Finding minimum value of function of several variables?

Hi everybody. I have a code to find minimum value of function of several variable (Mr) as below:

%                       GIOI THIEU
% Chuong trinh tinh toan toi uu hoa mat cat dam ban BTCT
% Tac gia: Le Tuan Dung
% Bo mon Cong trinh - Co so Thai Nguyen - DHCNGTVT
%-------------------*************--------------------------
% PHAN1: THAM SO DAU VAO CUA BAI TOAN
% 1.1. Cac gia tri trong luong rieng:
% Trong luong rieng be tong dam va ban
gamabt=25; % Don vi: kN/m3
% Trong luong rieng lop phu
gamalp=22.5; % Don vi: kN/m3
% 1.2. Cuong do cua vat lieu
% Cuong do be tong dam (f'c)
fcdam=30; % Don vi: Mpa
% Cuong do be tong ban (f'c)
fcban=30; % Don vi: Mpa
% Cuong do cot thep DUL (dieu 5.9.3)
fpu=1860; % Cuong do keo dut (Mpa)
fpy=0.9*fpu; % Cuong do chay deo (Mpa)
fpe=0.8*fpy; % Cuong do huu hieu (sau khi tru toan bo mat mat US) (Mpa)
% Modun dan hoi cua thep DUL (dieu 5.4.4.2)
Eps=197000; % (Mpa)
% 1.3. Tham so mat cat ngang cau:
Bcau=12; % % Be rong MCN cau (m)
hbbien=0.1; % Chieu day ban lien ket ben ngoai (m)
hbtrong=0.21; % Chieu day ban len ket ben trong (m)
Sodam=12; % Cau co 12 dam chu
daylp=0.08; % Chieu day lop phu mat cau (m)
Aclc=0.231; % Dien tich chan lan can
mlc=0.04;   % Khoi luong/chieu dai cua lan can (kg/m)
gtt=9.81; % Gia toc trong truong (m/s2)
Ldam=24; % Chieu dai dam (m)
agoi=0.35; % Dau dam cach goi cau (m)
% 1.4. Cac he so tinh toan mo men:
neta=0.95; % He so dieu chinh
gamatt1=1.25; % He so tai trong tinh tai giai doan 1
gamatt2=1.5;  % He so tai trong tinh tai giai doan 2
gamall=1.75;  % He so tai trong cua hoat tai
IM=0.25;      % He so xung kich
phif=0.9;     % He so suc khang uon
% He so quy quy khoi ung suat chu nhat:
if fcdam <= 28
   beta1=0.85;
elseif 28<fcdam<56
       beta1=0.85-0.05*(fcdam-28)/7;
else
   beta1=0.65;
end
% 1.5. Tai trong Hoat tai:
Ptruck=[145 145 35]; % (kN)
Kctruck=4.3;         % (m)
Ptandem=[110 110];   % (kN)
Kctandem=1.2;        % (m)
qlan=9.3;            % (kN/m)
% 1.6. Tham so kich thuoc cua mat cat ngang:
disp('MOI CAC BAN NHAP CAC THAM SO DAU BAI');
B1=input('Gia tri can duoi be rong B1(cm): ');
B2=input('Gia tri can tren be rong B2(cm): ');
bs1=input('Gia tri can duoi be rong suon bien bs1(cm): ');
bs2=input('Gia tri can tren be rong suon bien bs2(cm): ');
bsg1=input('Gia tri can duoi be rong suon giua bsg1(cm): ');
bsg2=input('Gia tri can tren be rong suon giua bsg2(cm): ');
H1=input('Gia tri can duoi chieu cao H1(cm): ');
H2=input('Gia tri can tren chieu cao H2(cm): ');
hnt1=input('Gia tri can duoi chieu cao nap tren hnt1(cm): ');
hnt2=input('Gia tri can tren chieu cao nap tren hnt2(cm): ');
hnd1=input('Gia tri can duoi chieu cao nap duoi hnd1(cm): ');
hnd2=input('Gia tri can tren chieu cao nap duoi hnd2(cm): ');
Bslk=input('Gia tri chieu rong ban be tong lien ket Bslk(cm): '); % Bslk= 1 (m)
hslk=input('Gia tri chieu day ban be tong lien ket hslk(cm): ');  % hslk=0.1 (m)
