I am trying to solve a profit maximization problem:
Revenue = R(X) = 10 * X
Cost = C(X) = X^3 - 6X^2 + 15X
Objective Function:
Maximize -> Profit = Revenue - Cost = 6X^2 - X^3 - 5X
I am wondering how this can be solved by using the Optimization Tool?
Solved! Go to Solution.
This article may help! Tool Mastery | Optimization - Alteryx Community
Hi @alexnajm ,
Thanks for the article.
The problem is with the quadratic and cubic parameters in the equation. All the Optimization applications, weekly challenges, and resources I've seen, including the link you shared, use one-degree equations (no powers raised), and I am really unsure about how to properly specify higher degree objective functions and constraints for the optimization
Hello. You can solve it without a tool, just with mathematics ;)
6X^2 - X^3 - 5X=Profit=> where is the max= where the derivative is 0 (well, one of the two, the other being the minimum),=>
12X-3X^2-5=0=>
Hi @simonaubert_bd,
Thanks for your reply. I know I can solve it using calculus, but the point is not to solve this particular problem.
My point is how do we (or can we) use the Optimization Tool to solve problems like this? There may be much more complex problems that are difficult to solve by hand, and I was just thinking if the Optimization Tool could be useful in those circumstances.
Does this solved quadratic problem help? https://community.alteryx.com/t5/Alteryx-Designer-Discussions/Quadratic-Programming-with-the-Optimiz...
@alexnajm Interesting, need to spend more time wrapping my head around this one.