revenue is 60 quatloos from which you can find the demand and revenue at the maximum.

CB
Thank you for the help, unfortunately I didn't understand so I asked my professor to break it down for me.

What I was doing wrong was using the incorrect part of the equation, i had to find the price in terms of quantity. So i use $q=2400-20p$

So here are my workings for anyone it can help out too:

$20p + q = 2400$
$20p = 2400 - q$
$p = 120 - 0.05q$

and then to find revenue

$R = p.q$
$(120 - 0.05q).q$
$R = 120q - 0.05q^2$