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calculus HELP URGENT

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Find the equation of the line through the point (3,4) which cuts from the first quadrant a triangle of minimum area.

  • calculus HELP URGENT - ,

    Draw a line through (3,4) cutting the x-axis at (x,0) and the y-axis at (0,y)

    Area = (1/2(x)(y)

    using slopes,
    (y-4)/-3 = +4/(x-3)
    xy - 3y - 4x + 12 = 12
    y(x-3) = 4x
    y = 4x/(x-3)

    then
    A = (1/2)x(4x)/x-3)
    = 2x^2/(x-3)

    dA/dx = [(x-3)(4x) - 2x^2]/(x-3)^2 = 0 for a max/min of A
    4x^2 - 12x - 2x^2 = 0
    2x^2 - 12x = 0
    2x(x-6)= 0
    x = 0 or x = 6, and y = 24/3 = 8

    using (0,8) and (3,4)
    slope = (8-4)/-3
    = -4/3

    line has equation
    y = (-4/3)x + 8
    slope of line =

  • calculus HELP URGENT - ,

    " using slopes,
    (y-4)/-3 = +4/(x-3)
    xy - 3y - 4x + 12 = 12 "


    can you explain that part

  • calculus HELP URGENT - ,

    nvr mind
    Anyways thanks Reiny

  • calculus HELP URGENT - ,

    Find the gradient of a line joining R(4,8) & S(5,-2).

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