Mathematica MathematicaSimultaneous Inequalities?

[Saladsamurai]

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 into Mathematica that will do this? I could write a code, but I am kind of sick of coding right now

If not..then code it is!

 

[Hepth]

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

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 that's positive where all three are positive).

It's not perfect, but its correct.

 

[Saladsamurai]

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.

 

[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):
<br /> y&gt;0\land \left((k&gt;0\land y&lt;15848. k\land k\leq 0.020819)\lor (k&gt;0.020819\land 11819. k+y&lt;576.\land k\leq 0.0486216)\lor \left(k&gt;0.0486216\land
<br /> \landk&lt;0.0486216\land y&lt;\frac{k (2.34677\times 10^{10} k-1.14104\times 10^9)}{59095. k-2880.}\right)\right)<br />

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

 

[Saladsamurai]

That is great Hepth! I think that will do the trick Thanks for your help.