ntao1=input('Gia tri can duoi cua so tao cap ntao1: ');
ntao2=input('Gia tri can tren cua so tao cap ntao2: ');
dtao=input('Nhap duong kinh cua mot tao thep DUL dtao(mm): '); 
B=(B1:B2);          % Be rong MCN (0,8 - 1,0 met)
bs=(bs1:bs2);       % Be rong suon bien (0,1 - 0,2 met)
bsg=(bsg1:bsg2);    % Be rong suon giua (0,1 - 0,2 met)
H=(H1:H2);          % Chieu cao MCN (0,8 - 1,0 met)
hnt=(hnt1:hnt2);    % Chieu day ban tren (0,1 - 0,2 met)
hnd=(hnd1:hnd2);    % Chieu day ban duoi (0,15 - 0,25 met)
ntao=(ntao1:2:ntao2); % So tao cap DUL la so chan(30 - 40 tao cap)
% Kiem tra tham so nhap vao:
Ktra=[B1 B2 bs1 bs2 bsg1 bsg2 H1 H2 hnt1 hnt2 hnd1 hnd2 Bslk hslk ntao dtao];
V={'B1','B2','bs1','bs2','bsg1','bsg2','H1','H2','hnt1','hnt2','hnd1','hnd2','Bs','hs','ntao','dtao'};
if find(Ktra<0)>0
    disp('GIA TRI NHAP PHAI LA SO DUONG');
end
for i=1:numel(Ktra)
    if Ktra(i)<0
       %Cach bieu dien ten bien so 
       fprintf('Nhap lai gia tri cua: %s \n',V{i}); 
    end
end
% PHAN 2: TOI UU HOA MAT CAT
% 2.1. Xac dinh tai trong tinh tai:
% Trong luong phan ban lien ket:
qlk=(2*(hbbien+hbtrong)*Bcau/2)/(2*Sodam); % Don vi: kN/m
% Tai trong do trong luong lop phu mat cau
qlp=(Bcau*daylp*gamalp)/Sodam;
% Tai trong do trong luong phan lan can
qlc=Aclc*gamabt+mlc*gtt; 
% 2.2. Xac dinh cac tai trong do hoat tai:
% Chieu dai nhip tinh toan
Ldamtt=Ldam-2*agoi; 
% Gia tri tung do duong anh huong:
y1=Ldamtt/4;
y2=((Ldamtt/2-4.3)*y1)/(Ldamtt/2);
y3=((Ldamtt/2-8.6)*y1)/(Ldamtt/2);
y4=((Ldamtt/2-1.2)*y1)/(Ldamtt/2);
Mrmin=Inf;
imin=1;
jmin=1;
kmin=1;
zmin=1;
lmin=1;
mmin=1;
nmin=1;
for i=1:numel(B)
    for j=1:numel(bs)
        for k=1:numel(bsg)
            for z=1:numel(H)
                for l=1:numel(hnt)
                    for m=1:numel(hnd)
                        for n=1:numel(ntao)
                            % Be rong cua lo rong:
                            br=((B(i)-2*bs(j)-bsg(k))/2)*0.01; % (m)
                            % Chieu cao phan lo rong
                            hr=(H(z)-hnt(l)-hnd(m))*0.01; % (m)
                            % TINH TOÁN MAT CAT QUY DOI
                            % Chieu cao canh tren:
                            ht=hnt(l)*B(i)/B(i)*0.01; % (m)
                            % Chieu cao canh duoi:
                            hd=hnd(m)*B(i)/B(i)*0.01; % (m)
                            % Chieu cao phan suon dam:
                            hsd=H(z)*0.01-ht-hd; % (m)
                            % Be rong phan suon dam:
                            bsd=(2*bs(j)*0.01*hr+bsg(k)*0.01*hr)/hsd; % (m)
                            % Tai trong do trong luong ban than dam chu:
                            qdc=(B(i)*0.01*hd+bsd*hsd+B(i)*001*ht)*gamabt; % (kN/m)
                            % TINH TOAN GIA TRI MOMEN UON
                            % Momen uon do tinh tai:
                            Mttai=neta*(gamatt1*qdc*Ldamtt*Ldamtt/8+gamatt1*qlk*Ldamtt*Ldamtt/8+...
