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How to use Solver?

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Rebecca Reyes
Rebecca Reyes on 13 Apr 2021
Edited: madhan ravi on 14 Apr 2021
clear dlc
%feed composition
z1=.5;
z2=.1;
z3=.15;
z4=.25;
%other given data
F=150;
PD=250;
T=10;
F=150;
%k-values from chart yi/xi
k1=56;
k2=0.65;
k3=0.175;
k4=0.055;
%total balances
syms x
eqn=solve(k1*((F*z1)/(k1*x+150-x))+k2*((F*z2)/(k2*x+150-x))+k3*((F*z3)/(k3*x+150-x))+k4*((F*z4)/(k4*x+150-x))==1)
I'm trying to solve for x in the eqn but all i get is this:
eqn =
150
root(z^3 - (151985*z^2)/231 + (2120672000*z)/17787 - 36140000000/5929, z, 1)
root(z^3 - (151985*z^2)/231 + (2120672000*z)/17787 - 36140000000/5929, z, 2)
root(z^3 - (151985*z^2)/231 + (2120672000*z)/17787 - 36140000000/5929, z, 3)
Very confused since I had defined my z constants and I am only looking for the x variable

Accepted Answer

madhan ravi
madhan ravi on 13 Apr 2021
Use vpasolve()
  6 Comments
madhan ravi
madhan ravi on 14 Apr 2021
Wow , you just completely ignored what I told you and simply asked the same question.
%feed composition
z1=.5;
z2=.1;
z3=.15;
z4=.25;
%other given data
F=150;
PD=250;
T=10;
F=150;
%k-values from chart yi/xi
k1=56;
k2=0.65;
k3=0.175;
k4=0.055;
%total balances
syms x
Eqn = k1*((F*z1)/(k1*x+150-x))+k2*((F*z2)/(k2*x+150-x))+k3*((F*z3)/(k3*x+150-x))+k4*((F*z4)/(k4*x+150-x))==1;
X = vpasolve(Eqn) % you can see 4 answers so if you're looking for the root in a certain range say 50 - 100 you can use the below line
X = 
X_in_the_range = vpasolve(Eqn, [50 100])
X_in_the_range = 
88.652321154384366746951471346875
subs(Eqn, X_in_the_range) % to check if it satisfies, ofcourse it's not an exact solution but a close match to what you're looking for
ans = 
Come on , was that hard for you ? The reason I suggested vpasolve() is that you can input the domain for the solver to look for. Please stop asking the same question multiple times just because you didn't understand. It's not encouraged in this forum. Volunteers spend some time in helping and instead of giving them a proper response OP's tend to ask another question simultaneously.

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