Posted by **Javier** on Friday, November 12, 2010 at 12:23am.

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 -
**Reiny**, Friday, November 12, 2010 at 9:55am
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 -
**Reiny**, Friday, November 12, 2010 at 9:57am
last area line should be

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

denominator

- Big correction - Math -
**Reiny**, Friday, November 12, 2010 at 10:18am
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.

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