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 November 13th, 2018, 04:05 AM #1 Senior Member   Joined: Dec 2015 From: somewhere Posts: 550 Thanks: 83 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$\displaystyle [\frac{x}{y}]=$?
 November 13th, 2018, 05:23 AM #2 Senior Member     Joined: Sep 2015 From: USA Posts: 2,500 Thanks: 1372 there is no maximum value with $x \neq 0$, as y approaches zero $\dfrac x y$ becomes arbitrarily large.
November 13th, 2018, 05:36 AM   #3
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Quote:
 Originally Posted by romsek there is no maximum value with $x \neq 0$, as y approaches zero $\dfrac x y$ becomes arbitrarily large.
$x,y$ seem to be integers so it should have a max.

@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.

 November 13th, 2018, 01:06 PM #4 Global Moderator   Joined: May 2007 Posts: 6,786 Thanks: 708 Since $y=0$ is allowed, there is no max, since dividing by $0$ will lead to trouble.
November 13th, 2018, 04:20 PM   #5
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Quote:
 Originally Posted by mathman Since $y=0$ is allowed, there is no max, since dividing by $0$ will lead to trouble.
Are you implying that $\frac{x}{0} = \infty$? I assume this is undefined and thus there is no trouble.

 November 14th, 2018, 01:31 PM #6 Global Moderator   Joined: May 2007 Posts: 6,786 Thanks: 708 For this question $\infty =$ trouble.
 November 14th, 2018, 05:22 PM #7 Senior Member   Joined: May 2016 From: USA Posts: 1,310 Thanks: 551 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?
 November 15th, 2018, 10:05 AM #8 Senior Member   Joined: Dec 2015 From: somewhere Posts: 550 Thanks: 83 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$\displaystyle [\frac{x}{y}]=\frac{max[|x|]}{min[|y|]}=3$
November 15th, 2018, 10:10 AM   #9
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Quote:
 Originally Posted by mathman For this question $\infty =$ trouble.
Yes, infinity is trouble. But division by zero doesn't equal infinity.

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