To find the extreme values of function
From your last statement, it is apparent that you understand the general technique for finding extrema: for a differentiable function $f(x,y,\cdots)$ on an open or closed domain $D$, we first look in the interior for critical points. This by finding common zeros of all partial derivatives. Once we have found the critical points, we examine their nature (suprema, infima, saddle points). Finally, if $D$ has a boundary, as in your first problem, we investigate the boundary separately. This can be done by Lagrange multipliers or by parameterizing the boundary and finding extrema of the resulting function on the boundary; the second method makes sense in your case.
So it seems that your problems are more computational than choosing the wrong techniques. I will try to help you with the questions without giving away the answers completely (this would be meaningless).
- Our function is $f(x,y) = \sin(x)+\sin(y)+\sin(x+y)$ and our domain is the square $[0,\pi/2]\times [0,\pi/2]$. In the interior of the square, look at the common zeros of $\partial_x f$ and $\partial_y f$. For instance $$\partial_x f = \cos(x) +\cos(x+y)=0$$ and $$\partial_y f = \cos(y)+\cos(x+y)=0.$$ If both these equations are to hold, we can subtract them equation by equation (thereby getting rid of the annoying $\cos(x+y)$ term) and get $$ \cos(x)-\cos(y)=0.$$ But on $[0,\pi/2]$ the cosine is monotonic, so the only way two cosines can be equal is if their arguments are so. Thus $x=y$. Then we return to either of the original equations to get $\cos(x)+\cos(2x)=0$ and we can use $\cos(2x) = 2\cos(x)^2 -1$ to conclude $$2\cos(x)^2+\cos(x)-1=0,$$ a quadratic that I trust you can solve for $\cos(x)$ finally giving $x$. You will find two values, but only one is of significance!
The boundary is easier. You just need to check four line segments: $x=0,\pi/2$ and $y=2,\pi/2$. For instance, on the line segment $y=\pi/2$ the variable $x$ runs from $0$ to $\pi/2$ and your function is $$f(x,\pi/2)=\sin(\pi/2)+\sin(x)+\sin(x+\pi/2) = 1+\sin(x)+\cos(x).$$ I trust you can find extrema for this function. After you are done with all four sides of the square (do not forget to check the corners!), gather all the values that you found along with the one in the interior and compare them! A theorem you know states that a continuous function always has a minimum and a maximum on a closed, bounded set, so you know the biggest of the values you found is the maximum and the smallest one the minimum.
Here there is no boundary, so it is a smaller computation. Do not be intimidated by the degree of the monomials that appear. Reason smartly: you want to find the maximum. If $x,y,z$ are all a billionth, you see that $f$ has a positive value. So $0$ cannot be the maximum. But if any of the three variables is zero, your function becomes zero. Since that cannot be the maximum, you can safely assume $x,y,z\neq 0$, which helps a lot because you can freely divide by them. Now just take partial derivatives. For instance, $$\partial_z f = xy-x^2y-xy^2-2xyz=0.$$ Dividing by $xy$ you get $$1-x-y-2z=0.$$ You will get similar equations from the other two derivatives, and you will easily conclude that $x,y,z$ must all be equal, and then going back to the equation above conclude that any one of them is $1/4$. Now, this is a critical point, but is it a maximum, a minimum or a saddle point? For this, you want to check the Hessian of $f$ at the critical point, and this is a straightforward computation now that you have its coordinates.
Milly settled this partially above. To show that a function has no extrema all you need to do is find points with higher and higher (resp. lower and lower) values. Can you understand that if you do this, you can conclude that your function has no global extrema? If yes, then all you need to do is find convenient points. And what is more convenient than just picking your favorite line and see if it works? Some of my favorite lines are the coordinate axes like $x=y=0$. If I restrict your function there I get $f(0,0,z)=2z^2$ and for big $z$ it gets as large as you want, larger than any hypothetical global extremum, so the latter cannot exist. Milly did this picking another line $x=y=-z$ that gave large negative values. In general, test your function on simple curves. Start with simple lines. If they don't work, why don't they work? Answering this question may give you a hint as to what other curve to use.
This is of the same form as 1. Can you see what, if anything, changes? In particular, what is the domain of the function? Is it open, or closed? Proceed as in problem 1.
Some general pointers about finding extrema: always begin by having a clear picture of the domain and its geometric properties. Is it open, closed? What is the boundary? The interior? If you do not have perfect command of the domain, you've lost the game before even getting the leather armor.
Then work systematically, but not blindly. Do not rush to Lagrange multipliers on a domain whose boundary has fifteen corners, reason that there must be an easier way. A symmetry, an easy parametrization or simply an observation about the function that allows you to deal with the boundary immediately (e.g. f is zero throughout the boundary). Pay attention to your function above the domain. Then treat the interior. The resulting system of equations may be tricky to do and there is no canonical method, of course, but always look for simplifications. If a term is just too complicated to work with, it may go away after some manipulation.
Do not be afraid to use your theorems. They are your arsenal, not your enemy! The fact that the Hessian allows you to understand the nature of a critical point in most cases gives you power and certainty.
Finally, treat the boundary; examine it, see if you need to use Lagrange multipliers and if so, double check if the equation describing the boundary is correct and that you have the right relations between the gradients. If not, parametrize the boundary carefully and solve the extrema problem one dimension down. At points where the boundary is not differentiable, differential methods fail and you need to check these points by hand. Usually they are not too many. If there are infinitely many of them, either you walked into the wrong class, or your instructor did.