Question

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

Answer 1

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.

Answer 2

last area line should be

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

denominator

Answer 3

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.