A triangle is drawn in a region bounded by the $x$-axis and a polynomial curve with all real roots, with one side of the triangle on the $x$-axis, as shown in this example:

What is the supremum of the ratio $R$ of the triangle's area to the region's area?

(We require that the polynomial has all real roots, because otherwise the answer would be trivial: the supremum would be $1$, for example with $y=x-x^n$ and letting $n\to\infty$.)

I considered the largest area triangle drawn in the region bounded by the $x$-axis and the polynomial $y=(-1)^{n-1}(x^2-1)\left(\left(\frac{x}{a}\right)^2-1\right)^n, a>1$, from $x=-1$ to $x=1$.

Numerical investigation suggests that for any given $a$ value, the ratio $R$ is maximized when $n\approx 1.880001a^2$.

$\color{blue}{a=10}$:

$\space R\approx0.8953810$ with $y=(x^2-1)\left(\left(\frac{x}{10}\right)^2-1\right)^{\color{red}{187}}$

$\space R\approx0.8953823$ with $y=-(x^2-1)\left(\left(\frac{x}{10}\right)^2-1\right)^{\color{red}{188}}$

$\space R\approx0.8953822$ with $y=(x^2-1)\left(\left(\frac{x}{10}\right)^2-1\right)^{\color{red}{189}}$

$\color{blue}{a=100}$:

$\space R\approx0.8955295$ with $y=-(x^2-1)\left(\left(\frac{x}{100}\right)^2-1\right)^{\color{red}{18799}}$

$\space R\approx0.8955296$ with $y=(x^2-1)\left(\left(\frac{x}{100}\right)^2-1\right)^{\color{red}{18800}}$

$\space R\approx0.8955295$ with $y=-(x^2-1)\left(\left(\frac{x}{100}\right)^2-1\right)^{\color{red}{18801}}$

$\color{blue}{a=1000}$:

$\space R\approx0.89553095$ with $y=(x^2-1)\left(\left(\frac{x}{1000}\right)^2-1\right)^{\color{red}{1880000}}$

$\space R\approx0.89553105$ with $y=(x^2-1)\left(\left(\frac{x}{1000}\right)^2-1\right)^{\color{red}{1880001}}$

$\space R\approx0.89553101$ with $y=(x^2-1)\left(\left(\frac{x}{1000}\right)^2-1\right)^{\color{red}{1880002}}$

I conjecture that $\text{sup}(R)$ is approached when $a\to\infty$ (for each value of $a$, we choose the value of $n$ that maximizes $R$), and that $\text{sup}(R)\approx 0.895531$.