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Math Help - Optimization problems

  1. #1
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    Optimization problems

    Imagine if you will a fenced-in area composed of a rectangle with one side being a semicircle. The perimeter of this fence is 498 feet. I need to maximize the area of this pen. How do I do this? (I already know the major steps: set up your two equations, relate variables, take the first derivative and set it to zero, etc. I need specifics.)
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  2. #2
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    Hello? A little help? Please?
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  3. #3
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by G-Rex View Post
    Imagine if you will a fenced-in area composed of a rectangle with one side being a semicircle. The perimeter of this fence is 498 feet. I need to maximize the area of this pen. How do I do this? (I already know the major steps: set up your two equations, relate variables, take the first derivative and set it to zero, etc. I need specifics.)
    the formula for the total area is the formula for the area of the rectangle plus the formula for the area of the semi circle. this will give the function you must maximize.

    call one side of the rectangle x and the other y. let the semicircle be attached to one of the x sides.

    can you continue? (by the way, is this for a calculus class?)
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  4. #4
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    No, that's not enough information for me to continue, seeing as I already know that, and yes, this is a homework assignment I'm trying to complete for a class. What? Do you think I just ask this stuff for the fun of it?
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  5. #5
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by G-Rex View Post
    No, that's not enough information for me to continue, seeing as I already know that
    ok, give me the formula we should work with then

    and yes, this is a homework assignment I'm trying to complete for a class. What? Do you think I just ask this stuff for the fun of it?
    the point of my question was that i wanted to know if this was calculus or precalculus, since we will use different methods depending on which class this is for. obviously you don't do this for fun
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  6. #6
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    Hello, G-Rex!

    Imagine a fenced-in area composed of a rectangle with one side being a semicircle.
    The perimeter of this fence is 498 feet.
    Maximize the area of this pen.
    Code:
                  * * *
              *           *
            *               *
           *                 *
    
          * - - - - * - - - - *
          |    r         r    |
          |                   |
        x |                   | x
          |                   |
          |                   |
          *-------------------*
                   2r

    The radius of the semicircle is r.
    Hence, the length of the rectangle is 2r.
    Let the width of the rectangle be x.

    The semicircle has perimeter \pi r.
    The rectangular portion of the pen has perimeter 2r + 2x

    We have: . \pi r + 2r + 2x \:=\:498\quad\Rightarrow\quad x \:=\:\frac{498-(\pi+2)r}{2}\;\;{\color{blue}[1]}


    The area of the semicircle is: . \frac{1}{2}\pi r^2
    The area of the rectangle is: . (2r)(x) \:=\:2rx

    The area of the pen is: . A \;=\;\frac{1}{2}\pi r^2 + 2rx\;\;{\color{blue}[2]}

    Substitute [1] into [2]: . A \;=\;\frac{1}{2}\pi r^2 + 2r\left(\frac{498-(\pi+2)r}{2}\right)

    . . which simplifies to: . A \;=\;498r - \left(\frac{\pi+4}{2}\right)r^2

    And that is the function we must maximize . . .

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  7. #7
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    Where does the .5pi(r^2) go? I worked through the problem following your steps, but I can't figure out what you did there.

    To Jhevron: It's a Calc class.
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  8. #8
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by G-Rex View Post
    Where does the .5pi(r^2) go? I worked through the problem following your steps, but I can't figure out what you did there.

    To Jhevon: It's a Calc class.
    when he substituted the expression for x and simplified, the (1/2)pi*r^2 got absorbed into the other r^2 term
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  9. #9
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    I'm...not seeing how. In fact, I'm not even sure how he simplified it like that in the first place. When I simplify, I get .5pi(r^2) + 498r + (pi + 2)r^2. If I add the two pi(r^2) terms together, I get 1.5pi(r^2) + 498r + 2r^2.

    Anyway, it's almost midnight here. I'm hittin' the hay If I haven't figured this out by tomorrow I'll be back. In the mean time, if you'd like, you can give me the next step and I may get a chance to check this in the morning before I leave. Anyway, good night.
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  10. #10
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    Quote Originally Posted by G-Rex View Post
    Imagine if you will a fenced-in area composed of a rectangle with one side being a semicircle. The perimeter of this fence is 498 feet. I need to maximize the area of this pen. How do I do this? (I already know the major steps: set up your two equations, relate variables, take the first derivative and set it to zero, etc. I need specifics.)
    Since you know what you say you know, how about showing us how you
    have tried to apply that knowlege, so we can see where you are going wrong
    and address the real problems you are having.

    RonL
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  11. #11
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    Alright guys, I managed to figure it out (mostly) on my own today. Thanks to those who tried to assist, even if wasn't that big a help. It 's the effort that counts.
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