calculus

a right triangle is formed in the first quadrant by the x- and y- axes and a line through the point (1,2). write the length of the hypotenuse as a function of x. find the vertices of the triangle such that its area is a minimum.

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  1. let P(x,0) be the x-intercept and Q(0,y) be the y-intercept
    label (1,2) as A(1,2)
    the slope AP = slope AQ
    -2/(x-1) = (2-y)/1
    2x - xy - 2 + y = -2
    y(1-x) = -2x
    y = -2x/(1-x) or 2x/(x-1)

    H^2 = x^2 + y^2
    = x^2 + ( (2x/(x-1) )^2 --- simplify if needed
    (don't know why we are finding the hypotenuse ? )

    Area of triangle OPQ
    = (1/2)base x height
    = (1/2)xy
    = (1/2)x(2x/(x-1))
    = (1/2) (2x^2/(x-1) )

    dA/dx = (1/2) [ (x-1)(4x) - 2x^2(1) ]/(x-1)^2
    = (1/2) [ 4x^2 - 4x - 2x^2 ]/(x-1)^2
    = 0 for a max of A
    2x^2 - 4x = 0
    x(2x-4) = 0
    x=0 ---- > makes no sense, look at diagram
    or
    x = 2
    then y = 2(2)/(2-1) = 4

    the triangle has (2,) and (0, 4) as its vertices for a maximum area

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  2. sticky keyboard, last line should obviously say

    the triangle has (2,0) and (0, 4) as its vertices for a maximum area

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