# Maximize objective function using fmincon with a limit

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Angel Ani on 27 Jun 2021
Commented: William Rose on 29 Jun 2021
Maximize P: (U*y1 - P*y1) + (P*y2 - V*y2)
where 0.4 <= P <= 0.3
f=@x-1*((U*y1 - (x)*y1) + ((x)*y2 - V*y2))
U=2;y1=3;y2=1;V=0.2;
A=[]; q=[]; lb=0.3; ub=0.4;
[ans,ans1]= fmincon(f,A,q,lb,ub);
ans; ans1

William Rose on 27 Jun 2021
Create the following function and save it as file parameterfun.m:
function y=parameterfun(x,U,V,y1,y2)
y=-((U*y1 - x*y1) + (x*y2 - V*y2)); %-profit
Then execute the following code:
U=2;y1=3;y2=1;V=0.2;
f = @(x)parameterfun(x,U,V,y1,y2);
x0=0.35; %initial guess
lb=0.3; ub=0.4;
[x,fval] = fmincon(f,x0,[],[],[],[],lb,ub);
fprintf('x=%.3f, Profit=%.3f\n',x,-fval);
This produces the following output on the console:
x=0.300, Profit=5.200
William Rose on 29 Jun 2021
I see now that the problem statement in your orignal post has issues. You stated the problem as:
Maximize P: (U*y1 - P*y1) + (P*y2 - V*y2),
where 0.4 <= P <= 0.3
The issues are:
1. Line 2 is impossible: P cannot be greater than .4 and less than .3.
2. Equation (U*y1 - P*y1) + (P*y2 - V*y2) does not constrain P, because is has no value on the right or left side.
3. There is no "x" in the equation above, and U,V,y1,y2 are specified (in the code), so there is nothing to adjust.
4. In maximization problems, one does not constrain the thing being maximized (Profit, in this case). One constrains the adjustable input(s) that determine the profit. Threfore the second line does not make sense, if we are to maximize P.
One possibility is that the original statement should have been
Maximize P: (U*y1 - x*y1) + (x*y2 - V*y2),
where 0.3 <= x <= 0.4.
The problem above is the one which my code solves, and which your originally posted code almost solves.
P at the solution is outside the bounds [0.3, 0.4] because [0.3, 0.4] are bounds for x, not for P.

R2016a

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