Advertisement
If you have a new account but are having problems posting or verifying your account, please email us on hello@boards.ie for help. Thanks :)
Hello all! Please ensure that you are posting a new thread or question in the appropriate forum. The Feedback forum is overwhelmed with questions that are having to be moved elsewhere. If you need help to verify your account contact hello@boards.ie
Hi there,
There is an issue with role permissions that is being worked on at the moment.
If you are having trouble with access or permissions on regional forums please post here to get access: https://www.boards.ie/discussion/2058365403/you-do-not-have-permission-for-that#latest

Some help for the old scientist!

  • 17-02-2010 9:19pm
    #1
    Registered Users, Registered Users 2 Posts: 378 ✭✭


    I expect to sell 100L of beer each night at 3$ per litre. For every 2$ increase price per litre, I expect to sell 5 less litres per night.

    I need an equation to the above problem so I can maximize my net revenue using differentiation.

    Thanks.


Comments

  • Closed Accounts Posts: 190 ✭✭Chewbacca.


    Einstein? wrote: »
    I expect to sell 100L of beer each night at 3$ per litre. For every 2$ increase per litre, I expect to sell 5 less litres per night.

    I need an equation to the above problem so I can maximize my net profit using differentiation.

    Thanks.
    for max profit you sell 3litres less per night
    a-100l=3
    a-95l=5
    solve simul==> L=2/5 a=43
    p(x)price function=a-lx
    = 43-2/5x
    put your price function into profit function B(x)=x[p(x)]
    set to 0 and x=53.75 max profit


  • Registered Users, Registered Users 2 Posts: 3,745 ✭✭✭Eliot Rosewater


    Well the question is flawed. What you're really maximizing is revenue, not profit. Revenue = profit - cost, where cost is based on the quantity sold.



    I made up an equation for this but as per the forum charter Im hesitant to just throw it up. Do you have any idea what to do?

    Hint: Your equation will have two parts: the maximum that can be sold less lost sales due to price increases. Given that revenue = price*quantity its best to just find the quantity sold first and then multiply by the price.


  • Registered Users, Registered Users 2 Posts: 378 ✭✭Einstein?


    Well the question is flawed. What you're really maximizing is revenue, not profit. Revenue = profit - cost, where cost is based on the quantity sold.



    I made up an equation for this but as per the forum charter Im hesitant to just throw it up. Do you have any idea what to do?

    Hint: Your equation will have two parts: the maximum that can be sold less lost sales due to price increases. Given that revenue = price*quantity its best to just find the quantity sold first and then multiply by the price.

    Hey I'm in first year DCU, I'm really good at maths but can't seem to get around this one, I've been doing a whole project for the past 6 hours and I'm mentally exhausted. I know some of the wording is flawed when it comes to revenue= profit - cost ( That's right ).

    Anyway, the way I see it is...
    Revenue = price x quantity
    300 = 3 x 100
    475 = 5 x 95
    ..etc

    But I can't see how to put it into an equation; I'll give it a shot:

    The revenue is quite irrelevant, the price and quantity are your variables and one is dependent on the other by something like:

    3+n =100 - 5 ( n-1 ) where n =2 but that doesn't do anything.

    I want something that has like an x and a y so that when I put in any value for one variable, I get the corresponding value for the other.


  • Registered Users, Registered Users 2 Posts: 3,745 ✭✭✭Eliot Rosewater


    I think you have to put it in terms of revenue.

    I dont think you can maximize the price just in terms of quantity.

    But youre on the right track. If you get the quantity in terms of the price, you can easily create a revenue/net profit equation and then maximize that.

    We will say the price is p and the quantity is q.

    q = initial quantity - quantity lost due to price increase.

    Initial quantity is obviously 100. The quantity loss is dependent on the price.
    For p=3 the quantity lost is 0.
    For p=5 the quantity lost is 5.
    For p=7 the quantity lost is 10.
    ...
    For p=n the quantity lost is ???


  • Registered Users, Registered Users 2 Posts: 378 ✭✭Einstein?


    I think you have to put it in terms of revenue.

    I dont think you can maximize the price just in terms of quantity.

    But youre on the right track. If you get the quantity in terms of the price, you can easily create a revenue/net profit equation and then maximize that.

    We will say the price is p and the quantity is q.

    q = initial quantity - quantity lost due to price increase.

    Initial quantity is obviously 100. The quantity loss is dependent on the price.
    For p=3 the quantity lost is 0.
    For p=5 the quantity lost is 5.
    For p=7 the quantity lost is 10.
    ...
    For p=n the quantity lost is ???

    How can you possibly do that, even if you get some q loss = ??? thats still not an equation. And I can't do that ???
    One increases in increments of +2 the other in +5 but there is no link between them if you know what I mean. Could you give me all your info you think I can work on please


  • Advertisement
  • Registered Users, Registered Users 2 Posts: 3,745 ✭✭✭Eliot Rosewater


    I meant that you should make the pattern into an equation :)

    So the quantity lost due to price is [latex]\displaystyle \left(\frac{p-3}{2}\right)5[/latex] where p is price I think.

    Sub in a few values and you will see it will work with the patter above.


  • Registered Users, Registered Users 2 Posts: 378 ✭✭Einstein?


    Ok that works but thats only for p and q;

    (p-3) / 5 = q will only tell you the price/loss in quantity by subbing in arbitrary values.


    Referring back to my original question, I need to increasing the price to its maximum possible value which will in turn make the most money. I know I can do that many different ways, but I want an equation which will let me do it with differentiation.


  • Registered Users, Registered Users 2 Posts: 3,745 ✭✭✭Eliot Rosewater


    I know Im doing this convolutedly but its probably a better way to learn.

    So we know that the quantity, [latex]\displaystyle q=100-\left(\frac{p-3}{2}\right)5[/latex]

    The price is p. Revenue is q*p.

    So revenue = [latex]\displaystyle \left[100-\left(\frac{p-3}{2}\right)5\right]p[/latex]

    And then maximize for p. Do you understand that?


  • Registered Users, Registered Users 2 Posts: 378 ✭✭Einstein?


    Brilliant, worked perfectly.

    Cheers.


    p.s admin can you delete this thread please thanks.


Advertisement