Well, I Bring you the following problem:
Determine the image of $f: A \rightarrow \mathbb{R}, f(x,y)= \log(x+4) + \log(y+4)$
$A=\{(x,y) \in \mathbb{R^{2}}: (x-2)^{2} + (y-2)^{2} \le1, y\le2\} $
attempt of solution: I put the conditions on general form:
$(x-2)^{2} + (y-2)^{2}-1 \le0$ $=$ $g(x,y)$
$y-2$ $\le$ $0$ $=$ $h(x,y)$
Then, the expression to work on will be: $T=f(x,y) - Lg(x,y) -Gh(x,y)$ were L and G are the Lagrange multipliers
at this point I do the partial derivatives:
$\frac{\partial{T}}{\partial{x}}= \frac{1}{x+4}-2L(x-2)$ $=0$
$\frac{\partial{T}}{\partial{y}}= \frac{1}{y+4}-2L(y-2)-G$ $=0$
$\frac{\partial{T}}{\partial{L}}$= $-(x-2)^{2} - (y-2)^2$ $=0$
$\frac{\partial{T}}{\partial{G}}$= $-(y-2)$ $=0$
For some reason I don't find any congruent solutions for this system.
Thanks for your help
PS: hint, maybe is necessary to use the reduction theorem to study the $\partial{A}$.