October
16, 2006
The
horizontal line $y=c$ intersects the curve $y=2x-3x^3$ in the first
quadrant as in the figure. Find $c$ so that the areas of the two shaded
regions are equal.
Solution. Let $(b,c)$
denote the rightmost intersection point. We wish to find $c$ such
that
$\int_0^b \left((2x-3x^3)-c\right)\,dx = 0.$
This leads to $b^2-(3/4)b^4-bc=0$. Since $c=2b-3b^3$, we get $b^2 -
(3/4)b^4 - b(2b-3b^3)=0 \Longrightarrow b^2-(3/4)b^4-2b^2 +3b^4 = 0
\Longrightarrow (9/4)b^4 - b^2 = 0 \Longrightarrow b=0, b=\pm 2/3$
So $b=2/3$ is the unique positive solution, leading to $c=4/9$.
We can double-check this answer by finding the zeros of $2x-x^3-4/9 =
(2/3-x)(3x^2+2x-2/3)$. The zeros of the last quadratic are
$x=(-1\pm \sqrt{3})/3 \approx -0.910684, 0.244017$. Therefore,
the point $(2/3, 4/9)$ is the rightmost intersection point of $y=4/9$
and $y=2x-3x^3$ as advertised.
