How to solve problem of maximum and minimum in two variables? I have function
$$f(x,y)=\frac{3(x+y)-2}{x^2+y^2}$$
and need maximum and minimum in domain
$$D=\left\{(x,y)\in\Bbb R^2 ; x^2+y^2=8\right\}\cup\left\{(x,y)\in\Bbb R^2;|x|+|y|=1\right\}$$
I thought of Lagrange multipliers but that square above is hard: it is enclosed by lines $\;y=x+1,\,y=-x+1,\,y=x-1,\,y=-x-1\;$ , so my question is:
Can I do only two constraints and form the function
$$H(x,y,\lambda,\mu)=f(x,y)-\lambda(x^2+y^2-8)-\mu(|x|+|y|-1)$$
or else I need five constraints, one for circle and each line above
$$G(x,y,\lambda_1,\lambda_2,\lambda_3,\lambda_4,\lambda_5)=f+\lambda_1(x^2+y^2-8)-\lambda_2(x-y+1)-\lambda_3(-x+y+1)-$$
$$-\lambda_4(x-y-1)-\lambda_5(-x-y-1)\;?$$
Both ways looks very hard, because in first taking derivatives of $\;x,\,y\;$ give two possible signs, for example:
$$H'_x=f'_x-2\lambda x\,\mp\mu\;,\;\;\text{and etc.}$$
whereas in second way, with $\;G\;$, I need to solve seven nonlinear equations...
Is there perhaps some other easier way to do? I thought: do first partial equations of $\;f(x,y)\;$ alone, with Hessian and find points inside the domain, and then do substitution for circle $\;x^2+y^2=8\;$ and then for square $\;|x|+|y|=1\;$ and find, if possible, maximal points by evaluation. For example: in $\;|x|+|y|=1\;$ , it is clear $\;-1\le x,\,y\le 1\;$ , so perhaps:
$$\frac{3(x+y)-2}{x^2+y^2}\le\frac{3-2}{1}=1\;\;\;\text{ and etc...}$$
This really confuses...Thanks for any help.

With $|x|+|y|=1$ we obtain:
$$\frac{3(x+y)-2}{x^2+y^2}\leq\frac{3(|x|+|y|)-2}{\frac{(|x|+|y|)^2}{2}}=2$$
The equality occurs for $x=y=\frac{1}{2}$.
With $x^2+y^2=8$ we get a smaller value.
In another hand, for $x^2+y^2=8$ we obtain:
$$\frac{3(x+y)-2}{x^2+y^2}\geq\frac{-3\sqrt{(x+y)^2}-2}{x^2+y^2}\geq$$
$$\geq\frac{-3\sqrt{2(x^2+y^2)}-2}{x^2+y^2}=-\frac{7}{4}.$$
The equality occurs for $x=y=-2$.
With $|x|+|y|=1$ we'll get for $x=y=-\frac{1}{2}$ a value $-10$.
Now, easy to show that it's a minimal value.
Indeed, we need to prove that
$$3(x+y)-2\geq-10(x^2+y^2)$$ or
$$3\left(\left(x+\frac{1}{2}\right)^2+\left(y+\frac{1}{2}\right)^2\right)+7(x^2+y^2)-\frac{7}{2}\geq0,$$
which is true because $$x^2+y^2\geq\frac{1}{2}(|x|+|y|)^2=\frac{1}{2}.$$
Done!
