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I need some help in plotting the following inequality using Mathematica:

$$\frac{1+(x/100+0.1)\times y/100}{1.15+1.15\times(x/100)\times(y/100)}\geq 1$$

assuming $x,y\in \mathbb{Q}$, the set of rational numbers and $5\leq x\leq 90,\ 50\leq y \leq 650$.

Copyable code for the above inequality:

(1 + (x/100 + 0.1)(y/100))/(1.15 + 1.15 (x/100)(y/100)) >= 1

I've tried the following in Maple (because it's what we had available) but it only simplifies the term and does not plot it and enclosing the inequality with plot() in Maple only generates errors.

assume(x, rational);
assume(y, rational);
assume(x>=5 and x<=90);
assume(y>=50 and y<=650);
(1+1*(x/100+0.1)*(y/100))/(1.15+1.15*(x/100)*(y/100))>=1;

Optional theoretical problem:

$x$ could theoretically (not in practice) be up to $100$, but $(x/100+0.1)$ on the left side would still always max out at $1$. Should the equation be changed to $x_1$ and $x_2$ with respective limits? But those limits would only apply for $x_1$ and $90< x_2 \leq 100$.

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  • 3
    $\begingroup$ The "optional" part is not really a question for this site. $\endgroup$ – rm -rf Jul 7 '12 at 17:06
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Something like

RegionPlot[(1 + 1*(x/100 + 0.1)*(y/100))/(1.15 + 1.15*(x/100)*(y/100)) >= 1,
           {x, 5, 90}, {y, 50, 650}]

kenshin's inequality

might be what you need.

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You may also get the inequalities:

p = Reduce[(1 + 1*(x/100 + 1/10)*(y/100))/(115/100 + 115/100 (x/100) (y/100)) >= 1 &&
            x >= 5 <= 90 && y >= 50 <= 650, {x, y}]
(*
5 <= x < 200/3 && y >= -(30000/(-200 + 3 x))
*)

And then

Show[RegionPlot[p, {x, 0, 100}, {y, 0, 700}], 
     Graphics[{Opacity[0.3], Red, Rectangle[{5, 50}, {90, 650}]}]]

enter image description here

Credit due to Heike (the torn image function)

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  • $\begingroup$ Interesting how Opacity[] makes the triangulation of the region apparent... $\endgroup$ – J. M. is away Jul 7 '12 at 15:26
  • $\begingroup$ @J.M. Now it doesn't :D $\endgroup$ – Dr. belisarius Jul 7 '12 at 16:24
  • $\begingroup$ @Dr.belisarius thank you for introducing me to torn image function +1 o/c :) $\endgroup$ – ubpdqn Nov 7 '16 at 23:31

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