Math 1520: Finding the domain in Max/Min Word problems
Published: Wed, 03/06/13
Many profs insist that you include the domain when solving a Max/Min Word Problem. I rarely do this in my book because I do not consider this very important at all. Sometimes it is too hard to find anyway, and not worth the effort. However, I regret not showing the domain in my solutions. Failure to include the domain in a max/min word problem could cost you marks.
Here are some tips, though, if you want to include the domain (or endpoints).
First, as I say in my steps, you can generally assume that x > 0. The difficult thing is to decide if the domain actually includes 0 or not.
When in doubt, say the domain is (0, ∞). That is often the case. If the Q equation is undefined at x = 0, then you can definitely say the left endpoint of the domain is up to but, not including 0. This would usually be because x = 0 causes the denominator to be 0.
If, on the other hand, the Q equation is defined when x = 0 (subbing x = 0 into it does result in a number of some sort, any number at all), then you know the domain includes 0. That would mean the domain is [0, something ...
We now need to establish the right endpoint of the domain. Again, assume the right endpoint is infinity, ∞, unless you clearly can find a finite right endpoint in the problem,
If you have a constraint equation, sub y = 0 in and solve x to get the right endpoint of x. In those cases, you can bet x can also equal 0, so your domain is [0, k] where k is the answer you got for x when you subbed y = 0 into the constraint. If subbing y = 0 into the constraint gives you an undefined answer, that almost certainly means x can be infinitely large (you are probably creating k/0). Then the domain is (0, ∞) (note the round brackets).
NEVER waste your time looking for the domain at the start. Solve the problem first. Only get the domain after you have found your critical points to help you eliminate critical points that are outside the domain. As I say in my notes, it is generally enough to know x can't be negative, to eliminate critical points.