What is the smallest possible value of their sum? The product of two positive numbers is 36. What is the smallest possible value of their sum?
so far I got
$$xy=36$$
$$y=\frac{36}{x}$$
 A: Hint: $x+y \geq 2\sqrt{xy}$. Can you finish it?
A: OK, so you have the two numbers $x$ and $\frac{36}{x}$. You want to minimise their sum, so you want to minimise the expression $x + \frac{36}{x}$. Remember that to minimise an expression you take the derivative and set it equal to zero. Do you think you can solve it from there?
A: HINT: Note that $4xy=(x+y)^2-(x-y)^2$ or $(x+y)^2=4xy+(x-y)^2$
A: Hint: Consider the function $f(x)=x+\frac{36}{x}$ (i.e. the sum of $x$ and $y$) and find the minimum for that function.
A: Note: there are many ways you can show this but i am assuming from your tag you need an answer via derivatives here it goes
let $m=x+y=x+\frac{36}x=x+36x^{-1}$
then $m'=1-36x^{-2}$
to have a minimum value of $m$ let $m'=0$
$1-36x^{-2}=0$
$36=x^2$
so $x=6$ and $y=6$
A: xy=36, so 2xy=72. By $x^2-2xy+y^2 \ge 0$, $x^2+y^2 \ge 72$, so $x^2+2xy+y^2 \ge 72+36\cdot2=144$. Thus $x+y \ge 12$, since they are positive.
A: First note that
$$f(x)=x+\frac{36}{x}=x+36x^{-1}$$
$$\frac{d}{dx}[f(x)]=1-36x^{-2}=1-\frac{36}{x^2}$$
So now let's find the critical points
$$\frac{d}{dx}[f(x)]=0 $$
$$ 1-\frac{36}{x^2}=0 $$
$$ 1=\frac{36}{x^2} $$
$$ x^2=36= (\pm 6)^2 $$
$$ x=\sqrt{(\pm 6)^2}=\pm 6 $$
Since we're only considering positive values, we'll omit the negative value. Next we must check if $x=6$ is a local minima of $f(x)$. So now
$$\frac{d^2}{dx^2}[f(x)]=72x^{-3}=\frac{72}{x^3}$$
$$\frac{d^2}{dx^2}[f(6)]=\frac{72}{6^3}\gt 0 $$
Therefore, $f(x)=x+\frac{36}{x}$ has a local minima at $x=6$. Which implies that the smallest sum of positive values is 
$$f(6)=6+6=12$$
