Math
posted by Javier .
Find the maximum area of a triangle formed in the first quadrant by the xaxis, yaxis, and a tangent line to the graph of y = (x+1)^2

Math 
Reiny
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 yintercept of the tangent
the base of the triangle is the xintercept of the equation
xintercept, let y = 0
ax = a^2 + b(a+1)^3
x = (a^2 + b(a+1)^3)/a
yintercept, 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 yintercept 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
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
Major slipup 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 
Pablo
yes
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