Need help with bisection of this function, any advise or help would be great! thanks in advance to all!

this is the code I used for the first part of the question and is correct:

h=0.1;

t=36;

N=(t/h);

t=zeros(1,N+1);

p=zeros(1,N+1);

t(1)=0;

p(1)=87;

for n=1:N

      t(n+1)=t(n)+h;

      u=sin((2*pi*t)/12);

      H=15*[nthroot(u,15)+1]

      p1(n+1)=p(n)+h*[(0.016*p(n)*(100-p(n)))-H(n)];

      p(n+1)=p(n)+(h/2)*[[(0.016*p(n)*(100-p(n)))-H(n)]+[(0.016*p(n+1)*(100-p(n+1)))-H(n+1)]];

end

What I need to find now is the largest value of H for which the solution for P(t) does not be 0 at t = 36.

I know some of the code :

H=@(t)15*[nthroot(u,15)+1];

upper-limit=100;
lower-limit=1;

mid-point=(upper-limit+lower-limit)/2;

while lower-limit-upper-limit>0
    if (?)*(?)<0
        upper-limit=mid-point;
    else
        lower-limit=mid-point;
    end
end

if anyone knows how to fill this parts of my code it would be a huge help!! just confused about how to account for H changing and P at the sametime!!