November 13th, 2018, 05:05 AM  #1 
Senior Member Joined: Dec 2015 From: Earth Posts: 276 Thanks: 32  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, 06:23 AM  #2 
Senior Member Joined: Sep 2015 From: USA Posts: 2,203 Thanks: 1157 
there is no maximum value with $x \neq 0$, as y approaches zero $\dfrac x y$ becomes arbitrarily large. 
November 13th, 2018, 06:36 AM  #3  
Senior Member Joined: Sep 2016 From: USA Posts: 521 Thanks: 293 Math Focus: Dynamical systems, analytic function theory, numerics  Quote:
@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, 02:06 PM  #4 
Global Moderator Joined: May 2007 Posts: 6,642 Thanks: 626 
Since $y=0$ is allowed, there is no max, since dividing by $0$ will lead to trouble.

November 13th, 2018, 05:20 PM  #5 
Senior Member Joined: Sep 2016 From: USA Posts: 521 Thanks: 293 Math Focus: Dynamical systems, analytic function theory, numerics  
November 14th, 2018, 02:31 PM  #6 
Global Moderator Joined: May 2007 Posts: 6,642 Thanks: 626 
For this question $\infty =$ trouble.

November 14th, 2018, 06:22 PM  #7 
Senior Member Joined: May 2016 From: USA Posts: 1,210 Thanks: 497 
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, 11:05 AM  #8 
Senior Member Joined: Dec 2015 From: Earth Posts: 276 Thanks: 32 
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, 11:10 AM  #9 
Senior Member Joined: Oct 2009 Posts: 631 Thanks: 193  

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