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Math Help - Complete the Square? Parabola?

  1. #1
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    Complete the Square? Parabola?

    Alright, so I recently started my new semester, and I've forgotten almost everything that I've learned the previous year regarding math. I was assigned to do various questions, and I'm not exactly sure how to do it. The question is:

    Rebecca has 600 m of fencing for her cornfield. The creek that goes through her farmland will form one side of the rectangular boundary. Rebecca considers different widths to maximize the area enclosed.

    A. What are the minimum and maximum values of the width of the field?

    I'm assuming that I will need to create a parabola for this question, due to the question asking a minimum and maximum. I also have the feeling completing the square will be involved, but I pretty much forgot how to do both. But here's what I got so far:

    P = L + 2W (the creek makes up 1 length?)
    L = 600 - 2W
    A = W X L

    That's all I got for now... any help, please?


    B. What equations describe each?
    i) The relationship between the length and width of the field.
    ii) The relationship between the area and width of the field.

    C. Copy and complete these table of values for widths that go from the least to greatest possible values in intervals of 50 m.

    Width (m) | Length (m) | Area (m^2)
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  2. #2
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    Quote Originally Posted by shadow6 View Post

    P = L + 2W (the creek makes up 1 length?)
    L = 600 - 2W
    A = W X L

    That's all I got for now... any help, please?
    You are doing very well up to here.

    From your working, A = W X L is really

    A = W \times (600 - 2W)

    A =  600W - 2W^2

    Do you know how to find the maximum of this parabola? You need to make A=0 to solve for W . The maximum value will be half way between the 2 solutions.
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  3. #3
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    P = L + 2W
    L = 600 - 2W
    A = W X L
    A = W X (600 - 2W)
    A = 600W - 2W^2
    A = -2W^2 + 600W
    A = -2(W^2 - 300W) -> (300/2)^2 = 22500
    A = -2(W^2 - 300W + 22500 - 22500)

    Alright, got this so far. But I'm not sure what to do from here. From what I remember, I will have to use the -2 again on the next line, and isolate the -22500, but how do I do this... if that's what I'm even supposed to do?
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    Quote Originally Posted by shadow6 View Post
    P = L + 2W
    L = 600 - 2W
    A = W X L
    A = W X (600 - 2W)
    A = 600W-2W^2
    You really don't need to complete the square from here, it would just be making life difficult and who has time for that?

    To solve for W make A=0

    0 = 600W-2W^2

    Take common factors from the RHS

     0= 2W(300-W)

    Using the null factor law W = 0,300

    Quote Originally Posted by pickslides View Post
    The maximum value will be half way between the 2 solutions.
    What is half way between 0 and 300? W=\dots

    Put this value into A = 600W-2W^2 for W to get max area.
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  5. #5
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    P = L + 2W
    L = 600 - 2W
    A = W X L
    A = W X (600 - 2W)
    A = 600W - 2W^2
    0 = 600W - 2W^2
    0 = 2W(300 - W)

    Wait, where did the 0,300 come from? In order for the right side of the equation to = 0, W would have to = 300.

    Half way between 0 and 300 = 150. Therefore W (min value?) = 150.

    A = 600W - 2W^2
    A = 600(150) - 2(150)^2
    A = 90000 - 2(22500)
    A = 90000 - 45000
    A = 45000

    Therefore max area is 45000. Though, the question is asking for the minimum and maximum of the width of the field. Is the max area the max value for the width?
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    Quote Originally Posted by shadow6 View Post
    P
    0 = 2W(300 - W)

    Wait, where did the 0,300 come from? In order for the right side of the equation to = 0, W would have to = 300.
    0 = 2W(300 - W) \implies 2W=0,300-W=0\implies W=0,300

    Quote Originally Posted by shadow6 View Post

    Therefore max area is 45000. Though, the question is asking for the minimum and maximum of the width of the field. Is the max area the max value for the width?
    Your answer for max width is 150 as W=150
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    Alright, thanks very much for the help.

    Before even attempting to complete the square, I guessed that the minimum width was 1, and the maximum width was 299. I guessed 1 because the length could be 598, and the two widths would be one each, equaling 600 (perimeter), but I guess that was wrong.

    The next question asks:

    B. What equations describe each?
    i) The relationship between the length and width of the field.

    A = W X L

    ii) The relationship between the area and width of the field.

    W = A/L

    These look correct?
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    Quote Originally Posted by shadow6 View Post

    These look correct?
    They are both worng




    Quote Originally Posted by shadow6 View Post

    B. What equations describe each?
    i) The relationship between the length and width of the field.

    You need someithng with only L and W in it. I like

    L = 600 - 2W


    Quote Originally Posted by shadow6 View Post

    ii) The relationship between the area and width of the field.


    You need someithng with only A and W in it. I like

     A = 600W - 2W^2
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  9. #9
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    Got it now, ty.

    Question C asks:

    C. Copy and complete these table of values for widths that go from the least to greatest possible values in intervals of 50 m.

    Width (m) | Length (m) | Area (m^2)


    Width (m)
    0
    50
    100
    150
    200
    250
    300

    Length (m)
    600
    500
    400
    300
    200
    100
    0

    Area (m^2)
    0
    25000
    40000
    45000
    40000
    25000
    0


    A friend shared these results with me. I don't even see any pattern within them at all. Are they correct? Will I have to somehow apply my results in questions A and B to complete this chart?
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  10. #10
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    I would suggest to used the information from the previous question

    Width (m) | Length (m) | Area (m^2)

    0 ~~~~~~ L = 600-2(0) ~~~~~~ A=600(0)-2(0)^2

    50 ~~~~~~ L = 600-2(50) ~~~~~~ A=600(50)-2(50)^2

    100 ~~~~~~ L = 600-2(100) ~~~~~~ A=600(100)-2(100)^2



    etc....
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