Minimum value for a multiplication in an equation with 2 variables

Question

How can I find out the minimum value of $mn$ where $m$ and $n$ must be natural numbers, bigger than $0$, from the equation $$60m+3n=2mn$$

Answer 1

By AM-GM $$2mn\geq2\sqrt{60m\cdot3n}=2\sqrt{180mn},$$ which gives $$mn\geq180.$$ the equality occurs for $60m=3n,$ which gives $$(m,n)=(3,60).$$ Id est, $180$ is the minimal value.

Answer 2

Assuming $m,n>0$

$$30/n+3/2m=1$$

Now use AM GM inequality

Answer 3

First we see that $n=2k$ so we have $$30m+3k=2mk$$ so $k=2l$ and now we have $$15m+3l=2ml$$Now we have $$l ={15m\over 2m-3}$$ so $$2m-3\mid 15$$ so $$2m-3\in \{1,3,5,15,-1,-3,-5,-15\}$$

so $$2m\in \{2,4,6,8,18\}$$ so $m\in \{1,2,3,4,9\}$ and respectively $l\in\{-15,30,15,12,9\}$. So the minimum value of $mn = 4ml = 180$.