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Find the maximum area of a triangle formed in the first quadrant by the x-axis, y-axis, and a tangent line to the graph of y = (x+1)^-2

  • Math -

    let the point of contact of the tangent be (a,b)
    dy/dx = -2(x+1)^-3 = -2/(x+1)^3
    at (a,b)
    slope of tangent = -2(a+1)^3

    so equation of tangent :
    ax + (a+1)^3 y = c
    but (a,b) lies on this
    a(a) + (a+1)^3 (b) = c
    equation of tangent:
    ax + (a+1)^3 y = a^2 + b(a+1)^3

    the height of the triangle is the y-intercept of the tangent
    the base of the triangle is the x-intercept of the equation

    x-intercept, let y = 0
    ax = a^2 + b(a+1)^3
    x = (a^2 + b(a+1)^3)/a

    y-intercept, let x = 0
    y = (a^2 + b(a+1)^3)/(a+1)^3

    but remember since (a,b) is on the curve
    b = 1/(a+1)^2

    so the x intercept reduces to (a^2 + a + 1)/a
    and the y-intercept reduces to (a^2 + a + 1)/(a+1)^3

    Area of triangle = A
    = (1/2)(a^2+a+1)^2 / (2a(a+1)^3)

    .... long way from being done.... What a question!

    Ok, now take the derivative of A using the quotient rule, set that equal to zero and solve for a
    sub that value of a back into the area equation.

  • small correction - Math -

    last area line should be

    (1/2)(a^2+a+1)^2 / (a(a+1)^3) , change is in the


  • Big correction - Math -

    Major slip-up near the top

    "slope of tangent = -2(a+1)^3

    so equation of tangent :
    ax + (a+1)^3 y = c " should say

    slope of tangent = -2/(a+1)^3

    so equation of tangent :
    2x + (a+1)^3 y = c

    Notice the two changes.
    Let me know if you are even reading this.
    If you are, I will continue the solution.

  • Math -


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