can you help me to find the minimum R in this code. What kind of iteration code could i use in this case?

This is the code which i am working. I need to find the the correct m, s, and Lr in order to get R apporaximately zero.

please help.

rp = 5.89;     
m = rp;       
s = 3.84;      
Lr = 5000;       
lint = 0.36;   

function f = unaccess(rs,m,s);

my = log(m/sqrt(1+s^2/m^2));
sig = sqrt(log(1+s^2/m^2));

%x = log(rs/m/(1+(s/m)^2)^3.5)/sqrt(2*log(1+(s/m)^2));
%f = (1+ erf(x))/2;

%x = (4*sig^2-log(rs)+my)/(sqrt(2)*sig);
%f = (1- erf(x))/2;

fun = @(r) r.^4./(r.*s.*sqrt(2.*pi)).*exp(-(log(r)-my).^2./(2.*sig.^2));
format long;
q1 = integral(fun,0,Inf,'RelTol',1e-8,'AbsTol',1e-13);

fun = @(r) r.^4./(r.*s.*sqrt(2.*pi)).*exp(-(log(r)-my).^2./(2.*sig.^2));
format long;
q2 = integral(fun,0,rs,'RelTol',1e-8,'AbsTol',1e-13);

f=q2/q1;

end

function f = access(rs,rp,m,s)

my = log(m/sqrt(1+s^2/m^2));
sig = sqrt(log(1+s^2/m^2));

fun = @(r) r.^4./(r.*s.*sqrt(2.*pi)).*exp(-(log(r)-my).^2./(2.*sig.^2));
format long;
q1 = integral(fun,0,Inf);

fun = @(r) r.^4.*(1-rs./r).^2.*(1+2.*(rs./r)-(1./3).*(rs./r).^2-(4./9).*(rs./r).^3)./(r.*s.*sqrt(2.*pi)).*exp(-(log(r)-my).^2./(2.*sig.^2));
format long;
q2 = integral(fun,rs,Inf);

f=q2/q1;

end

dj=[0.886/rp 1.032/rp 1.568/rp 2.179/rp 3.165/rp]; 

dCli=[0.997 0.991 0.881 0.468 0.016];          

for i=1:5

dr(i) = dj(i)*rp   ;                   

fnsi(i) = unaccess(dr(i),m,s) ;         

fai(i) = access(dr(i),rp,m,s) ;         

dCl(i)= (1-fnsi(i))*exp(-fnsi(i)*Lr);  

R(i) = (dCl(i)-dCli(i))^2;
end

Rsum = sum( R);

thank you