How do you prove the following maximization task?

Question

If $x + y = a$, then the maximum value of $x * y$ is when $x=y=a/2$. ($x, y$ are positive numbers)

Answer 1

$$ xy=\frac{(x+y)^2-(x-y)^2}4=\frac{a^2-(x-y)^2}4\le\frac {a^2}4$$ with equaloity iff $x-y=0$.

Answer 2

Calculus Approach

Since $x+y=a$, let $y=x-a$ and you are optimizing $$ xy = x(x-a) \quad \forall x \in (0,a). $$

Note that $f'(x) = 2x-a$, and $f''(x) = 2>0$. Can you finish this?

Geometric Approach

Consider a rectangle with sides $x$ and $y$. You are fixing the perimeter $P = 2a$ and asking for a fixed perimeter, which values of $x,y$ maximize the area of the rectangle. It's clear the problem is symmetric in $x,y$, so only possible answers are $x=y$ or $x=0,y=a$, and the first one yields maximum and the second yields minimum.

Answer 3

With $$y=a-x$$ we get $$x(a-x)=ax-x^2=-\left(x^2-ax+\frac{a^2}{4}-\frac{a^2}{4}\right)=-\left(x-\frac{a}{2}\right)^2+\frac{a^2}{4}\le \frac{a^2}{4}$$