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Math Help - The basics

  1. #1
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    The basics

    I've got this rather simple problem; I get the right solution, the trouble is that I reach it by intuition rather than understanding why I should do as I do.
    I should also point out that I know how to do this by simply taking the derivative of the function, but right now I'm rehearsing a chapter that haven't reached that point yet and I want to actually understand this too.

    So, to boil it down I want to find the integer x > 0 for which
    T = 100 - 30x + 3x^2

    reach an absolute minimum. What I do is this:

    T = 3*(x-5)^2 + 75

    3*(x-5)^2 + 75 = 0

    (x-5)^2 + 25 = 0

    And here simply observe that for x > 0 T has an absolute minimum at 25, which it reaches when the squared expression is zero, i.e. x = 5, which is the correct answer according to the key.
    Now I only want to understand why this works. A little help here would be greatly appreciated!
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  2. #2
    Senior Member Twig's Avatar
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    Hi

    hi

    I donīt think there is anything wrong with this solution. I like it.
    You simply write the expression in quadratic form so to speak, so it will be easy to see its minimum value and for which x value.

    There are several solutions to this problem.
    Your method was one way.
    The derivative is another.

    Another I am thinking about is:
     3(\frac{100}{3} -10x +x^2)

    Here we know that for quadratic polynomials written in this way, the max/min are at the line of symmetry. Which is the cofficient in front of the x-term divided by -2. Which gives x=5.
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  3. #3
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    Yea, well, what bothers isn't so much wether this is a good solution method, it's that I get this unpleasant feeling of being detached from reality. Like, for example, in this case it's a matter of how the number T days to complete a task depends on the number x of workers. Now, I have no problem seeing how equaling the derivative to zero - i.e. when the rate of change of days needed is no longer changing - and then check wether it's a max or min using the second derivative, gives me when the efficiency is no longer improving. But simply equaling the original function to zero seems to me like I'm asking when the work takes zero days to complete, which is obviously not true. It's this logical little loop I want to get out of.
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  4. #4
    Senior Member Twig's Avatar
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    hi

    well, I dont think a quadratic polynomial is a good model for number of workers needed to complete a building in T days. I guess it would be something more like a decaying function or something.
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  5. #5
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    I don't write the exercises, and frankly I don't care wether it's a matter of the number of clouds on a Sunday afternoon in Wales as a function of the variances in the orbit of Pluto or anything else. What I don't like is the feeling that when I equal the original function to zero it seems like I'm just asking it when the work takes zero days, or there are zero clouds, which can't be the case as the solution - confirmed by taking the derivative - says something else. You might rephrase my question that way - I don't understand what I'm asking the function when I equal it to zero. Why does it work to do it this way? It must somehow be possible to relate it to the method of taking the derivatives, since both give the same correct solution, but how?
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  6. #6
    Super Member Showcase_22's Avatar
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    What's the actual question?

    When you set T=0, are you setting the derivative or something else to zero?
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  7. #7
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    Quote Originally Posted by Sputnik View Post
    I've got this rather simple problem; I get the right solution, the trouble is that I reach it by intuition rather than understanding why I should do as I do.
    I should also point out that I know how to do this by simply taking the derivative of the function, but right now I'm rehearsing a chapter that haven't reached that point yet and I want to actually understand this too.

    So, to boil it down I want to find the integer x > 0 for which
    T = 100 - 30x + 3x^2

    reach an absolute minimum. What I do is this:

    T = 3*(x-5)^2 + 75

    3*(x-5)^2 + 75 = 0

    (x-5)^2 + 25 = 0

    And here simply observe that for x > 0 T has an absolute minimum at 25, which it reaches when the squared expression is zero, i.e. x = 5, which is the correct answer according to the key.
    Now I only want to understand why this works. A little help here would be greatly appreciated!
    You're overthinking things - a top ten reason for confusion and trouble in all aspects of life.

    It works because you have found the minimum turning point of a quadratic. It's that simple. Draw the graph and see it.
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  8. #8
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    Quote Originally Posted by Sputnik View Post
    I've got this rather simple problem; I get the right solution, the trouble is that I reach it by intuition rather than understanding why I should do as I do.
    I should also point out that I know how to do this by simply taking the derivative of the function, but right now I'm rehearsing a chapter that haven't reached that point yet and I want to actually understand this too.

    So, to boil it down I want to find the integer x > 0 for which
    T = 100 - 30x + 3x^2

    reach an absolute minimum. What I do is this:

    T = 3*(x-5)^2 + 75

    3*(x-5)^2 + 75 = 0...This step is wrong!

    (x-5)^2 + 25 = 0

    And here simply observe that for x > 0 T has an absolute minimum at 25, which it reaches when the squared expression is zero, i.e. x = 5, which is the correct answer according to the key.
    Now I only want to understand why this works. A little help here would be greatly appreciated!
    Sputnik,

    Your confusion seems to lie at the step marked in red above. Instead, you should have stopped at

    T = 3*(x-5)^2 + 75.

    Now simply note that (x-5)^2 is at a minimum when x-5 = 0.
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