Mathematica Stack Exchange is a question and answer site for users of Mathematica. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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);

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 $90< x_2 \leq 100$.

share|improve this question
The "optional" part is not really a question for this site. – R. M. Jul 7 '12 at 17:06

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.

share|improve this answer

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)

share|improve this answer
Interesting how Opacity[] makes the triangulation of the region apparent... – J. M. Jul 7 '12 at 15:26
@J.M. Now it doesn't :D – Dr. belisarius Jul 7 '12 at 16:24

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.