Following the guideline contained in
you can easily solve the problem as follow:
First you need to define an objective or cost function. This function must describe a minimization problem.
Let's assume you want to maximize the profit or minimize it negate as a function of the number of Bucket x(1) and Kettle x(2).
Your cost function then is:
f'*x := f(1) * x(1) + f(2) * x(2)
hence
>> f = [-2; -1.3];
let's define the inequality constraints for the production costs
7 * x(1) + 4 * x(2) <= 15000
and for the size of the warehouse:
x(1) + x(2) <= 4000
in matrix form
>> A = [7 4; 1 1];
>> b = [15000; 4000];
also x(1) and x(2) can assume value in the range 0 <= x <= 4000, therefore the lower bound and upper bound constrains are
>> lb = [0 0];
>> ub = [4000 4000];
you don't have any equality constrains therefore
>> Aeq = [];
>> beq = [];
Now you are ready to call the LP solver:
>> options = optimoptions('linprog','Algorithm','dual-simplex');
>> [x,fval,exitflag,output] = linprog(f,A,b,Aeq,beq,lb,ub,options);
the optimal solution should be
>> x
x =
0
3750
so the maximum profit within the given cost is given by producing 3750 Kettle only. You still have 250 storage capacity available in the warehouse, so this constraint wasn't hit.
The fact that we have 75 machines to utilize effectively is irrelevant for the above solution. Perhaps additional information such the production type: continuous or batch, time scheduled for producing B or K, etc. would make this additional information necessary/useful.
