# Optimization and Modeling

• Nov 14th 2006, 02:08 PM
Jacobpm64
Optimization and Modeling
The hypotenuse of a right triangle has one end at the origin and one end on the curve y = (x^2)*e^(-3x), with x > 0. One of the other two sides is on the x-axis, the other side is parallel to the y-axis. Find the maximum area of such a triangle. At what x-value does it occur?

All right, I know that I'll probably need A = (1/2)bh. I know I have to in some way find out where the curve intersects either the vertical side of the triangle or the hypotenuse to maximize the area? I think? But I don't know if i'm thinking the right way, and I don't know how to go about setting this up.
• Nov 14th 2006, 03:07 PM
Soroban
Hello, Jacobpm64!

It's easier than you think . . .

Quote:

The hypotenuse of a right triangle has one end at the origin
and one end on the curve $\displaystyle y = x^2e^{-3x}$, with $\displaystyle x \geq 0.$
One of the other two sides is on the x-axis, the other side is parallel to the y-axis.
Find the maximum area of such a triangle. At what x-value does it occur?

Code:

        |         |                *         |            *         |        *  |         |      *    |y         |    *      |       --*------------+---         |      x

The area of a triangle is: $\displaystyle A \,= \,\frac{1}{2}bh$

The area of this triangle is: .$\displaystyle A \:=\:\frac{1}{2}xy$ . . . where $\displaystyle y \:=\:x^2e^{-3x}$

So we have: .$\displaystyle A \:=\:\frac{1}{2}x\cdot x^2e^{-3x}\quad\Rightarrow\quad A \:=\:\frac{1}{2}x^3e^{-3x}$

Differentiate: .$\displaystyle A' \;= \;\frac{1}{2}x^2\!\cdot\!e^{-3x}(-3) + \frac{1}{2}\!\cdot\!3x^2\!\cdot\!e^{-3x} \;= \;\frac{3}{2}x^2e^{-3x}(-x + 1)$

Equate to zero: .$\displaystyle \frac{3}{2}x^2e^{-3x}(-x + 1)\;=\;0$

Set each factor equal to zero and solve.

. . $\displaystyle \frac{3}{2}x^2\,=\,0\quad\Rightarrow\quad x\,=\,0$ . . . This gives minimum area ... ha!

. . $\displaystyle e^{-3x} \,=\,0$ . . . but $\displaystyle \frac{1}{e^{3x}}$ can never equal zero.

. . $\displaystyle -x + 1 \:=\:0\quad\Rightarrow\quad x\,=\,1$ . . . There!

Maximum area occurs when $\displaystyle \boxed{x \,= \,1}$ and $\displaystyle y \,= \,1^2\!\cdot\!e^{-3} \,= \,\frac{1}{e^3}$

The maximum area is: .$\displaystyle A \:=\:\frac{1}{2}(1)\left(\frac{1}{e^3}\right) \:=\:\boxed{\frac{1}{2e^3}\text{ square units}}$

• Nov 14th 2006, 03:29 PM
Jacobpm64
thanks!

Is that LaTeX that you used to format the text?

If so, what's the tag for it in these forums? I know the other forums I use are , and they don't seem to work here :eek:
• Nov 14th 2006, 04:32 PM
ThePerfectHacker
Quote:

Originally Posted by Jacobpm64
thanks!

Is that LaTeX that you used to format the text?

If so, what's the tag for it in these forums? I know the other forums I use are , and they don't seem to work here :eek:

This is what you use.