Being given the area $S>0$, maximize the upper bound of a definite integral $\tau>0$ of a continuous, differentiable, one-variable function $f(t)$.
$$ \int_{0}^\tau f(t)\,dt=S.$$
Assume that the function $f(t)$ satisfies the following properties:
1) $f(0)=a$, where $a>0$ and is a given constant;
2) $f(t)>0$ for $t \in [0,\tau)$;
3) $f(\tau)=0$.
EDIT: differentiable.
The answer is $\frac S \tau$. Clearly $S=\int_0^{\tau} f(x)dx \leq M \tau$ where $M$ is the supremum of f. So $M \geq \frac S \tau$. Now let $\epsilon >0$ and $f$ be the function with $f(0)=a,f(\tau)=0,f(x)=a_{\epsilon} \frac S \tau$ on $[\epsilon, \tau -\epsilon]$ and linear in $[0,\epsilon]$ as well as $[\tau -\epsilon, \tau]$. We can choose $a_{\epsilon}$ so that $\int_0^{\tau} f(x)dx=S$ As $\epsilon \to 0$ $a_\epsilon \to 1$ and the result follows from this. Note that the restrictions on $f(0)$ and $f(\tau)$ have no effect on the answer. I just noticed that you have imposed differentiability now, but the answer is the same since we can 'smoothly' join the graph on $[\epsilon, \tau -\epsilon]$ to the points $(0,a)$ and $(\tau ,0)$ and still choose $a_{\epsilon}$ to make teh integral $S$.