# 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.

-
Try $36 +1$, then $360 + 0.1$, then $3600 + 0.01$ and then spot the pattern –  Henry Aug 1 '12 at 20:53
No maximum. Take $x= 1000000, \; y = \frac{36}{1000000}$ for example –  Will Jagy Aug 1 '12 at 20:54
Perhaps you want $x,y$ to be integers? –  Alex Becker Aug 1 '12 at 20:55
By the way, if $x,y$ are natural numbers, I suppose the optimum is $x=36, y=1$ or the reverse. –  Will Jagy Aug 1 '12 at 20:56
You tagged this as belonging to elementary-number-theory. That makes many of us think that you want $x$ and $y$ to be integers. Please confirm/deny! –  Jyrki Lahtonen Aug 1 '12 at 20:57
show 2 more comments

## 5 Answers

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.

-
add comment

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 $x=y=6$.

-
add comment

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 $F$?

-
Thanks. But I actually wanted the minimum value o $x+y$. Sorry for the inconvenience –  MistyD Aug 1 '12 at 20:59
@MistyD: this method still works. –  akkkk Aug 1 '12 at 21:00
add comment

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.

-
To have $x$ and $y$ unequal, you need $\alpha$ to be a non-zero minimum to make everything an integer, which means that $x$ and $y$ have to be as close together as possible, since the difference between them is $2\sqrt{\alpha^2-36}$ –  Mark Bennet Aug 1 '12 at 21:32
add comment

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 $\min_{x \in \mathbb{N}, \frac{36}{x} \in \mathbb{N}, x \neq 6} f(x) = 13$.

-
add comment