Is there an alternative way to solve this trivial problem?

Question

I dealing with a trivial problem:

Let $f(x) = xy$ be some area. Given $1500 = 15x + 6y$, find $x$ and $y$ so that the area is maximized.

Usually pretty easy when you can just plug in $(1500 - 15x)/6$ for $y$ and then set the derivative equal to zero. However, is there an alternative way to solve something like this without setting the derivative to zero?

Answer 1

Just use GM-AM $$xy= \frac 1{90}(15x \cdot 6y) \leq \frac 1{90}\left(\frac{15x+6y}2\right)^2 = 6250$$ Equality is reached for $15x=6y$ which gives together with the other constraint $x=50$ and $y=125$.

Answer 2

$f(x)=xy=\frac{x(1500-15x)}{6}=:z\\\Rightarrow 6z=1500x-15x^2\\\Rightarrow 5x^2-500x+2z=0\\\Rightarrow x=\frac{500 \pm \sqrt {500^2-40z}}{10}\\\because x \in \mathbb R;\\ 500^2-40z \ge 0\\\Rightarrow z\le 6250\\\text{This is a good alternative way to find maxima and minima.}$