how I Found the minimum of function using the fminbnd ??

f(x)=3x1+2x1x2+x2 such as 0.1≤ x1 ≤1 0.25 ≤ x2 ≤ 4 help me please thanks in advance

Answer 1

[Walter Roberson]

You cannot do that. fminbnd() can only be used for a function of a single variable.

[Walter Roberson]

fun = @(x) 3*x(1) + 2*x(1)*x(2) + x(2)
x0 = [.5 2.5];
A = [];
b = [];
Aeq = [];
beq = [];
lb = [0.1 0.25];
ub = [1 4];
x12 = fmincon(fun, x0, A, b, Aeq, beq, lb, ub)
x1 = x12(1); x2 = x12(2);

Answer 2

[Torsten]

You can't. fminbnd is for functions of one variable - your f is a function of two variables.

Use "fmincon" instead.

Best wishes

Torsten.

[ahlem sellami]

thanks for your help please how I found the minimum of the function f by using fmincon ?? can you help me thanks in advance

[Torsten]

lb = [0.1 0.25];
ub = [1 4];
fun = @(x) 3*x(1)+2*x(1)*x(2)+x(2);
x0 = [0.5 2];
sol = fmincon(fun,x0,[],[],[],[],lb,ub)

If your problem is quadratic (like this test example), MATLAB's "quadprog" is an alternative to "fmincon".

Best wishes

Torsten.

[ahlem sellami]

excuse me c'est quoi x0???

[Torsten]

initial guess for the solution "sol"

Here is the relevant documentation:

https://de.mathworks.com/help/optim/ug/fmincon.html

[ahlem sellami]

another question please !!!!

if 90≤ x1 ≤120
2900 ≤ x2 ≤ 5800

the x0 is equal what??

[Torsten]

In principle, x0 is arbitrary.

But if you know an x0 that is feasible and near to the optimum, you should use it since you'll most probably get faster and more stable convergence towards the solution.

[ahlem sellami]

thanks a lot and best regards