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Math Help - I hate these problems..

  1. #16
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by xxlvh View Post
    Ahah
    aggghh
    I see -- for some reason I thought 200/3 was negative... so okay, fixed equation. Or at least I hope this is fixed.

    a = -3(y2+(100/3))2 - (10 000/9)
    it's not fixed. (you could have checked by expanding what you have here. you'd realize you do not get the original quadratic). there was no problem with the part you changed. the problem was with your constant. i suspect the -3 in front of the y^2 messed you up. remember for completing the square, the coefficient of the squared term must be one.

    so you would have to rewrite A = -3y^2 + 200y as A = -3 \left( y^2 - \frac {200}3y \right) and complete the square with what's inside the brackets, then multiply out.

    the answer is: A = -3 \left( y - \frac {100}3\right)^2 + \frac {10000}3

    moving on.

    what is the y-value that gives the maximum?
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  2. #17
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    Looks like I need to do much more review on this unit.

    Okay, (10,000/3) is the max y-value

    Happy holidays, by the way =)
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  3. #18
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by xxlvh View Post
    Looks like I need to do much more review on this unit.

    Okay, (10,000/3) is the max y-value

    Happy holidays, by the way =)
    it seems you do need review. 100/3 is the max y-value. at that value, the term being squared is zero. so we are subtracting the least amount we possibly can to leave a maximum behind. note that the squared term is always greater than or equal to zero

    (i hope you're not getting confused with the whole "y-value" thing. technically here, we are treating y as our input variable, so you kind of think of it as you would usually the x-value in that sense. our quadratic is in y, not x)

    anyway, that is our y-value. what is the corresponding x-value?

    Happy Holidays! (not so much holidays for me, i have make-up finals. got sick during finals time)
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  4. #19
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    *sigh*

    Ahah,. Okay: I am now officially lost...
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  5. #20
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by xxlvh View Post
    *sigh*

    Ahah,. Okay: I am now officially lost...
    when we complete the square of a quadratic, we end up with something of the form: a(x - h)^2 + k. the vertex occurs at (h,k). so just take the h-value, and that is your vertex point.

    that's what we did. we can talk about why this is later, if you're interested, now let's just take that as a given and move on. we're making the problem look complicated by dragging it out this long

    so that's our y-value. which is one of our dimensions. now we wish to find the x-dimension. we know that x = 200 - 3y. so just plug in the y-value and solve for x

    so what are the dimensions that maximize our area?
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  6. #21
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    x = 100

    so dimensions are... 100 by 100/3
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  7. #22
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by xxlvh View Post
    x = 100

    so dimensions are... 100 by 100/3
    yes.

    so let's summarize what happened here.

    we had:

    3y + x = 200 ..................(1), our constraint equation

    A = xy ............................(2), our objective equation

    from (1), x = 200 - 3y. plug this into our objective equation and we get:

    A = 200y - 3y^2

    this is a downward opening parabola. it's maximum occurs at it's vertex. we can find it's vertex by completing the square, or by using the vertex equation, which i prefer in this case.

    by the vertex equation, the y-value that gives the vertex is: y = \frac {-200}{-6} = \frac {100}3

    so, \boxed{y = \frac {100}3}

    but x = 200 - 3y

    \Rightarrow \boxed{x = 100}

    so the dimensions are: length = 100, width = 100/3


    now, any questions?
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