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Thread: Optimization problem - maximizing profits

  1. #1
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    Question Optimization problem - maximizing profits

    Hi,

    I'm trying to solve for the following optimization problem:
    Optimization problem - maximizing profits-screen-shot-2017-08-12-9.37.30-am.png

    I'm hoping that someone can confirm with me whether or not my calculations are right:
    Optimization problem - maximizing profits-20795199_1499709063418516_320220845_o.jpg

    The answer according to the textbook is $1100 or $1125. I was able to get the answer of of $1125, but I have no clue where the alternative answer of $1100 came from. Can someone help? Can someone also confirm the domain restriction for this question? I just said that x had to be greater than 0, but I feel like I may not be entirely correct.

    Hope someone can help!

    - Olivia
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    Re: Optimization problem - maximizing profits

    I gather that your teacher wants you to define constraints in terms of domain or range. Seems a bit weird to me, but you have to do what is required.

    x = the number of rent increases in 25 dollar increments, r(x) = revenue, c(x) = cost, and p(x) = profit.

    $r(x) = (50 - x)(900 + 25x) = 45000 + 350x - 25x^2.$ Good.

    $c(x) = 75(50 - x) = 3750 - 75x.$

    The problem does not exactly say that, but it makes practical sense. So

    $p(x) = r(x) - c(x) = 45000 + 350x - 25x^2 - (3750 - 75x) = 41250 + 425x - 25x^2 \implies$

    $p'(x) = 425 - 50x \implies p'(x) = 0 \iff 50x = 425 \text { or } x = 8.5.$

    That SEEMS to imply that the rent should be $900 + 25 * 8.5 = 900 + 212.50 = 1112.50,$

    Which is what you got. You did NOT get the answers in the book.

    But x must be an integer.

    $\therefore x = 8 \text { or } x = 9.$

    $x = 8 \implies rent = 900 + 25 * 8 = 900 + 200 = 1100.$

    $x = 9 \implies rent = 900 + 25 * 9 = 900 + 225 = 1125.$

    But this does not end the problem. Which level of rent gives the maximum profit? How do you determine that?

    The domain restriction on x is that it must be an integer. (Or perhaps an integer between 0 and 36.)
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    Re: Optimization problem - maximizing profits

    Okay thanks... so when I get a number (8.5 in this case) that is in-between two integers (8 and 9) I don't "round up" as I normally would, but instead consider both the values in-between 8 and 9? That doesn't seem intuitive to me.
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    Re: Optimization problem - maximizing profits

    Quote Originally Posted by otownsend View Post
    Okay thanks... so when I get a number (8.5 in this case) that is in-between two integers (8 and 9) I don't "round up" as I normally would, but instead consider both the values in-between 8 and 9? That doesn't seem intuitive to me.
    Rounding has nothing to do with it. Without getting lost in the technical weeds, you need to test both 8 and 9 to see whether one or the other gives a higher profit. If they give the same profit you can choose either.

    $p(8) = 41250 + 425 * 8 - 25 * 64 = 43050.$

    $p(9) = 41250 + 425 * 9 - 25 * 81 = 43050.$
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    Re: Optimization problem - maximizing profits

    To expand a bit. Calculus doesn't know about integers, which are discrete. It deals with continuous functions. So calculus tells us that there may be an integer answer close to 8.5, but it cannot say which integer gives the optimum answer. It is very similar to the situation involving boundary conditions, where calculus does not let you exclude the boundary.
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    Re: Optimization problem - maximizing profits

    My point that I'm making here is what if x = 8.8 instead of 8.5? Would I still test for p(8) even though x is clearly closer to 9?
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    Re: Optimization problem - maximizing profits

    Yes, that is what JeffM just said: "So calculus tells us that there may be an integer answer close to 8.5, but it cannot say which integer gives the optimum answer." If x is very close to an integer, that integer may be more likely to be the correct one but you cannot be sure of that.
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    Re: Optimization problem - maximizing profits

    Interesting... okay well thank you for clarifying!
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