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Thread: Find pt. on 6x + y = 9 closest to (-3, 1) in terms of y

  1. #1
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    Find pt. on 6x + y = 9 closest to (-3, 1) in terms of y

    Find the point on the line 6x + y = 9 closest to the point (-3, 1) in terms of y

    [tex]y=9-6x[/tex]

    Distance Formula:

    [tex]d=sqrt{(x+3)^2 + (y-1)^2}[/tex]

    [tex]d(x)=sqrt{(x+3)^2 + (9-6x-1)^2}[/tex]

    [tex]d(x)=sqrt{37x^2-90x+73}[/tex]

    [tex]f(x)=d(x)^2=37x^2-90x+73[/tex]

    derivitive:

    [tex]f'(x)=74x-90[/tex]

    Set the derivative equal to 0:

    [tex]0=74x-90[/tex]

    [tex]x=45/37[/tex]

    That is how far I can get by looking at the examples in the book. Then I get kinda lost... Any ideas? Thank you!

  2. #2
    Senior Member skeeter's Avatar
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    6x + y = 9

    slope of this line is m = -6

    a perpendicular line has slope m = 1/6

    the line passing through (-3,1) with slope m = 1/6 is

    y - 1 = (1/6)(x + 3) edit mistake ... I wrote y + 1 instead of y - 1 on the left side of the linear equation
    y = (1/6)x + (3/2)

    the point of intersection of the line y = (1/6)x + (3/2) and 6x + y = 9 is ...

    9 - 6x = (1/6)x + (3/2)
    54 - 36x = x + 9
    45 = 37x

    x = 45/37, y = 9 - 6(45/37) = 63/37 now we should all agree

  3. #3
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    that was simple

    thanks

  4. #4
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    I get the same thing as gopher.

    When you are maximizing or minimizing a distance, there is a trick you can use to avoid radicals. The distance and the square of the distance have their max and min at the same point. Therefore, the minimum occurs at

    [tex]\L\\S=L^{2}=(x+3)^{2}+(\underbrace{(9-6x)}_{\text{y}}-1)^{2}[/tex]

    [tex]\L\\S=37x^{2}-90x+73[/tex]

    [tex]\L\\S'=74-90x[/tex]

    [tex]\L\\x=\frac{45}{37}, \;\ y=\frac{63}{37}[/tex]


    Excuse me, skeet, but I think the y-intercept of that line should be 3/2, not 1/2.
    Hence the discrepancy. [tex]y=\frac{1}{6}x+\frac{3}{2}[/tex]
    gives the same result as the calc way.

  5. #5
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    very good thank you both

  6. #6
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    ok im stuck again
    Find the dimensions of a rectangle with area 1000m^2 whose perimeter is small as possible.


    [tex]Area=x*y[/tex]
    [tex]1000=x*y[/tex]
    [tex]y=1000x[/tex]

    [tex]Permiter=2x*2y[/tex]

    im stuck on how these releate to each other

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