                                gamatt1*qlc*Ldamtt*Ldamtt/8+gamatt2*qlp*Ldamtt*Ldamtt/8);
                            % Tinh he so phan bo ngang dam giua:
                            K=2.5*power(Sodam,-0.2);
                            de=0;
                            e=1.04+de/7600;
                            gg1lan=K*power((B(i)*0.01)/(2.8*Ldamtt),0.5)*power(1,0.25); % He so pbn 1 lan dam giua
                            gg2lan=K*power(B(i)*10/7600,0.6)*power((B(i)*0.01)/Ldamtt,0.2)*power(1,006); % He so pbn 2 lan dam giua
                            % Tinh he so phan bo ngang dam bien:
                            gb1lan=0.2; % Tinh theo PP don bay
                            gb2lan=e*gg2lan;
                            glay=[gg1lan gg2lan gb1lan gb2lan];
                            % He so phan bo ngang de tinh toan:
                            gpbn=max(glay);
                            % Momen uon do xe tai thiet ke:
                            Mtruck=neta*gpbn*(1+IM)*gamall*(Ptruck(1)*y1+Ptruck(2)*y2+Ptruck(3)*y3); % kN.m
                            % Momen uon do xe 2 truc:
                            M2truc=neta*gpbn*(1+IM)*gamall*(Ptandem(1)*y1+Ptandem(2)*y4); % kN.m
                            Mxetk=[Mtruck M2truc]; % kN.m
                            % Momen uon do tai trong lan xe
                            Mlanxe=neta*gpbn*gamall*9.3*0.5*(1/4)*Ldamtt*Ldamtt; % kN.m
                            % Momen uon do hoat tai
                            Mhtai=max(Mxetk)+Mlanxe; % kN.m
                            % Momen uon su dung tinh toan
                            Mutt=Mttai+Mhtai; 
                            % TINH TOAN DAC TRUNG HINH HOC MAT CAT QUY DOI
                            % Dien tich mat cat ngang:
                            Aqd=B(i)*0.01*hd+bsd*hsd+B(i)*0.01*ht; %(m2)
                            % Momen tinh cua mat cat so voi day mat cat:
                            Sqd=(1/2)*B(i)*0.01*hd*hd+(0.5*hsd+hd)*hsd*bsd+(H(z)*0.01-0.5*ht)*ht*B(i)*0.01; % (m3)
                            % K/cach tu trong tam cua tiet dien quy doi toi day
                            % mat cat:
                            yd=Sqd/Aqd; %(m)
                            % K/cach tu tho tren cung toi truc trung hoa:
                            yt=H(z)*0.01-yd; %(m)
                            % Momen quan tinh cua mat cat doi voi truc trung
                            % hoa:
                            Jqd=(B(i)*0.01*ht*ht*ht)/12+power((yt-0.5*ht),2)*ht*(B(i)*0.01)+bsd*power(hsd,3)/12+...
