Hello, I have been given this task, which I couldn't solve properly:
There are given 2 functions:
$f(x) = 3xe^{-x}$
$g(x) = \frac{x}{2}$
and number $v$ is the x coordinate of their second intersection (not 0).
The task is to find parameter $u$, where $0 < u < v$, such that triangle formed by vertices $[0, 0], [u, f(u)], [u, g(u)]$ has maximal area.
Please take in mind, this is supposed to be a high school level solution.
My idea was to substract the area of triangle $T2=\{[0,0], [u, 0], [u, g(u)]\}$ from the area of triangle $T1 = \{[0, 0], [u, 0], [u, f(u)]\}$ which took me to this fuction: $\frac{3}{2} x^{2} e^{-x} - \frac{x^{2}}{4}$ and it's derivation $-\frac{1}{2}e^{-x} x (-6 + e^{x} + 3 x)$. Then searching for extremes:
$-\frac{1}{2}e^{-x} x (-6 + e^{x} + 3 x) = 0$
$-6 + e^{x} + 3 x = 0$
...which is probably not elementary solvable and I would have to use something like Lambert W function, which is not high school level Math.
Is there a better way to approach this problem or did I make a mistake?
The area of the triangle is: $$S=\frac12\cdot u\cdot 3ue^{-u}-\frac12\cdot u\cdot \frac u2\\ S'=3ue^{-u}-\frac32u^2e^{-u}-\frac u2=0 \Rightarrow u\approx 1.04869.$$