I know that you can use derivatives to find the minimum/maximum of a function. However, sometimes it is very difficult or too long. For example,
In the problem above, there are 3 variables, and on the AMC 10 you don't exactly have the time for fancy calculus. Is there another general technique for cases like this? Thanks!
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For this specific problem, I just thought, that amc + am + mc + ac thing looks a lot like multiplication, I bet it behaves like that. And when splitting an interval to make a product (choose a + b = 1 to maximize ab) the maximal split is half-and-half, so I figured for splitting it into 3 parts to make a product the maximal split would be around third-third-third. So choose 3-3-4. And lo and behold, it gives one of the answers.
With multiple choice, you can kinda just guess.
Let $S = AMC+AM+AC+MC$, and let $x=A+1$, $y=M+1$, $z=C+1$. We notice that $$xyz = S+A+M+C+1 = S+10+1=S+11$$ so we can maximize $S$ by maximizing $xyz$ subject to the constraint $x+y+z =13$, since $A+1+M+1+C+1=13$.
For the moment let's suppose $x,y,z$ can be non-integer. Then by the AM-GM inequality, we know $xyz$ is maximized when $x=y=z$. Since $x+y+z=13$, this yields $x=y=z=13/3$, with $xyz=(13/3)^3 \approx 81.4$, corresponding to $S =xyz-11\approx 70.4$. So with integer constraints, we know we can't do better than $S=70$; that eliminates choices (D) and (E).
If we look for integer values near $13/3$ with sum $13$, it's natural to try $x=y=4$, $z=5$, which yields $xyz=80$, corresponding to $S=80-11=69$. So the answer is (C).