# How to find the minimum of a function and plot with it

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JACINTA ONWUKA on 9 Jul 2019
Answered: Charles DeLorenzo on 10 Jul 2019
My mission is to find Mycp such that Jcp is minimum and then plot a graph with minimum of Mycp as a function of rho.
I just started using matlab so dont understand why the graph is not showing? Can someone please help me?
close all
L=1;
T=100;
r=0.03;
I1=0.1;
p=0.05;
epsilon=0.3;
beta=0.1;
rho=1000;
Myrho=0:100:1000
MycpRho=zeros(numel(Myrho),1);
for j = 1:numel(Myrho)
MyrhoCurrent=Myrho(j);
Mycp = 0:1:100
n = zeros(numel(Mycp),1 );
n2 = zeros(numel(Mycp),1 );
n3 = zeros(numel(Mycp),1 );
Jcp = zeros(numel(Mycp),1 );
for i = 1:numel(Mycp )
MycpCurrent=Mycp(i);
delta = 1-MycpCurrent/100 ;
tau = (1/(beta*(L+delta*p)))*log((L*(I1+delta*p))/(delta*p*(L-I1 )));
t05 =(1/(beta*(L+delta*p)))*log((L*(0.05*L+delta*p))/(delta*p*(L-0.05*L )));
I2= @(t)(L*delta*p*(exp (beta*(L+delta*p)*t)-1)) ./ (L + delta*p* exp(beta*(L+delta*p)*t ));
I3= @(t)(L*(I1+delta*p)*exp((epsilon*beta)*(L+delta*p)*(t-tau))-...
delta*p*(L -I1))./(L-I1+(I1+delta*p)*exp(epsilon*beta*(L+delta*p)*(t-tau)));
fun = @(t,MycpCurrent) MycpCurrent*L*exp(-r*t);
fun2=@(t)rho*I2(t).*exp(-r*t);
fun3=@(t)rho*I3(t).*exp(-r*t);
n(i) = integral(@(t)fun(t,MycpCurrent),0,100, 'ArrayValued',1);
n2(i)= integral(fun2,t05,tau);
n3(i)= integral(fun3,tau,100);
JCp(i)= n(i)+n2(i)+n3(i);
end
MycpBest=Mycp(JCp==min(JCp));
JCpBest=min(JCp);
plot(Myrho,MycpBest)
hold on
scatter(MycpBest,JCpBest)
hold off
end

Charles DeLorenzo on 10 Jul 2019
I changed your plot to this and it seems to work fine. All the values seem to show up on the plot
JCpBest=min(JCp);
figure (1)
plot(Myrho,MycpBest,'x',MycpBest,JCpBest,'x');
However, i changed your second for loop to
for i = 1:numel(Mycp )-1
to match
Mycp = 0:1:100
or else the code gives a warning.
I hope this helps.