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 idontknow November 13th, 2018 04:05 AM

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$\displaystyle [\frac{x}{y}]=$?

 romsek November 13th, 2018 05:23 AM

there is no maximum value

with $x \neq 0$, as y approaches zero $\dfrac x y$ becomes arbitrarily large.

 SDK November 13th, 2018 05:36 AM

Quote:
 Originally Posted by romsek (Post 602332) 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.

 mathman November 13th, 2018 01:06 PM

Since $y=0$ is allowed, there is no max, since dividing by $0$ will lead to trouble.

 SDK November 13th, 2018 04:20 PM

Quote:
 Originally Posted by mathman (Post 602345) 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.

 mathman November 14th, 2018 01:31 PM

For this question $\infty =$ trouble.

 JeffM1 November 14th, 2018 05:22 PM

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?

 idontknow November 15th, 2018 10:05 AM

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$\displaystyle [\frac{x}{y}]=\frac{max[|x|]}{min[|y|]}=3$

 Micrm@ss November 15th, 2018 10:10 AM

Quote:
 Originally Posted by mathman (Post 602377) For this question $\infty =$ trouble.
Yes, infinity is trouble. But division by zero doesn't equal infinity.

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