Let $f$ be a smooth positive monotonically increasing real function which is defined and finite in $[0,1]$, and define the following two quantities (see the figure below):
- $F=\int_{x=0}^1{f(x)dx}$ = the area under $f$.
- $G=\max_{x=0}^1{f(x)*(1-x)}$ = the largest rectangle bounded by $f$ and the point (0,1). Usually the maximum is attained when $x$ satisfies the first-order condition: $f(x)=(1-x)f'(x)$.
What function $f$ maximizes the ratio $F/G$?
The main challenge here is that this is a maximization problem on the set of all functions.
I managed to solve the problem for specific families of functions:
Consider the family of functions $f_d(x)=x^d$, where $d>0$. Then:
$F_d = \frac{1}{d+1}$
$G_d$ is attained when $x^d=(1-x) d x^{d-1}$, i.e. $x=\frac{d}{d+1}$ and $G_d=(1-x)x^{d}=\frac{1}{d+1} (\frac{d}{d+1})^d$
$F_d/G_d = (\frac{d+1}{d})^d$
When $d \to \infty$, the ratio goes to $e\approx2.7$, so this is the supremum of $F_d/G_d$ for this family.
But maybe there are functions for which $F/G$ is larger? Here is another example I tried:
Consider the family of functions $f_d(x)=e^{xd}$, where $d>0$. Then:
$F_d = \frac{e^d-1}{d}$
$G_d$ is attained when $e^{x d}=(1-x) d e^{x d}$, i.e. $x=1-\frac{1}{d}$ and $G_d=(1-x)e^{x d}=\frac{1}{d} e^{d-1}$
$F_d/G_d = (e^d-1)/(e^{d-1})$
When $d \to \infty$, the ratio again goes to $e\approx2.7$.
But maybe there are functions for which $F/G$ is larger than $e$? How can I find the function which maximizes the ratio among all possible functions?
Consider the function $f_t:[0,1]\to\Bbb R$, with $t>1$.
$$f_t(x)=\frac1{t(1-x)+t-1}\\ \begin{align} \implies F_t &= \int_0^1f_t=\frac 1t\log\left(\frac{2t-1}{t-1}\right)\\ \implies G_t &= \max\{f_t(x)(1-x)\mid x\in[0,1]\}=\frac 1{2t-1}\\ \end{align}$$
As $t$ approaches $1$ from the right, $f_t(x)$ approaches $\frac1{1-x}$. The maximum area of a rectangle bounded by $f_t$ approaches $1$, but the area under the entire curve grows without bound.
$$\lim_{t\to1^+}\frac{F_t}{G_t}=\infty$$