What is generally the best way to approach optimization problems in Calculus 1? I'm having trouble even getting started on the basic ones like the fence ones or the rectangle ones. All I know is that they deal with minimizing and maximizing values. I would appreciate any tips and/or advice!

Here is a typical question:

"A can is to hold 680 cm^3 and it costs twice as much for the material for the circular base and the top as it does to make the "sleeve" of the can.
What must be the radius and the height of the can for a minimum cost"

First thing you do is to visualize the problem, usually by making a sketch.
It asks for the radius and the height, so let's define them as r cm and h cm
mark them on your sketch
It also asks for "a minimum cost" , so our main equation will be
cost = .....
Well, what determines the cost?
It costs twice as much for the material forming the sleeve of the can, than for the two circles.
Imagine taking apart the can, you would have two circles and a rectangle.
the area of the rectangle = 2πrh and its cost is 1(2πrh) units of money
the area of the two circles = 2(πr^2) and its cost is 2(2πr^2) = 4πr^2 units of money

Main equation:
Cost = 2πrh + 4πr^2
but the right side contains 2 variables, eventually we will have to differentiate it, and we can only have one kind of variable.
So ...., how to we get rid of one of the variables?
Well, what information haven't we used yet?
the volume must be 680 cm^3
volume of cylinder = πr^2 h
πr^2 h = 680
h = 680/(πr^2) , replace h in our cost equation
cost = 2πrh + 4πr^2
=2π r (680/(πr^2) + 4πr^2
= 1360/r + 4πr^2
I now have a function equation for cost in terms of a single variable, you usually want this!
So let's take the derivative with respect to r
d(cost)/dr = - 1360/r^2 + 8πr
= 0 for a max/min of cost
so 8πr = 1360/r^2
r^3 = 170/π
r = (170/π)^(1/3) = 3.78 cm
then h = 15.13

So the minimum cost is obtained when the radius is 3.78 cm and the height is 15.13 cm

If it had asked for the actual minimum cost, you would then sub in those values into
cost = 2πrh + 4πr^2

Hope this shows you the way I usually go about these kind of problems.
The difficulty is that they are all different, but you will find some common properties
1. Look for what it is that will be max or minimized
2. Form that main equation
3. If the right side contains more than one variable, you must somehow replace one of them
--- so look for information that somehow relates your different variables
4. Once you have done that, take the derivative , set it equal to zero and solve that equation

Good luck