Maximum value x and y integers $\displaystyle 2\leq x \leq 3$ $\displaystyle 3\leq y \leq 6$ Find maximum value of $\displaystyle \frac{x}{y}$ or max$\displaystyle [\frac{x}{y}]=$? 
there is no maximum value with $x \neq 0$, as y approaches zero $\dfrac x y$ becomes arbitrarily large. 
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@OP: Where are you stuck? This problem is pretty simple to just check every possible value of $x,y$. If you give it a moments thought you can also see that you can maximize the value of $x/y$ by maximizing the value of $x$ and minimizing the value of $y$. Do this also for $x/y$ and take the larger of the two. 
Since $y=0$ is allowed, there is no max, since dividing by $0$ will lead to trouble. 
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For this question $\infty =$ trouble. 
I think the problem is underspecified. It is unclear whether zero as a denominator is excluded or included from consideration. If it is included, a rational maximum does not exist. If it is excluded, a rational maximum does exist. What is the question? 
Let $\displaystyle y\neq 0$ $\displaystyle z=\frac{x}{y}$ , symmetry $\displaystyle z(x,y)=z(x,y)$ defines the nature of $\displaystyle x,y$ , so x and y must be both negative or positive So max$\displaystyle [\frac{x}{y}]=max[\frac{x}{y}]$ or we can work with the new function $\displaystyle z_1=\frac{x}{y}$ max$\displaystyle [\frac{x}{y}]=\frac{max[x]}{min[y]}=3$ 
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