math problem

This is a discussion on math problem within the A Brief History of Cprogramming.com forums, part of the Community Boards category; I am having trouble on a math problem. I have tried everything I know to do, but I cannot seem ...

  1. #1
    l'Anziano DavidP's Avatar
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    Question math problem

    I am having trouble on a math problem. I have tried everything I know to do, but I cannot seem to find the answer. (Well, I get an answer, but I do not think it is correct).

    I will write out the problem, show you my work, and my answer. Tell me your thoughts.

    The Problem:

    The cost per unit for the production of a certain radio model is $60. The manufacturer charges $90 per unit for orders of 100 or less. To encourage large orders, the manufacturer reduces the charge by $0.15 per radio for each unit ordered in excess of 100.

    a. Write the profit as a function of x.
    b. How many radios should he sell to maximize profit?

    My work:

    So there are a few different functions here.

    The cost to produce a radio is $60. Therefore:

    c ( x ) = 60 x

    The revenue the guy gets from selling radios is dependent on how many radios he sells. If it is less than 100 radios, it is $90 per radio, otherwise it is $90 per radio minus .15 cents per every radio over 100.

    Therefore:

    f ( x ) =

    if x <= 100
    then return 90x
    else if x > 100
    then return 90x - .15(x - 100)

    Profit is revenue minus cost. Therefore:

    p(x) = f(x) - c(x)

    Therefore, if we are under 100 radios, p(x) is this:

    90x - 60x

    If we are above 100 radios:

    (90x - .15(x-100)) - 60x

    Simplify:

    (90x - .15x + 15) - 60x

    (89.85x + 15) - 60x

    p(x) = 29.85x + 15

    This function is completely linear. It has no max and no min. Profit never maximizes. It approaches infinity.

    Am I doing something wrong?
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  2. #2
    Wen Resu
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    NOt sure on the anwer itself but the fact that if he does sell enough radios, he will be making no money as after the 300th, hes selling the ones in excess of the first 100, at cost.
    so to maxmize, i think he might have to sell under 300.
    If you go high enough he will eventualy be makign no cash at all

  3. #3
    Pursuing knowledge confuted's Avatar
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    I'll do it when I get home
    Away.

  4. #4
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    Re: math problem

    Originally posted by DavidP
    If we are above 100 radios:

    (90x - .15(x-100)) - 60x
    This is the source of your difficulty. Try:

    (90 - 0.15(x-100))x - 60x

    Simplify:

    45x - 0.15x^2

    If you factor, you can see the zeroes. As Iamien suggests, 300 is a zero. We'd expect 0 to be the other one.

    x(-0.15x + 45)

    Then just find the maximum. Piece of cake.
    If I did your homework for you, then you might pass your class without learning how to write a program like this. Then you might graduate and get your degree without learning how to write a program like this. You might become a professional programmer without knowing how to write a program like this. Someday you might work on a project with me without knowing how to write a program like this. Then I would have to do you serious bodily harm. - Jack Klein

  5. #5
    l'Anziano DavidP's Avatar
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    why would you put the x outside of the parantheses?

    Under my understanding, you would calculate the full cost of the radios (90x) and then subtract the discount (.15(x-100))
    My Website

    "Circular logic is good because it is."

  6. #6
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    The cost per unit for the production of a certain radio model is $60. The manufacturer charges $90 per unit for orders of 100 or less. To encourage large orders, the manufacturer reduces the charge by $0.15 per radio for each unit ordered in excess of 100.
    There are lots of ways to interpret this question.

    1: For each radio ordered over 100, the manufacturer reduces the price of all radios ordered by $0.15.
    To me, this is the most likely scenerio. If you order 200 radios, the manufacturer charges you $75 per radio for a total of $15,000. For orders of over 100, this is modeled by:
    (90 - 0.15(x-100))x

    2: For each radio ordered over 100, the manufacturer reduces the price of each radio ordered over 100 by $0.15.
    Now, if you order 200 radios, the manufacturer charges you $90 per radio for the first 100, and $75 per radio for the last 100, for a total of $16,500. For orders of over 100, this is modeled by:
    9000 + (90-0.15(x-100))(x-100)

    Your expression, (90x - .15(x-100)), basically says that you get a $0.15 discount on every radio past the 100th.

    To answer your question, the x is on the outside of the parentheses because you first have to calculate the price per unit, and then multiply by the number of units.

    Hope that helps.
    If I did your homework for you, then you might pass your class without learning how to write a program like this. Then you might graduate and get your degree without learning how to write a program like this. You might become a professional programmer without knowing how to write a program like this. Someday you might work on a project with me without knowing how to write a program like this. Then I would have to do you serious bodily harm. - Jack Klein

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