# MathematicaSimultaneous Inequalities?

1. Apr 22, 2010

Hey guys,

I have some variables A,B, and C such that all must be greater then zero. A,B and C are all functions of K. I would like to find the range of values of K such that the 3 inequalities are satisfied.

Is there a function built in to Mathematica that will do this? I could write a code, but I am kind of sick of coding right now :tongue2:

If not..then code it is!

2. Apr 22, 2010

### Hepth

Well, as a quick response I know Reduce can sort of find the limits if you know what you're looking for.

Code (Text):
a[k_] = (-1) k^3 + k^2 + 6;
b[k_] = Sin[k];
c[k_] = Cos[k];
Plot[{a[k], b[k], c[k]}, {k, 0, 3}]
Reduce[a[k] > 0 && b[k] > 0 && c[k] > 0, k] // FullSimplify
So the result :
c1 elem of Z OR c1 <= 0 OR 0 <k - 2 pi c1 < pi/2

Means its true if (looking at the second two) if c1(some constant) is negative, so that

k - 2 pi n < pi/2
basically the first solution being
0< k < pi/2 (which is the region thats positive where all three are positive).

It's not perfect, but its correct.

3. Apr 22, 2010

Hey there Hepth,

this is what I have:

A = 576 - 11819*K
B = 15848*K
C = C = 397119*K - 33.6*B/A

and I need to find the range of K such that A,B, and C all remain greater than 0.

4. Apr 22, 2010

### Hepth

A = 576 - 11819*k;
B = 15848*k;
CC = 397119*k - 33.6*B/A;
Plot[{A, B, CC}, {k, 0, 0.04862}, PlotRange -> {0, 1000}]
Reduce[A > 0 && B > 0 && CC > 0, k] // FullSimplify
Reduce[A > y && B > y && CC > y && y > 0, y] // FullSimplify

Gives for k : 0<k<0.0486216 (WHERE they're all positive)

And for Y (bounded by this region):
$$y>0\land \left((k>0\land y<15848. k\land k\leq 0.020819)\lor (k>0.020819\land 11819. k+y<576.\land k\leq 0.0486216)\lor \left(k>0.0486216\land$$
$$\landk<0.0486216\land y<\frac{k (2.34677\times 10^{10} k-1.14104\times 10^9)}{59095. k-2880.}\right)\right)$$

Which, while ugly, describes that range and domain bounded by the functions. I think you're only asking for the first one though.

Last edited: Apr 22, 2010
5. Apr 22, 2010