Question

The hypotenuse of a right triangle has one end at the origin and one end on the curve y = x^5*e^(−5x), with x ≥ 0. One of the other two sides is on the x-axis, the other side is parallel to the y-axis.

a) Express the area of the triangle in terms of x.
b) Find the critical point for x>0.
c) Find the maximum area of the triangle. Give an exact answer.

Answer 1

The curve is: \[y=x^{5}e^{-5x}\]

Answer 2

For part a)\[A=\frac{ 1 }{ 2 } (x*x^{5}e^{-5x})\]

Answer 3

I am stuck on how to proceed with part b.

Answer 4

First derivative test: first find what values of \(x\) make
\[\dfrac{d}{dx}[x^5~e^{-5x}]=0\]

Answer 5

\[\frac{ d }{ dx }(x^{5}e^{-5x})=e^{-5x}(5x^4-5x^5)\]

Answer 6

I was getting x=1

Answer 7

\(e^{-5x}>0\) for all \(x\), so solving \(5x^4-5x^5=0\) indeed gives you \(x=0,1\) (disregarding 0).

Now check to see if \(x=1\) is a critical point. The sign of the derivative to either side of the point determines whether the curve is increasing or decreasing:

On the interval \((0,1)\), \(y'>0\), so \(y\) is increasing on this interval.
On \((1,\infty)\), \(y'<0\), so \(y\) is decreasing on this interval.

The increase-decrease indicates a maximum at \(x=1\). And since \(y\) is continuous at \(x=1\), that makes \(x=1\) a critical point.

Answer 8

Okay, thank you!
The website kept saying x=1 was wrong so I'll be sure to contact my instructor about that.

Answer 9

Maybe the answer is the point itself, which would be \((1,y(1))\) ?

Answer 10

The box has it marked as x= and then I am supposed to type in the answer. I had wondered if I had to type the entire point but it says it only wants the x-value. Oh well.
Thank you again!

Answer 11

You're welcome