Hello I like Serena,
Thanks for providing me such a great solution to tackle the problem using trigonometric approach and I appreciate it and I also like it!
Your patience and willingness to answer my "request" touched me very much. And I know I approached it differently than you (because I didn't introduce another variable ($\theta$ in your case) in my attempt.). If I happen to be able to solve it again, I will quickly come to this thread and add the solution here so that I could always check back to my original line of thought, and I will keep my fingers crossed for that to occur!
Also, since I am new to the topic of Lagrange Multipliers, MarkFL offered to help guide me through using the technique on this problem.
First, he told me to state the objective function $f(x,y)$ and constraint function $g(x,y)=0$ and so I wrote:
$$f(x,y)=x+y$$
$$g(x,y)=x+y-3\left(\sqrt{x+1}+\sqrt{y+2} \right)=0$$
Next, he said to construct the following system:
$$f_x(x,y)=\lambda g_x(x,y)$$
$$f_y(x,y)=\lambda g_y(x,y)$$
And so I found:
$$1=\lambda\left(1-\frac{3}{1\sqrt{x+1}} \right)$$
$$1=\lambda\left(1-\frac{3}{1\sqrt{y+2}} \right)$$
And so I was then asked what this implies, and I could see that this implies:
$$x+1=y+2$$
$$y=x-1$$
So, Mark told me to substitute for $y$ in the constraint, and solve for $x$:
$$x+x-1-3\left(\sqrt{x+1}+\sqrt{x+1} \right)=0$$
$$2x-1=6\sqrt{x+1}$$
Squaring both sides (keeping in mind extraneous solutions may be introduced):
$$4x^2-4x+1=36x+36$$
$$4x^2-40x-35=0$$
The quadratic formula gives us:
$$x=\frac{10\pm3\sqrt{15}}{2}$$
A check reveals that the smaller root is extraneous, so we are left with:
$$x=\frac{10+3\sqrt{15}}{2}$$
And so:
$$x+y=2x-1=9+3\sqrt{15}$$
And so we may conclude that one extremum is:
$$f\left(\frac{10+3\sqrt{15}}{2},\frac{8+3\sqrt{15}}{2} \right)=9+3\sqrt{15}$$
I asked Mark how we know this is a maximum, since we have found only one extremum. He said this is where the boundary extrema you suggested come into play. He told me to look at the constraint to see if it implies any further constraints on $x$ and $y$, and so I noticed we must have:
$$x+1\ge0\implies x\ge-1$$
$$y+2\ge0\implies y\ge-2$$
Letting $x=-1$, we find the constraint function implies:
$$-1+y-3\sqrt{y+2}=0$$
Solving for $y$, we then find:
$$y-1=3\sqrt{y+2}$$
$$y^2-2y+1=9y+18$$
$$y^2-11y-17=0$$
$$y=\frac{11\pm3\sqrt{21}}{2}$$
A check reveals that the smaller root is extraneous, so we are left with:
$$y=\frac{11+3\sqrt{21}}{2}$$
And so another critical point is:
$$\left(-1,\frac{11+3\sqrt{21}}{2} \right)$$
And we find:
$$f\left(-1,\frac{11+3\sqrt{21}}{2} \right)=\frac{3\left(3+\sqrt{21} \right)}{2}$$
Now, checking the other boundary, we may let $y=-2$ and the constraint implies:
$$x-2-3\sqrt{x+1}=0$$
$$x-2=3\sqrt{x+1}$$
$$x^2-4x+4=9x+9$$
$$x^2-13x-5=0$$
$$x=\frac{13\pm3\sqrt{21}}{2}$$
A check reveals that the smaller root is extraneous, so we are left with:
$$x=\frac{13+3\sqrt{21}}{2}$$
And so another critical point is:
$$\left(\frac{13+3\sqrt{21}}{2},-2 \right)$$
And we find:
$$f\left(\frac{13+3\sqrt{21}}{2},-2 \right)=\frac{3\left(3+\sqrt{21} \right)}{2}$$
Now, since:
$$\frac{3\left(3+\sqrt{21} \right)}{2}<9+3\sqrt{15}$$
we may then conclude:
$$f_{\min}=f\left(-1,\frac{11+3\sqrt{21}}{2} \right)=f\left(\frac{13+3\sqrt{21}}{2},-2 \right)=\frac{3\left(3+\sqrt{21} \right)}{2}$$
$$f_{\max}=f\left(\frac{10+3 \sqrt{15}}{2},\frac{8+3\sqrt{15}}{2} \right)=9+3\sqrt{15}$$
