Minimum value of $x+y$ when $xy=36$ How would I calculate minimum value of $x+y$ when $xy=36$ and x and y are unequal positive integer numbers. I don't even know the answer. Any help would be appreciated. 
Edit:
Sorry It was the minimum value. 
 A: There are not many pairs $(x,y)$ of distinct positive integers such that $xy=36$. We can try them all and pick the "winner." That is undoubtedly the simplest approach, but we describe a more structured way of looking at things.
Note that 
$$(x+y)^2=(x-y)^2+4xy=(x-y)^2+144.$$
We want $(x+y)^2$ to be as small as possible. We can do this by making $|x-y|$ as small as possible.
We were asked to use distinct positive integers $x$ and $y$ whose product is $36$. The ones that minimize $|x-y|$ are $4$ and $9$. 
So the minimum value of $(x+y)^2$ under our constraints is $25+144$. That gives $x+y=13$. 
Remark: If we remove the condition that the numbers are distinct, then $(x-y)^2$ is minimized when $x=y$.  Indeed the smallest value of $x+y$ as $(x,y)$ ranges over all pairs of positive real numbers such that $xy=36$ is given by taking $x=y=6$. 
A: Now that you have specified $x$ and $y$ to be integers, this problem is pretty trivial.  There are only a few possible choices.  To further reduce the possibilities, if the values of $x$ and $y$ are switched, we get the same value for $x + y$, so we might as well consider $x < y$.  Now, we can factor 36 as
$\begin{align*}
  1 &* 36 \\
  2 &* 18 \\
  3 &* 12 \\
  4 &* 9 \\
  6 &* 6
\end{align*}$
But, the last has $x = y$, so we exclude that.  So, the question is reduced to:  Find the minimum value from the set $\{1 + 36, 2 + 18, 3 + 12, 4 + 9\} = \{37, 20, 15, 13\}$.  So, the answer is 13.
A: You're trying to determine the minimum (if there is one) of the function
$$f(x,y)=x+y$$
given the constraint that $xy=36$. In particular this means we can write $y=\dfrac{36}x$ and consider a single variable function:
$$F(x)=f \left(x,\frac{36}{x} \right)=x+\frac{36}{x}$$
Now, what does $F'(x)$ tell you about the maximums and minimums of $F$?
A: Here is a different approach. Let $x+y=2\alpha$ so that $x$ and $y$ are the roots of $$z^2-2\alpha z+36=0$$ these roots being $$x,y=\frac{2\alpha \pm \sqrt {4\alpha^2-144}}2=\alpha \pm \sqrt {\alpha^2-36}$$
You are looking for the minimum possible positive value of $2\alpha$ hence of $\alpha$, and $\alpha$ cannot be less than 6 if $x$ and $y$ are to be real numbers.
A: This is the answer to the question as originally posed:
If $xy=36$, then both $x,y$ are non zero, and you have $y = \frac{36}{x}$, so the problem becomes finding the maximum of $f(x) = x+\frac{36}{x}$. 
The answer is there is no maximum, or it is $+\infty$, depending on your range, since $\lim_{x \downarrow 0} f(x) = +\infty$.
Similarly, there is no minimum, because $\lim_{x \uparrow 0} f(x) = -\infty$.
This is the answer after the positivity constraint was added:
We have $f''(x) = \frac{73}{x^3}$, which is positive on $(0,\infty)$, so $f$ is strictly convex. Furthermore, $f'(x_o) = 0$ imples $x_0 = 6$, and since $f$ is strictly convex, this is the minimum. However, then you have $y=6$ which equals $x$. So, the answer to your question is that there is no minimum with $x$ and $y$ not equal, but there in an infimum, and that is $\inf_{x > 0, x \neq 6} f(x) = 12$.

This is the answer after the integer requirement was added:
$x$ and $y$ are now positive different integers. Clearly, we need only consider $x \in \{1,\cdots, 5 \}$, since $f(x) = f(\frac{36}{x})$. We can exclude $x=5$ since $\frac{36}{5}$ is not an integer. From evaluation or from the graph, we see that $x=4$ is the minimum value, and so we have $\min_{x \in \mathbb{N}, \frac{36}{x} \in \mathbb{N}, x \neq 6} f(x) = 13$.
