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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 $n\approx 1.88a^2$.

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

$\space R\approx0.8953810$ 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 $y=-(x^2-1)\left(\left(\frac{x}{100}\right)^2-1\right)^{\color{red}{18801}}$

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

$\space R\approx0.8955309$ with $y=(x^2-1)\left(\left(\frac{x}{1000}\right)^2-1\right)^{\color{red}{1879990}}$

$\space R\approx0.8955311$ with $y=(x^2-1)\left(\left(\frac{x}{1000}\right)^2-1\right)^{\color{red}{1880000}}$

$\space R\approx0.8955310$ with $y=(x^2-1)\left(\left(\frac{x}{1000}\right)^2-1\right)^{\color{red}{1880010}}$

I conjecture that the supremum of $R$ is approached when $a\to\infty$ (for each value of $a$, we choose the value of $n$ that maximizes $R$), and that $R\approx 0.8955$.

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 $\text{sup}(R)\approx 0.895531$.

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 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 polynomials of the form $y=(-1)^{n-1}(x^2-1)\left(\left(\frac{x}{a}\right)^2-1\right)^n, a>1$, from $x=-1$ to $x=1$.

It seems that for any given $a$ value, the ratio is maximized with a certain $n$ value.

For example, when $a=5$, the ratio $R$ is maximized when $n=47$:

$R\approx0.894915$ 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 $y=-(x^2-1)\left(\left(\frac{x}{5}\right)^2-1\right)^{\color{red}{48}}$

I conjecture that the supremum of the ratio is approached when $a\to\infty$ (and for each value of $a$, we choose the value of $n$ that maximizes the ratio).

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 $n\approx 1.88a^2$.

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

$\space R\approx0.8953810$ 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 $y=-(x^2-1)\left(\left(\frac{x}{100}\right)^2-1\right)^{\color{red}{18801}}$

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

$\space R\approx0.8955309$ with $y=(x^2-1)\left(\left(\frac{x}{1000}\right)^2-1\right)^{\color{red}{1879990}}$

$\space R\approx0.8955311$ with $y=(x^2-1)\left(\left(\frac{x}{1000}\right)^2-1\right)^{\color{red}{1880000}}$

$\space R\approx0.8955310$ with $y=(x^2-1)\left(\left(\frac{x}{1000}\right)^2-1\right)^{\color{red}{1880010}}$

I conjecture that the supremum of $R$ is approached when $a\to\infty$ (for each value of $a$, we choose the value of $n$ that maximizes $R$), and that $R\approx 0.8955$.

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 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 polynomials of the form $y=(-1)^{n-1}(x^2-1)(x^2-a^2)^n, a>1$, from $x=-1$ to $x=1$.

It seems that for any given $a$ value, the ratio $R$ is maximized with a certain $n$ value.

For example, when $a=5$, the ratio $R$ is maximized when $n=47$:

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

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

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

I conjecture that the supremum of the ratio is approached when $a\to\infty$ (and for each value of $a$, we choose the value of $n$ that maximizes the ratio).

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 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 polynomials of the form $y=(-1)^{n-1}(x^2-1)\left(\left(\frac{x}{a}\right)^2-1\right)^n, a>1$, from $x=-1$ to $x=1$.

It seems that for any given $a$ value, the ratio is maximized with a certain $n$ value.

For example, when $a=5$, the ratio $R$ is maximized when $n=47$:

$R\approx0.894915$ 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 $y=-(x^2-1)\left(\left(\frac{x}{5}\right)^2-1\right)^{\color{red}{48}}$

I conjecture that the supremum of the ratio is approached when $a\to\infty$ (and for each value of $a$, we choose the value of $n$ that maximizes the ratio).