Math

posted by .

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

    denominator

  • 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 -

    yes

Respond to this Question

First Name
School Subject
Your Answer

Similar Questions

  1. Calc.

    Find the area of the region bounded by the parabola y=x^2, the tangent line to this parabola at (1,1) and the x-axis. I don't really get what this question is asking. It looks like the area of right triangle to me...try the graph, …
  2. Calculus

    I am doing the AP calculus review, these are the questions I have no Idea on how to do: 1. if 0<= k <=pi/2 and the area under the curve y-cosx from x=k to x=pi/2 is 0.2, then k= 2. let F(x) be an antiderivative of (ln x)^4/x …
  3. Calc

    A line tangent to y=(x^2)+1 at x=a, a>0, intersects the x-axis at point P. a) Write an expression for the area of the triangle formed by the tangent line, the x-axis, and the line x=a. b) For what value of 'a' is the area of the …
  4. math

    the point P lies in the first quadrant on the line y=10-4x from P, perpendicular lines are drawn to the x and y axis. What is the maximum area of the rectangle formed?
  5. Calculus 151

    At what point in the first quadrant on the parabola y=4-x^2 does the tangent line, together with the coordinate axis, determine a triangle of minimum area?
  6. calculus

    The graph of f'(x) is shown for 0=< x =<10. The areas of the regions between the graph of f' and the x-axis are 20, 6, and 4, respectively. I'm going to describe the graph of f' since I can't post pictures. The first section …
  7. math

    a) What is the area of the tri-angle determined by the lines y= − 1/ 2x + 5,y =6x and they-axis?
  8. calculus

    5. Let for and f(x)=12X^2 for x>=0 and f(x)>=0 a. The line tangent to the graph of f at the point (k, f(k)) intercepts the x-axis at x = 4. What is the value of k?
  9. calculus

    Let R be the region in the first quadrant that is enclosed by the graph of y = tanx, the x-axis, and the line x = π/3 h. Find the area of R i. Find the volume of the solid formed by revolving R about the x-axis
  10. calculus

    Find the point on the curve y =2/3 〖√(18-x^2 )〗 (first quadrant) where a tangent line may be drawn so that the area of the triangle formed by the tangent line and the coordinate axes is a minimum.

More Similar Questions