calculus

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The base of a triangle is on the x-axis, one side lies along the line y=3x, and the third side passes through the point (1,1). What is the slope of the third side if the area of the triangle is to be a minimum.

  • calculus -

    Interesting problem. Check to make sure I get this right.

    So, we have a triangle with base k, and vertex where the line through (1,1) and (0,k) intersects y=3x.

    The line through (1,1) and (0,k) is
    (y-1)/(x-1) = -1/(k-1)
    y = (x-1)/(1-k) + 1

    So, that line intersects y=3x when

    3x = (x-1)/(1-k) + 1
    x = k/(3k-2)
    y = 3k/(3k-2)

    The area a of the triangle is thus

    a = 1/2 k * 3k/(3k-2) = 3k^2/(6k-4)

    da/dk = (6k(6k-4) - 3k^2 * 6)/(6k-4)^2
    = 6k(3k-4)/(6k-4)^2

    da/dk = 0 when k=0 or k = 4/3
    Thus, the vertex of the triangle is where the line through (1,1) and (4/3,0) intersects y=3x

    (y-1)/(x-1) = 1/(-1/3)
    y = -3x + 4

    That line has slope -3

    . . .

    -3x + 4 = 3x
    6x = 4
    x = 2/3
    y=2

    So, the triangle has base k=4/3, height h=2, area a=4/3

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