How to find a function $y:[0,1] \rightarrow [0, +\infty), y=y(x), x \in [0, 1]$ to make $\int_0^1y(x)dx$ minimum with the following restrictions:
$\begin{array}{cc} y(0) &=& 0, \\ \int_0^1y(t)r(1-t)dt &=& C \end{array}$
where $C>0$ is a constant, and $r(x)$ is a known rational function, $r(x) = \frac{ax + b}{x + b}$, where $0<a<1$ and $b>1$ have been known.
I think the problem may be applied with some method from the calculus of variations. But I don't know how to do it.
I presume here (as there is some lack of precision in the question) that $r(1-t)$ is continuous on $[0,1]$, that $M=\max\{r(1-t): 0\leq t\leq 1\}$ is positive and also that $C$ is positive.
When $y\geq 0$ the inequality:
$$ C=\int_0^1 y(t)r(1-t)\; dt \leq M \int_0^1 y(t)\; dt $$ shows that $I=\int_0^1 y(t)\; dt \geq C/M$. Let the max of $r$ be attained at $t_0$. Then for $\epsilon>0$ there is $\delta>0$ so that $r(1-t)\geq M-\epsilon$ for $t\in J=[t_0-\delta,t_0+\delta]\cap [0,1]$. If $y$ has support in $J$ then we also have the lower bound: $$ C=\int_0^1 y(t)r(1-t)\; dt\geq (M-\epsilon) \int_0^1 y(t)\; dt $$ So $I\geq C/(M-\epsilon)$. The infimum value is thus $C/M$ but it need not be attained unless there is some interval on which $r$ attains its maximum. The upshot is that we concentrate the 'mass' of $y$ around the point $t_0$ so as to satisfy the constraint in the most economical way (smallest integral of $y$).
If $C$ is negative then $r$ has to negative for some values (or else the constraint can not be satisfied). Letting $m<0$ be the minimum of $r$, the inf of $\int_0^1 y\;dt$ is then $(-C)/(-m)>0$ (seen as above by concentrating the mass of $y$ near the minimum of $r$).