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Math Help - optimization of a social gathering using N(t)=30t-t^2

  1. #1
    Junior Member surffan's Avatar
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    optimization of a social gathering using N(t)=30t-t^2

    The question is
    the interactions at a social gathering follow the mathematical progression N(t)=30t-t^2, where t is the time in minutes since the party began, and N is the number of separate conversations occuring. what is the maximum number of interactions? thats all the information i was given. I'm not sure what to do with this problem
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  2. #2
    Super Member craig's Avatar
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    Quote Originally Posted by surffan View Post
    The question is
    the interactions at a social gathering follow the mathematical progression N(t)=30t-t^2, where t is the time in minutes since the party began, and N is the number of separate conversations occuring. what is the maximum number of interactions? thats all the information i was given. I'm not sure what to do with this problem
    To find the maximum or minimum value of your function you need to differentiate it and set it equal to zero.

    In the case above we differentiate wrt t.

    N(t) = 30t - t^2

    N'(t) = 30 - 2t = 0

    So t = 15

    You then substitute this value of t back into your original equation to find the maximum value.
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  3. #3
    Super Member craig's Avatar
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    Sorry I should probably mention about minimum values.

    If you differentiate your original function, set equal to zero and solve, you may get 1, 2 or inifite solutions, depending on the type of your equation.

    To check whether it is a maximum or minimum value you that you have found, you take the second derivative and then substitute in your value(s) again.

    If the function is less than 0, then it is a maximum, if it is greater than zero then it is a minimum.

    Hope that wasn't too confusing
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  4. #4
    Junior Member surffan's Avatar
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    thank you so much, i dont know why i found it so difficult, i kept forgetting to carry the 2 when a function is squared to calculate the derivative. I have another question that is similar except my derivative doesnt leave me with much. its to maximize revenue and the function is 1200(1.50-x)
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  5. #5
    Super Member craig's Avatar
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    Quote Originally Posted by surffan View Post
    thank you so much, i dont know why i found it so difficult, i kept forgetting to carry the 2 when a function is squared to calculate the derivative. I have another question that is similar except my derivative doesnt leave me with much. its to maximize revenue and the function is 1200(1.50-x)
    If you have a new question please in future start a new thread.

    However as this is extremely similar to your original I'll just answer it here.

    As you will have noticed when you differentiate it, the variable (x) disappears, in this case, there is no maximum as the function is unbounded, so the answer is infinity or undefined.
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  6. #6
    Junior Member surffan's Avatar
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    for the function 1200(1.50-x)
    this is what i got so far, i dont know if its right
    (1800-1200x)x
    1800x-1200x^2
    f'(x) = 1800-2400x
    x=0.75
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  7. #7
    Super Member craig's Avatar
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    Quote Originally Posted by surffan View Post
    for the function 1200(1.50-x)
    this is what i got so far, i dont know if its right
    (1800-1200x)x
    1800x-1200x^2
    f'(x) = 1800-2400x
    x=0.75
    Where has this extra x magically appeared from?
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  8. #8
    Junior Member surffan's Avatar
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    I used revenue =price x quantity
    the equation represents quantity and the (x) represents price, so i multiplied the two to get an equation for revenue and took the derivative to solve for max and min, and i got 0.75
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  9. #9
    Super Member craig's Avatar
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    In that case then you're correct.

    In future though please create new threads for new questions.

    And if you are going to ask a question, post all the relevant information, not just bits of it
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  10. #10
    MHF Contributor

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    By the way, both of these problems involve quadratics so you could have found max or min by completing the square rather than differentiating.
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