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Math Help - Optimization problem

  1. #1
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    Optimization problem

    Alright, I've been stuck on this one for about an hour. This is the question verbatim. Any help would be appreciated.
    'You are designing a rectangular poster to contain 100in^2 of printing with a 4-in margin at the top and bottom and a 1-in margin at each side. what overall dimensions will minimize the amount of paper used?'
    All I need is a second equation, or even the correct first would be a help. As it is I trial-errored out an estimate at 15x6, or 23x8 with margins.Thanks for taking the time to read this far,
    -Grizzly
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  2. #2
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    Quote Originally Posted by GrizzlyAdamz View Post
    Alright, I've been stuck on this one for about an hour. This is the question verbatim. Any help would be appreciated.
    'You are designing a rectangular poster to contain 100in^2 of printing with a 4-in margin at the top and bottom and a 1-in margin at each side. what overall dimensions will minimize the amount of paper used?'
    All I need is a second equation, or even the correct first would be a help. As it is I trial-errored out an estimate at 15x6, or 23x8 with margins.Thanks for taking the time to read this far,
    -Grizzly
    Let the area of paper with margins included be

     A_1 = x \times y
    where x is the width of the paper and y is the height.

    Now taking into account of the margins, the area required is:

    100 = (x-2)(y-8)

    What you want to do is to express x as a function of y or vice versa from the second equation, and then substitute into the first equation. Then you can differentiate to find the minimum area required.
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  3. #3
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    So it really is an area-area kind of relation? Damn, I've been banging my head against those two equations for an hour. I was certain that wasn't the correct pair. Hmm, just a moment.
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  4. #4
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    I'm not sure how the bumping system works so I apologize for the double-post. When I go to substitute the new value for height, I have a straight function. -42x+100.
    I got that from 100/(x-2)=(y-8)
    going to (100/x)-50=(y-8)
    then I brought the -8 over and substituted into the first equation for x * ((100/x) - 42)
    Where did I mess up?
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  5. #5
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    Quote Originally Posted by GrizzlyAdamz View Post
    I got that from 100/(x-2)=(y-8)
    going to (100/x)-50=(y-8)
     \frac{100}{x-2} \neq \frac{100}{x} - 50
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  6. #6
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    /Facepalm Alright. Another moment, thanks for your time.
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  7. #7
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    I am feeling a bit like a lummox. I hit another problem.
    x*((100/(x-2))-8)
    (100x/(x-2))-8x
    ((100x-200-100x)/(x^2-4x+4))-8
    (-200/(x^2-4x+4))-8
    I avoided the -200 up top for lack of effort; as it were all the positive x values are negative.
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  8. #8
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    Quote Originally Posted by GrizzlyAdamz View Post
    I am feeling a bit like a lummox. I hit another problem.
    x*((100/(x-2))-8)
    (100x/(x-2))-8x
    ((100x-200-100x)/(x^2-4x+4))-8
    (-200/(x^2-4x+4))-8
    I avoided the -200 up top for lack of effort; as it were all the positive x values are negative.
    You don't have to worry about simplifying it. I just did:

     A = x \left(\frac{100}{x-2} + 8 \right)
     A = 8x + \frac{100x}{x-2}

    and then I just differentiated that equation using the quotient rule for the fraction.

    Note: It's +8, not -8.
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  9. #9
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    heh, that's what I had done. Lol, that positive 8 makes all the difference. as it is I'm at there being a 0 at x=7, but with my head pulverized from all that head-bashing I'm having trouble remembering the rules that let me ace the test earlier today. If I substitute that into the second equation and solve for y, that should give me the dimensions, correct?
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  10. #10
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    I think that's a win. It matches up with the correct lowest estimation, (23x8 had an error in it), Thanks for the help! This place is book marked, and I'm gonna try giving you kudos or whatever this site uses.
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  11. #11
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    Red face web designing

    I think that's a win. It matches up with the correct lowest estimation, (23x8 had an error in it), Thanks for the help! This place is book marked, and I'm gonna try giving you kudos or whatever this site uses.
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