1. ## AGM ineqaulity

AGM ineqaulity

prove that $\displaystyle x(c-x)$ is maximized when $\displaystyle x=c/2$

I know how to do this using derivatives but i don't really understand where to start this using the AGM inequality my prof isn't much help either this is what my text book says but it don't really get how to apply it

2. Originally Posted by treetheta
AGM ineqaulity

prove that $\displaystyle x(c-x)$ is maximized when $\displaystyle x=c/2$

I know how to do this using derivatives but i don't really understand where to start this using the AGM inequality my prof isn't much help either this is what my text book says but it don't really get how to apply it

The "AGM inequality" (arithmetic-geometric mean inequality) says that the geometric mean of a set of numbers is always less than or equal to the arithmetic mean or, as applied to two numbers, $\displaystyle \sqrt{ab}\le \frac{a+ b}{2}$, and they are equal if and only if the numbers in the set are all the same.

To apply that to this situation, Let a= x and b= c-x. Then the geometric mean is $\displaystyle \sqrt{x(c- x)}$ and the arithmetic mean is $\displaystyle \frac{x+ c- x}{2}= \frac{c}{2}$. The AGM inequality says that $\displaystyle \sqrt{x(c-x)}\le \frac{c}{2}$. The right side is a constant, independent of x, so that is the "maximum" possible value of the left side and the left side takes that value when x= c-x.

Of course, $\displaystyle x(c-x)= \left(\sqrt{x(c-x)}\right)^2$ will be maximum when $\displaystyle \sqrt{x(c- x)}$ is.

Here, by the way, is yet another way to prove that: $\displaystyle x(c- x)= cx- x^2= -(x^2- cx+ c^2/4)+ c^2/4= c^2/4- (x- c/2)^2$. Since a square in never negative, that can never be larger than $\displaystyle c^2/4$ and will only be equal when x-c/2= 0.

3. yes, I did this question very similar to the way you did, well pretty much exactly the same

thank you so much for your help in clarifying =D