                            power((yt-ht-hsd/2),2)*hsd*bsd+(B(i)*0.01)*power(hd,3)/12+power((yd-0.5*hd),2)*hd*(B(i)*0.01);
                            % TINH TOAN DAC TRUNG HINH HOC MAT CAT LIEN HOP
                            % Modun dan hoi dam:
                            Edam=0.043*power(gamabt*100,1.5)*sqrt(fcdam); % Mpa
                            % Modun dan hoi dam:
                            Eban=0.043*power(gamabt*100,1.5)*sqrt(fcban); % Mpa
                            % He so quy doi tu be tong ban sang be tong dam:
                            n1=Eban/Edam;
                            % Dien tich mat cat ngang:
                            Alh=Aqd+n1*Bslk*0.01*hslk*0.01; %(m2)
                            % Mome tinh cua mat cat so voi day mat cat:
                            Slh=Sqd+n1*Bslk*0.01*hslk*0.01*(H(z)*0.01+0.5*hslk*0.01); % (m3)
                            % K/cach tu trong tam cua tiet dien lien hop toi
                            % day ban:
                            ydlh=Slh/Alh; % (m)
                            % K/cach tu tho tren cung toi truc trung hoa:
                            ytlh=H(z)*0.01+hslk*0.01-ydlh; % (m)
                            % Momen quan tinh cua mat cat lien hop:
                            Jlh=(B(i)*0.01)*power(ht,3)/12+power((ytlh-hslk*0.01-0.5*ht),2)*ht*(B(i)*0.01)...
                                +bsd*power(hsd,3)/12+power((ytlh-hslk*0.01-ht-0.5*hsd),2)*hsd*bsd+...
                                (B(i)*0.01)*power(hd,3)/12+power((ydlh-0.5*hd),2)*hd*(B(i)*0.01)+...
                                n1*(Bslk*0.01*power(hslk*0.01,3)/12+Bslk*0.01*hslk*0.01*power((ytlh-0.5*hslk*0.01),2)); % (m4)
                            % TINH TOAN DAC TRUNG HINH HOC MAT CAT CO COT THEP
                            % He so quy doi tu cot thep DUL sang be tong dam:
                            n2=Eps/Edam;
                            % Dien tich mat cat ngang:
                            Act=Alh+(n2-1)*pi*dtao*0.001*dtao*0.001*ntao(n)*(1/4); %(m2)
                            % K/cach tu tim hang cot thep ngoai cung toi tho
                            % duoi:
                            ctduoi=0.06; % (m)
                            % K/cach tim toi tim giua cac hang cot thep:
                            timct=0.045;  % (m)              
                            % K/cach tu trong tam cot thep toi day mat cat:
                            dpsduoi=ctduoi+timct/2; %(m)
                            % K/cach tu trong tam dien tich cot thep toi tho tren cung cua MC:
                            dps=H(z)*0.01+hslk*0.01-dpsduoi;
                            % Momen tinh cua mat cat so voi day mat cat:
                            Sct=Slh+(n2-1)*pi*dtao*0.001*dtao*0.001*ntao(n)*(1/4)*(H(z)*0.01+hslk*0.01-dps); % (m3)
                            % K/cach tu trong tam tiet dien lien hop toi day MC
                            ydct=Sct/Act; %(m)
                            % K/cach tu tho tren cung toi TTH:
                            ytct=H(z)*0.01+hslk*0.01-ydct; % (m)
                            % Momen quan tinh cua mat cat co cot thep:
                            Jct=(B(i)*0.01)*power(ht,3)/12+power((ytct-hslk*0.01-ht/2),2)*ht*(B(i)*0.01)+...
                                bsd*power(hsd,3)/12+power((ytct-hslk*0.01-ht-hsd/2),2)*hsd*bsd+...
                                (B(i)*0.01)*power(hd,3)/12+power((ydct-hd/2),2)*hd*B(i)*0.01+...
                                n1*(Bslk*0.01*power(hslk*0.01,3)/12+Bslk*0.01*hslk*0.01*power((ytct-(hslk*0.01)/2),2))+...
