Posted by miley on Wednesday, March 31, 2010 at 9:22am.
The hypotenuse of a right triangle has one end at the origin and one end on the curve y+x^2e^(3x), with x > or = 0. One of the other two sides is on the xasis, the other side is parallel to the yaxis. Find the maximum area of such a triangle. At what xvalue does it occur?

math/calc  bobpursley, Wednesday, March 31, 2010 at 9:52am
Your curve
y+x^2 e^(3x)
what is that equal to ? A curve has to follow an equation.
If you mean the curve
y=x^2 e^(ex)
then the area is yx/2 which is equal to
1/2 x^3 e^(3x). Right?
dArea/dx= 3x^2e^(3x)3x^3e^(3x)=0
solve for x.
x=1 check that I did it in my head. Mornings is not a great time for me to try that.

math/calc  Reiny, Wednesday, March 31, 2010 at 9:55am
You appear to have a typo,
I will assume your curve is y = x^2e^(3x) or y = x^2/e^(3x)
let the point of contact of the hypotenuse be (x,y) on the curve.
Then the right angle will be at (x,0)
and the
Area = xy/2
= (1/2)(x)x^2e^(3x)
2A = (x^3)(e^(3x))
2dA/dx = x^3(3e^(3x)) + 3x^2(e^(3x))
= 3x^2(e^(3x))[x  1}
= 0 for a max of A
3x^2 = 0 >x=0, little sense, since no triangle
or
e^(3x) = 0 > no solution
or
x = 1, yeahh!
if x = 1 then
A = (1/2)1^3(e^3) = .0249
( I tested for the area with x = .99 and x = 1.01 and they were both smaller than .0249 by a "smidgeon")

math/calc  miley, Wednesday, March 31, 2010 at 11:13am
thank you and yes i had atypo i apologized it was y = x^2e^(3x
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