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Calculus-Applied Optimization Problem:

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Find the point on the line 6x + 3y-3 =0 which is closest to the point (3,1).

Note: Your answer should be a point in the xy-plane, and as such will be of the form (x-coordinate,y-coordinate)

  • Calculus-Applied Optimization Problem: - ,

    what a stupid note. Of course it will be a point with (x,y) coordinates! This is a calculus class - you know all that already.

    Given a point (x,y) on the line, y=1-2x

    The distance from (x,1-2x) to (3,1) is

    d^2 = (x-3)^2 + (1-(1-2x))^2 = 5x^2-6x+9

    so, we want minimum d, when dd/dx = 0:

    2d dd/dx = 10x-6
    dd/dx = 10x-6/2sqrt(blah blah)

    dd/dx=0 when x = 3/5.

    So, minimum d^2 is

    5(3/5)^2 - 6(3/5) + 9 = 36/5
    minimum d is 6/√5

    Or, as we all know, the distance from a point (x,y) to a line ax+by+c=0 is

    |ax+by+c|/√(a^2+b^2) = (6(3)+3(1)-3)/√(36+9) = 18/√45 = 6/√5

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