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:

[![enter image description here][1]][1]

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

(We require that the polynomial has all real roots, because [otherwise][2] the answer would be trivial: the supremum would be $1$, for example with $y=x-x^n$ and letting $n\to\infty$.)

I hypothesize that the supremum of the ratio is approached with a region bounded by the $x$-axis from $x=-1$ to $x=1$, and $y=(\pm)(x^2-1) \prod\limits_{k=1}^n \left(\left(\frac{x}{r_k}\right)^2-1\right), r_k\ge1$, $r_k$'s not necessarily unique.  

[![enter image description here][3]][3]

Using this hypothesis, the largest ratio $R$ that I have found is

$R\approx0.894934$ with $y=(x^2-1)\left(\left(\frac{x}{5}\right)^2-1\right)^{47}$

Strangely, if we adjust the power of the $\left(\left(\frac{x}{5}\right)^2-1\right)$ factor, the ratio has a maximum when the power is $47$.

$R\approx0.894915$ with $y=-(x^2-1)\left(\left(\frac{x}{5}\right)^2-1\right)^{\color{red}{46}}$

$R\approx0.894934$ with $y=(x^2-1)\left(\left(\frac{x}{5}\right)^2-1\right)^{\color{red}{47}}$

$R\approx0.894933$ with $y=-(x^2-1)\left(\left(\frac{x}{5}\right)^2-1\right)^{\color{red}{48}}$

And if we adjust the value of the repeated roots, the ratio has a maximum when the repeated root is about $5$.  

$R\approx0.892010$ with $y=(x^2-1)\left(\left(\frac{x}{\color{red}{4}}\right)^2-1\right)^{47}$

$R\approx0.894934$ with $y=(x^2-1)\left(\left(\frac{x}{\color{red}{5}}\right)^2-1\right)^{47}$

$R\approx0.890975$ with $y=(x^2-1)\left(\left(\frac{x}{\color{red}{6}}\right)^2-1\right)^{47}$

  [1]: https://i.sstatic.net/JWnw4.png
  [2]: https://math.stackexchange.com/q/4648043/398708
  [3]: https://i.sstatic.net/Z4JtN.png