For an input $x$ and output $y$ of a system it is know that $x,y$ always satisfy
$$ F(x,y) = 10 $$
At a certain point, $x=1$ and $y=1$. The question is how $y$ responds to a small decrease in $x$, e.g. to $ 0.999$ (that is, what is $y'(x)$ of the function $y(x)$ near $x=1$.
Now this somehow asks for an application of the Implicat Function Theorem, of a Taylor Polynomial, or both of them.
Intuitively, as $F(x,y) = 10$, can I not simply conclude that each decrease in $x$ has to be compensated by a change in $\Delta y =-x$, in order to keep $F(x,y)$ constant again? However, on economic grounds, a decrease in inputs resulting in an increase in output sounds kind of awkward.
So I summed up what I know again:
$$F(1,1) = 10 $$
$$ F(0.999, 1 + y'(1)) = 10 $$
$$ y'(1) = - \frac{\frac{\partial{F}}{x}(1,1)}{\frac{\partial{F}}{y}(1,1)} $$
But there I'm stuck - how to solve when both $F$ and $y$ are unknown?
Thanks for all hints!
You can use total differential. If $F(x,y)$ is constant it means that $$dF=\frac{\partial F}{\partial x}dx+\frac{\partial F}{\partial y}dy=0$$ $$\Rightarrow \frac{dy}{dx}=y'=-\frac{\frac{\partial F}{\partial x}}{\frac{\partial F}{\partial y}}$$ In economics it makes sense since it is an implicit function. Assume that the production function is given such as $$y=x^2+10 \Rightarrow F(x,y)=y-x^2=10$$ and $$y'=-\frac{\frac{\partial F}{\partial x}}{\frac{\partial F}{\partial y}}=-\frac{-2x}{1}=2x$$ which shows output will increase as a result of an input increase.