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 x-asis, the other side is parallel to the y-axis. Find the maximum area of such a triangle. At what x-value 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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