I came across this problem while playing around with some physics :
Define a function which passes through $(0,0)$ and $(x,0)$ for which the area under the curve from $0$ to $x$ is some constant $A$, such that $|m|$ (the slope of the graph) is the least possible.
Given $x > 0$ and $A$, minimise $\max_{\xi \in [0, x]} \lvert f'(\xi) \rvert$ for differentiable functions $f \colon [0, x] \to \mathbb{R}$ satisfying $\int_0^x f(\xi) \, d\xi = A$.
For example :
Let the area under the graph from $0$ to $1$ is $1$, $|m| > M$ find $M$. I don't really have an approach, but I made a graph using straight lines with slopes $m1$ and $m2$. This give a triangle with height $2$, and we can easily see that the minimum slope would be $4$. How would this change if one sketches any graph without the compulsion of it being a first-degree polynomial function.
Let's rename the quantity $x$ to $a$ and assume without loss of generality that $A \ge 0$. (The same reasoning with $A$ replaced by $-A$ applies when $A < 0$.)
We have a differentiable function $f \colon [0, a] \to \mathbb{R}$ with $f(0) = f(a) = 0$ and $$ \int_0^a f(y)\,dy = A > 0. $$ Let $$ m = \max_{x \in [0, a/2]} \lvert f'(x) \rvert \quad \text{and} \quad n = \max_{x \in [a/2, a]} \lvert f'(x) \rvert. $$ Using the fundamental theorem of calculus, we split $A = A_1 + A_2$ where $$ A_1 = \int_0^{a/2} f(y)\,dy = \int_0^{a/2} \int_0^y f'(x) \, dx \, dy \le \int_0^{a/2} \int_0^y m \, dx \, dy = \frac{1}{8} m a^2 $$ and $$ A_2 = \int_{a/2}^a f(y)\,dy = \int_{a/2}^a \biggl( -\int_y^a f'(x) \, dx \biggr) \, dy \le \int_{a/2}^a \int_y^a n \, dx \, dy = \frac{1}{8} n a^2. $$ Hence, $$ \max\{m, n\} \ge \frac{m + n}{2} \ge \frac{1}{2} \cdot \frac{A}{\frac{1}{8} a^2} = \frac{4A}{a^2}, $$ where equality holds if and only if $m = n = f'(x_1) = - f'(x_2)$ for almost all $x_1 \in [0, a/2]$ and $x_2 \in [a/2, a]$, which happens precisely when the graph of $f$ is the polygonal chain $$ (0, 0) - \biggl(\frac{a}{2}, \frac{2A}{a}\biggr) - (a, 0) $$ by Darboux's theorem. This optimal case is not admissible since $f$ would not be differentiable at $\frac{a}{2}$, but it can be approximated to arbitrary precision by slightly raising the graph and rounding the corner. Therefore, we conclude that $$ \inf_{f} \max_{x \in [0, a]} \lvert f'(x) \rvert = \frac{4 \lvert A \rvert}{a^2}. $$