Calculus

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Find an equation of the line through the point (3,5) that cuts off the least area from the first quadrant.

How do I solve? I am stuck. The only thing I can come with is this:

y-5=m(x-3)

How to use this I don't know.

  • Calculus -

    draw a line through (3,5) cutting the positive x and y axes at (a,0) and (0,b)

    then the area of the triangle is
    A = ab/2

    but the slope of the two segments must be equal, so 5/(3-a) = (b-5)/-3
    solving for b gave me b = 5a/(a-3)

    then A = 5a^2/(2a-6)
    using the quotient rule
    dA/da = [(2a-6)(10a) - 5a^2(2)]/(2a-6)^2
    = 0 for a max/min of A

    simplifying the top and setting it equal to zero gave me
    a^2 - 6a = 0
    a(a-6)=0
    a = 0 or a = 6
    clearly a=0 does not give me a triangle, so a = 6 and b= 10
    so slope = 5/-3
    and your equation would be
    y-5 = (-5/3)(x-3)

    take it from there.

  • Calculus -

    how did you get the slope of the two lines? I get everything else though.

  • Calculus -

    there is only one line
    you were given the point (3,5) and I found
    the x-intercept to be (6,0)
    so slope = (5-0)/(3-6)
    = 5/-3 = - 5/3

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