Register to reply 
Solving system of inequalities 
Share this thread: 
#1
Jun2014, 09:44 AM

PF Gold
P: 19

(This isn't homework, so I guess it'd go here...)
Okay, so I'm trying to solve a system of equations with a bunch of ranges: (Some number between 1.2 and 2.0) = (Some number between 5.5y and 9.1y) = (Some number between 5.4x and 10x) Where X and Y are ranges made up of two percentages (or ranges of percentages) each. I know one of the percentages for X is a range, and I know that range's approximate value (between 20 and 25%) and I also know one of the percentages for Y (21%). But I want to find the second percentage (or range of percentages) for each one. How the heck do I do this? Math with ranges makes my head hurt for some reason. I tried looking it up, but... I just got a bunch of stuff for summing numbers in Excel. :/ 


#2
Jun2014, 10:25 AM

HW Helper
P: 3,562

At the moment what I think you're trying to say is that there exists some distinct value of n in which [tex]1.2\leq n \leq 2.0[/tex] and also we want to find the range of x and y such that the above inequality still holds and these inequalities are [tex]5.4x \leq n \leq 10x[/tex] [tex]5.5y\leq n\leq 9.1y[/tex] The range that x and y can take such that n still falls in each range just needs to be chosen at the extreme points. 5.4x is the lowest value, hence if this value is the highest n value, then 5.4x=2.0, x=37% will force us to choose n=2.0. Similarly, if we let 10x=1.2, x=12% then we force n to be the minimum of n=1.2. So the range of values that x can take such that the first two statements still hold true is 12% < x < 37% The same can be done for y. I'm sure this isn't what you're looking for though, so you'll have to be a little more clear. 


#3
Jun2014, 10:31 AM

HW Helper
P: 3,562

Thinking about it some more, you might instead be looking for a range of x (and similarly, y) values such that the entire range of 1.2<n<2.0 is satisfied.
For example, if we take 1.2 = 5.4x, x=22% then 10x = 2.2 which is beyond 2.0 so for this x value, all of the range of n is satisfied. But we have "leftovers" between 2.0 and 2.2 so the other extreme that x can take is 2.0 = 10x, x=20%. For this value, 5.4x = 1.08 which of course we expect to be lower than 1.2. So the range of x values you can take is between 2022%. Are we on the right track? 


#4
Jun2014, 11:33 AM

PF Gold
P: 19

Solving system of inequalities
The first component of X is 2025%, and the first component of Y is 21%. So 2025% and ____% make up X (2022%), and 21% and ____% make up Y (22%). I've no idea how to get from X and Y to their unknown component. 


#5
Jun2114, 04:01 AM

HW Helper
P: 1,808

Do everything one step at a time and at each step, find the maximum and the minimum possible value.
So you start with ##1.2 ≤ n ≤ 2.0##, we have a range for ##n##. Now look at x: ##5.4x ≤ n ≤ 10x##, we know the maximum and minimum of ##x## will occur when ##n## is a maximum or minimum, we don't need to use calculus. Find the max and min for those two extreme values of ##n##, then ##x##'s range will be the highest and lowest of those 4 numbers. Proceed like this one step further and find the range of ##y##. 


#6
Jun2114, 07:32 AM

Mentor
P: 12,037

x=20% would fail at 5.4x=1.08 < 1.2, x=22% would fail at 10x=2.2 > 2. It would work if your calculated 20% would be larger than the calculated 22%. I think it would help a lot if you can give more context to the equations. 


#7
Jun2114, 07:59 PM

PF Gold
P: 19

Oh god. Running numbers again from the start, I realize I didn't even need help. Just a break. I ran numbers that weren't even... ugh.
Running the numbers again and arriving at T, the solution became mindnumbingly simple: T = 0.105 = 0.5x = 0.5y 2T = x = y = 0.21 !!! C = 0.5/U = 0.5/I C = 2.38 X and Y ended up equal, instead of slightly different and being inequalities. ======= I had 3 circles overlapping with eachother. Starting with Circle A's Area = Circle B's Area = 1, and knowing the A/B overlap (D), the B/C overlap (E), and the A/C overlap (F), I wanted to find the overlap between all three (T) and the area of circle C. Which ended up actually being ridiculously easy. A=B=1 D=0.21 E=F=0.5 T=D*E=D*F=0.21*0.5=0.105 And then I just set 0.105 equal to E*x and F*y where x and y were the fractions of circle C that were also part of circle A and B, respectively. This made D, y, and x the same number, and I knew then that 0.5/0.21 would get me the area of C. I feel really dumb right now. But obviously I'll take a break next time before I make a thread about it and feel dumb later. :/ 


#8
Jun2114, 08:09 PM

Mentor
P: 12,037

That could be true in some special cases, but certainly not in general. I don't think there is a general, unique solution to the problem you described. The circle C has 3 degrees of freedom (two for the position of the center and one for the radius), but you just have two constraints on them  E and F. Simple extreme example: Let E=F=1. That means circle C encloses both unit circles. How large is it? You have no way to tell. 


#9
Jun2114, 09:51 PM

PF Gold
P: 19

Now what is true for overlap D must also be true for E and F. However, we need one of the two overlaps in terms of C, not in terms of A or B (unit circles). Since this is unknown, this can be represented by variables: D(E+F)/2 = E(D+x)/2 = F(D+y)/2 Which means x=y and comes out to: 0.105 = 0.105/2 + 0.5x/2 0.105 = 0.0525 + 0.25x 0.0525 = 0.25x 0.21 = x = y Remember, since x and y are percents of C, we can take 0.21 and divide E or F by that to get C=0.5/0.21 (multiplied by B if B isn't equal to one) Erg, I need a fudging whiteboard to explain this properly. But it definitely SHOULD hold true for any three overlapping circles, if I did this right. 


#10
Jun2214, 05:27 AM

Mentor
P: 12,037

I go with the first one. I think I misread your definitions of x and y. x=F/C, y=E/C? Then F=E directly gives x=y. DE = E(D+E/C)/2 = E(D+E/C)/2 D = (D+E/C)/2 D = E/C Why should this be true? 


#11
Jun2314, 02:12 PM

PF Gold
P: 19

I'm defining everything in terms of A and B, as in, if A is really 1,000 and B is 1,000, and the overlap is 500, then it's 50% of either circle. The math is made a hundred times more simple by this fact, and the fact that the overlaps with circle C are the same. If this wasn't the case, it'd become much more complicated. Let's take the AC overlap in example. The overlap of this and B would be A's overlap with B plus C's overlap with B over two, and multiplying this by the AC overlap would give the ABC overlap. Make sense? ... You know what? I'll attach a diagram I drew this morning. One sec. In the diagram, A B and C are the circle areas, while T is the area of all three's overlap and X Y and Z are the areas of the overlaps between A&B, A&C, and A&B. 


#12
Jun2314, 05:30 PM

Mentor
P: 12,037

Example 1: The overlap AB plus the overlap BC, divided by two, is some nonzero number, but the overlap of (AC overlap) with B is empty. Example 2: The overlap of (AC) and B is identical to the (AB) overlap, if you multiply this by the (AC) overlap (which is smaller than 1), you don't get the right result. 


Register to reply 
Related Discussions  
Solving a system of Inequalities  Calculus & Beyond Homework  4  
EASY QUESTION... solving system of linear inequalities  Linear & Abstract Algebra  1  
Solving trigonometric inequalities  Precalculus Mathematics Homework  3  
Solving Inequalities  Precalculus Mathematics Homework  3  
Solving Inequalities  Calculus  3 