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Math Help - Word problems, For Functions

  1. #1
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    Word problems, For Functions

    Is it possible to build a fence on 3 sides of a rectangular piece of land with an area of 100m^2, so that the total length of the fence is 25 m ?

    I created the two equations l+2w=25m (same as l=25-2w) and l*w=100m^2 . Then I put them together to make (25-2w)(w)=100 , and then I distribute the "w" and complete the square. The back of the book says it isn't possible though. Anyone able to do this and tell me what you get? Because I AM getting a final answer, but its not working out.
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  2. #2
    Junior Member roy_zhang's Avatar
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    Quote Originally Posted by mike_302 View Post
    (25-2w)(w)=100
    Please double check to see whether this quadratic equation has real solutions.

    Roy
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  3. #3
    Behold, the power of SARDINES!
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    Quote Originally Posted by mike_302 View Post
    Is it possible to build a fence on 3 sides of a rectangular piece of land with an area of 100m^2, so that the total length of the fence is 25 m ?

    I created the two equations l+2w=25m (same as l=25-2w) and l*w=100m^2 . Then I put them together to make (25-2w)(w)=100 , and then I distribute the "w" and complete the square. The back of the book says it isn't possible though. Anyone able to do this and tell me what you get? Because I AM getting a final answer, but its not working out.
    Well lets think of this in a different way!

    lets find the max area that can be enclosed with 25 meters of fence.

    we know that

     A=l \cdot w \mbox{ and } l=25-2w

    subbing the 2nd equation into the first we get

    A=(25-2w)w if we complete the square on this we get

    A=-2 \left( w-\frac{25}{4}\right)^2+\frac{625}{8}

    \frac{625}{8}=78.125

    So the maximum area that can be enclosed is 78.125 square feet.

    So the answer is no.
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  4. #4
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    thats the thing.. I'm getting the following:

    -2w^2+25w=100
    -2(w^2-25w+625/4-625/4)=100
    -2(w-25/2)^2 + 625/2 = 100
    ....
    ....
    ....
    where am i going wrong?
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  5. #5
    Junior Member roy_zhang's Avatar
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    Should this

    Quote Originally Posted by mike_302 View Post
    -2(w^2-25w+625/4-625/4)=100
    be -2\left(w^2-\frac{25}{2}w + \frac{625}{16}-\frac{625}{16}\right)=100 ?

    Roy
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  6. #6
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    OH! Ok, now I see what I am not doing. When I see that 25/2 , I always assume that I already divided by 2, but i forget I have to divide by 2 again (mulitply 25/2 by 1/2) and THEN square.

    Thanks! Took a bit of thinking to realize my error.

    On a side note, I'm doing another question. I'm fairly confident that I can complete the square this time, no problems, but I can't figure out the equation to begin with!

    The word problem:
    The hypotenuse is 3 cm greater in length than the next longest side of a right triangle. The line perpendicular to THAT line is 3 cm shorter . What are the side lengths. (so, you have your hypotenusse, c, which is 3 cm longer than side b, which is 3 cm longer than side c )
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  7. #7
    Junior Member roy_zhang's Avatar
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    Let c be the hypotenuse, let a be the longer leg, let b be the shorter leg.
    Base on the given, we have a = c-3, b = a-3 = c-6, by Pythagorean theorem, we have
    c^2=(c-3)^2+(c-6)^2
    From here, you can find all 3 sides.

    Roy
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