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Math Help - equation of line ...

  1. #1
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    equation of line ...

    Find the equation of line that passes through (6,8)
    and make with both axes a triangle with minimum area.

    My try:
    let the equation be y=mx+b
    area = 1/2 xb
    m=(b-8)/-6

    then .....
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  2. #2
    MHF Contributor Unknown008's Avatar
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    y = mx + b

    A = \dfrac{xb}{2}

    m = \dfrac{b-8}{-6}

    Good! Now, you need to find A' to get the minimum area. But for that, you'll need to get A in terms of either x alone, or b alone.

    y = \dfrac{8-b}{6} x + b

    At point (x, y) = (6, 8), find x in terms of b, or vice versa.

    Then, substitute this into your expression for the area.

    Differentiate the area, and solve for A' = 0.

    You'll get a value of b or x depending on what you chose previously.

    Substitute this value into the equation y = mx + b and use the point (6, 8) again to find the other value you need.
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  3. #3
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    Quote Originally Posted by Unknown008 View Post
    y = mx + b

    A = \dfrac{xb}{2}

    m = \dfrac{b-8}{-6}

    Good! Now, you need to find A' to get the minimum area. But for that, you'll need to get A in terms of either x alone, or b alone.

    y = \dfrac{8-b}{6} x + b

    At point (x, y) = (6, 8), find x in terms of b, or vice versa.

    Then, substitute this into your expression for the area.

    Differentiate the area, and solve for A' = 0.

    You'll get a value of b or x depending on what you chose previously.

    Substitute this value into the equation y = mx + b and use the point (6, 8) again to find the other value you need.
    when substitue x=6 ---> y=8 no relation to use it in A=1/2xb

    how can I find a relation ?
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  4. #4
    MHF Contributor Unknown008's Avatar
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    Oops, right. I tried to go from where you went and failed to see something more.

    m = \dfrac{8-b}{6} = \dfrac{b-0}{0-x}

    There, can you continue now?
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  5. #5
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    \frac {8-b}{6}=\frac {-b}{x}

    x=\frac {-6b}{8-b}

    area = A =\frac {-3b^2}{8-b}

    \frac {dA}{db}=0

    3b^2-48=0

    x=\pm 4

    Then when x=4 or -4 we can find x and then the Area.

    IS THIS RIGHT ??
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  6. #6
    MHF Contributor Unknown008's Avatar
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    Not quite.

    I'll change A to make things simpler. A = \dfrac{3b^2}{b-8}

    \dfrac{dA}{db} = \dfrac{(b-8)6b - 3b^2}{(b-8)^2}

    This gives:

    \dfrac{dA}{db} = \dfrac{3b^2 - 48b}{(b-8)^2}

    You missed the 'b' in 48b.
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  7. #7
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    you are right , thanks

    b=16
    x=12
    m=-4/3

    y=-(4/3)x+16

    thanks a lot
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