How to solve

parul sotobon — Thu, 12 Aug 2010 06:01:33 +0000

Hi all !
I want to solve following nonlinear equation.
My variables are c, q and f.
I have got values of c but I can not solve for q and f.
Can any body suggest me please?

a=1000;
b=2.5;
s=2;
t=0.5;
hb= 0.5 ;
Ab = 5 ;
Av = 10 ;
y=0.5;
w=0.001;
h=0.05;
r=10;
g=5;
tv=3;
l=0.5;
N=3
c=(a*b*(Ab+Av/N)+2*a*b*s+a*b*t)/(a*b-a);
x1=a*c^(-b)*(Ab+Av/N);
x2=y*w+y*w/N+y*h/N;
x3=hb/2+l*w+tv*w+l*h;
x2*q^3-x3*q^2-x1=0;
f=c*a-q*h ;
c, q, t ???

Thanks in advanced
Parul

Re: How to solve

Roger Stafford — Thu, 12 Aug 2010 06:16:22 +0000

The 'roots' function is your friend.

Roger Stafford

Re: How to solve

Walter Roberson — Thu, 12 Aug 2010 20:58:13 +0000

parul sotobon wrote:

> I want to solve following nonlinear equation.
> My variables are c, q and f.
> I have got values of c but I can not solve for q and f.
> Can any body suggest me please?
>
>
> a=1000;
> b=2.5;
> s=2;
> t=0.5;
> hb= 0.5 ;
> Ab = 5 ;
> Av = 10 ;
> y=0.5;
> w=0.001;
> h=0.05;
> r=10;
> g=5;
> tv=3;
> l=0.5;
> N=3
> c=(a*b*(Ab+Av/N)+2*a*b*s+a*b*t)/(a*b-a);
> x1=a*c^(-b)*(Ab+Av/N);
> x2=y*w+y*w/N+y*h/N;
> x3=hb/2+l*w+tv*w+l*h;
> x2*q^3-x3*q^2-x1=0;
> f=c*a-q*h ;
> c, q, t ???

Above you want to solve for c, q, and f, but here you ask about c, q, and t.

c = 385/18
t = 1/2

Now, let W = RootOf(18*_Z^3-42889*_Z^2-129600000*385^(1/2)*2^(1/2)
That is, W are the the three values _Z that satisfy the cubic equation
18 * _Z^3 - 42889 * _Z^2 - 129600000 * sqrt(385) * sqrt(2) = 0
then,

f = 192500/9 - 1/1540 * W
q = 1 / 77 * W

There are analytic solutions for the three routes but they are messy, Their
numeric values are approximately,

2416.924245, -17.1010110 + 287.0039229 i, -17.1010110 - 287.0039229 i

Note that in deriving these solutions, I assumed that each number given as a
decimal (e.g., 0.001) was intended as a compact way to write an exact rational
number (e.g., 1/1000), rather than being intended as indicated an
approximation (e.g., 95/100000 inclusive to 105/100000 exclusive). Interval
arithmetic is much more tiresome.

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