I have no idea how to solve an $\arg \max$ mathematically, for example $\arg \max(x(10-x))$. I know the solution is $x=5$, but how do I get there (for more difficult exercises I will have to solve). Can someone help?
1 Answer
If you have a quadratic then you could complete the square. $$x(10-x) \equiv 10x-x^2 \equiv -(x^2-10x) \equiv -((x-5)^2-25) \equiv -(x-5)^2+25$$ This tells us that the maximum is $x=5$ and $y=25$.
If you have something of higher degree, you could differentiate to find the turning points. $$\frac{\operatorname{d}\!y}{\operatorname{d}\!x} = 10-2x$$ We see that $10-2x=0$ if and only if $x=5$. When $x=5$, $y=25$. This is a maximum since when $x=0$, $y=0$ and when $x=10$, $y=0$.
-
$\begingroup$ so, f(x)=(10x-x^2), f'(x)=10-2x=0. great, thanks $\endgroup$– peteOct 29, 2013 at 15:24
-
3$\begingroup$ Yeah, that's right. But you must check it's a maximum. If $f'(x)=0$ then it could be a minimum (or even an inflexion). The formal test for a maximum is $f'(x)=0$ and $f''(x)<0$. P.S. Don't forget to vote for helpful answers. $\endgroup$ Oct 29, 2013 at 15:26
-
$\begingroup$ thanks fly by night for the answer. helped me out a lot $\endgroup$– peteOct 29, 2013 at 15:28