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Math Help - Finding dimensions

  1. #1
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    Finding dimensions

    I'm supposed to find the dimensions of a rectangular region with a maximum area that can be enclosed by 80 feet of fencing. this is what I have so far...


    2x+ 2y=80
    -2x -2x

    2y=80-2x

    y= 80-2x
    ------ --------> y= 40- (-1x)
    2

    Can somebody telling me if this is right so far?I got the -1x from the -2x
    ---
    2

    Is that correct so far? If not, could you please set me on the right track?
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  2. #2
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    Quote Originally Posted by suzdisp View Post
    I'm supposed to find the dimensions of a rectangular region with a maximum area that can be enclosed by 80 feet of fencing. this is what I have so far...


    2x+ 2y=80
    -2x -2x

    2y=80-2x

    y= 80-2x
    ------ --------> y= 40- (-1x)
    2

    Can somebody telling me if this is right so far?I got the -1x from the -2x
    ---
    2

    Is that correct so far? If not, could you please set me on the right track?
    "y = 40 - (-1x)"
    is not correct. (I presume it's just a typo.) It should be
    y = 40 - (1x)
    or
    y = 40 + (-1x)

    Do you know how to continue?

    -Dan
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  3. #3
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    well, I'll try. you tell me if I'm right or not

    since you get y=40 -(1x)

    you put that in for y, so: A=x(40-(1x))

    A=40x-x^2?
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  4. #4
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by suzdisp View Post
    well, I'll try. you tell me if I'm right or not

    since you get y=40 -(1x)

    you put that in for y, so: A=x(40-(1x))

    A=40x-x^2?
    Yep.

    Now what?

    -Dan
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  5. #5
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    well, I know you're supposed to find the derivatives and use it to determine the max. value of the area, but I have no idea how. I've started out by doing:


    dA
    ---= 40x-(x^2)
    dx

    after that, I'm lost
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  6. #6
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by suzdisp View Post
    well, I know you're supposed to find the derivatives and use it to determine the max. value of the area, but I have no idea how. I've started out by doing:


    dA
    ---= 40x-(x^2)
    dx

    after that, I'm lost
    You have the right idea, but took your derivative wrong:
    A = 40x - x^2

    \frac{dA}{dx} = 40 - 2x

    Now, what does the derivative of a function mean? It is the slope of the line tangent to the function at a given value of x. What happens to the derivative of a function when the function is at a relative maximum or minimum?

    -Dan
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  7. #7
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    I have no idea.
    but I think the finally answer is x and y both = 20

    that is what I have worked out. Am I right at all?
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  8. #8
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by suzdisp View Post
    I have no idea.
    but I think the finally answer is x and y both = 20

    that is what I have worked out. Am I right at all?
    That is correct, but the question is "how did you find this out?"

    When a function reaches either a relative max or min, the derivative vanishes. Thus
    40 - 2x = 0

    x = 20

    Then you can find that y = 20 also.

    -Dan
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  9. #9
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    thats what I did to find 20. I found an example in my book.

    then I solved for y. y=40-x



    y=40-20
    y=20


    then, I put 20 in for y

    20=40-1x
    -40 -40


    -20/ -1 =-1x/-1


    20=x



    YES!!!!!!!!!!!!!
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