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Math Help - Calculus Word Problem

  1. #1
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    Post Calculus Word Problem

    Hi, everyone. I'm new to the site and I need help setting up a calculus problem. Here it is:

    The stadium vending company finds that sales of hot dogs average 34000 hot dogs per game when the hot dogs sell for $2.50 each. For each 50 cent increase in the price, the sales per game drop by 5000 hot dogs. What price per hot dog should the vending company charge to realize the maximum revenue?

    Thanks in advance.
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  2. #2
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    Try defining some variables, such as revenue, quantity sold, and cost per hot dog... then try to express the revenue as a function of a single variable.

    Once you start doing some of the work it is much easier to give a reply. Welcome to the site btw
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  3. #3
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    Thanks for the reply, but I'm just having trouble setting up the actual equation.
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  4. #4
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    Well try this

    let p be the price per hot dog
    let q be the quantity of hot dogs sold
    let R be the revenue

    I think it should be very clear that R=pq

    but this is a function of two variables. We have a constraint, however. What you should do is eliminate q from the expression for R. This should be easy because they told you that there is a linear relationship between q and p (you know a point and the slope, so you know every point on the line).



    For the R function, after you make your substitution, you will get a 2nd degree polynomial.
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  5. #5
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    Not to sound ungrateful, but I'm aware of all that (should have made that more clear). It's just I'm having trouble plugging everything in and need the first line and could take it from there.
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  6. #6
    Member billa's Avatar
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    Not to sound ungrateful
    no worries

    It's just I'm having trouble plugging everything in
    Alright, if you give me the expression that relates p and q, then I will happily plug it in for you.

    To be honest, I can't really figure out what you are having a problem with
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  7. #7
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    Quote Originally Posted by railroadrider View Post
    Hi, everyone. I'm new to the site and I need help setting up a calculus problem. Here it is:

    The stadium vending company finds that sales of hot dogs average 34000 hot dogs per game when the hot dogs sell for $2.50 each. For each 50 cent increase in the price, the sales per game drop by 5000 hot dogs. What price per hot dog should the vending company charge to realize the maximum revenue?

    Thanks in advance.
    let x be the quantity of hot dogs sold
    let p(x) be the price function
    let R(x) be the revenue function

    The stadium vending company finds that sales of hot dogs average 34000 hot dogs per game when the hot dogs sell for $2.50 each so p(x) has the ordered pair (34000, 2.50)

    For each 50 cent increase in the price, the sales per game drop by 5000 hot dogs so p(x) has slope .50/-5000

    So p(x) - 2.50 = -.50/5000 (x - 34000)

    Solve for p(x) then find R(x)=xp(x)

    You can find the value of x that gives the maximum revenue by finding the derivative of R(x), setting it equal to zero and solving for x. If the derivative of R(x) changes from positive to negative at that value of x, you know that R(x) is maximum at that value of x.
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  8. #8
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    Quote Originally Posted by ione View Post
    let x be the quantity of hot dogs sold
    let p(x) be the price function
    let R(x) be the revenue function

    The stadium vending company finds that sales of hot dogs average 34000 hot dogs per game when the hot dogs sell for $2.50 each so p(x) has the ordered pair (34000, 2.50)

    For each 50 cent increase in the price, the sales per game drop by 5000 hot dogs so p(x) has slope .50/-5000

    So p(x) - 2.50 = -.50/5000 (x - 34000)

    Solve for p(x) then find R(x)=xp(x)

    You can find the value of x that gives the maximum revenue by finding the derivative of R(x), setting it equal to zero and solving for x. If the derivative of R(x) changes from positive to negative at that value of x, you know that R(x) is maximum at that value of x.
    Thank you kindly. That was very illuminating and just what I needed.
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