                                (n2-1)*pi*dtao*0.001*dtao*0.001*(1/4)*ntao(n)*power((ydct-dpsduoi),2); % (m4)
                            % XAC DINH SUC KHANG UON CUA MAT CAT NGANG
                            % Chieu cao khoi ung suat hinh chu nhat:
                            acn=beta1*ytct; %(m)
                            if ytct>(ht+hslk*0.01)
                            % Suc khang uon danh dinh:
                               Mn=pi*dtao*0.001*dtao*0.001*(1/4)*ntao(n)*fpe*1000*(dps-acn/2)+...
                                   0.85*fcdam*1000*(B(i)*0.01-bsd)*ht*(acn/2-ht/2)+0.85*fcdam*1000*(Bslk*0.01-bsd)*hslk*0.01*(acn/2-(hslk*0.01)/2); % kN.m                              
                            else
                               Mn=pi*dtao*0.001*dtao*0.001/4*ntao(n)*fpe*1000*(dps-acn/2)+...
                                   0.85*fcdam*1000*(Bslk*0.01-B(i)*0.01)*hslk*0.01*(acn/2-(hslk*0.01)/2); % kN.m
                            end
                            % Suc khang uon tinh toan:
                            Mr=phif*Mn; %kN.m
                            if (Mr<=Mrmin)&&(Mr>=Mutt)
                                Mrmin = Mr;
                                imin = i;
                                jmin = j;
                                kmin = k;
                                zmin = z;
                                lmin = l;
                                mmin = m;
                                nmin = n;
                            end
                        end
                    end
                end
            end
        end
    end
end
disp('DA CHAY XONG CHUONG TRINH');
if ytct>(ht+hslk*0.01)
   disp('TRUC TRUNG HOA DI QUA SUON DAM');
else
   disp('TRUC TRUNG HOA DI QUA BAN CANH TREN');
end
% CAC GIA TRI KICH THUOC MAT CAT NGANG VA SO TAO THEP TOI UU
disp(Mrmin);
disp(B(imin));
disp(bs(jmin));
disp(bsg(kmin));
disp(H(zmin));
disp(hnd(lmin));
disp(hnt(mmin));
disp(ntao(nmin));

And when program finished, it gives me:

    Mrmin=Inf;
    80
    10
    10
    80
    10
    15
    30
of cause they (above values) are lower bonds of values that i must enter in paragraph:

    B1=input('Gia tri can duoi be rong B1(cm): ');
    B2=input('Gia tri can tren be rong B2(cm): ');
    bs1=input('Gia tri can duoi be rong suon bien bs1(cm): ');
    bs2=input('Gia tri can tren be rong suon bien bs2(cm): ');
    bsg1=input('Gia tri can duoi be rong suon giua bsg1(cm): ');
    bsg2=input('Gia tri can tren be rong suon giua bsg2(cm): ');
    H1=input('Gia tri can duoi chieu cao H1(cm): ');
    H2=input('Gia tri can tren chieu cao H2(cm): ');
    hnt1=input('Gia tri can duoi chieu cao nap tren hnt1(cm): ');
    hnt2=input('Gia tri can tren chieu cao nap tren hnt2(cm): ');
    hnd1=input('Gia tri can duoi chieu cao nap duoi hnd1(cm): ');
    hnd2=input('Gia tri can tren chieu cao nap duoi hnd2(cm): ');
    Bslk=input('Gia tri chieu rong ban be tong lien ket Bslk(cm): '); % Bslk= 1 (m)
    hslk=input('Gia tri chieu day ban be tong lien ket hslk(cm): ');  % hslk=0.1 (m)
    ntao1=input('Gia tri can duoi cua so tao cap ntao1: ');
    ntao2=input('Gia tri can tren cua so tao cap ntao2: ');
    dtao=input('Nhap duong kinh cua mot tao thep DUL dtao(mm): ');

So, who can explanin to me why?

[Jan]

Why what? Why you have to enter the values or why they are the lower bounds? I do not understand the comments and have no idea, what's going on inside the code.