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November 13th, 2018, 05:05 AM   #1
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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$\displaystyle [\frac{x}{y}]=$?
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November 13th, 2018, 06:23 AM   #2
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there is no maximum value

with $x \neq 0$, as y approaches zero $\dfrac x y$ becomes arbitrarily large.
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November 13th, 2018, 06:36 AM   #3
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Quote:
Originally Posted by romsek View Post
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.
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November 13th, 2018, 02:06 PM   #4
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Since $y=0$ is allowed, there is no max, since dividing by $0$ will lead to trouble.
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November 13th, 2018, 05:20 PM   #5
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Originally Posted by mathman View Post
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.
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November 14th, 2018, 02:31 PM   #6
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For this question $\infty =$ trouble.
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November 14th, 2018, 06:22 PM   #7
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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?
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November 15th, 2018, 11:05 AM   #8
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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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November 15th, 2018, 11:10 AM   #9
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Quote:
Originally Posted by mathman View Post
For this question $\infty =$ trouble.
Yes, infinity is trouble. But division by zero doesn't equal infinity.
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