I'm working on an optimization problem involving a single decision variable p1, subject to four inequality constraints:

• Two upper bound constraints:

  p1 < b1,p1 < b2

• Two lower bound constraints:

  c1 ≤ p1,c2 ≤ p1

Effectively, the feasible region for p1 is:

max⁡(c1,c2)  ≤  p1  ≤  min⁡(b1,b2)

I have already formulated the Karush-Kuhn-Tucker (KKT) conditions for this setup, and now I'm trying to determine:

1. The optimal value p1∗​

2. The corresponding feasibility conditions

3. A case-wise breakdown depending on which constraints are active or inactive

Sheet:  Q_P1_Optimum_condition.mw

How can I rearrange the inequality to isolate p1​ on one side?
I'm currently using the 'isolate' function, but it's returning expressions with the signum function.

Q_isolate.mw

Given two optimality conditions defined as inequalities — a lower bound condition and an upper bound condition — under the assumptions that all parameters are positive and satisfy the relationships w>Pu>Ce>U[0]>0, and w>Pu>Ce,   and upsilon,varphi, tau0, eta are between (0,1), where 1>=varphi>=tau0>=0. and upsilon>eta

Under what conditions on the parameters does the lower bound condition become violated while the upper bound condition is also violated?

Detailed question in sheet attached: Q_Condition.mw

[ see also Q_Const.mw ]

I am working on a script (attached sheet) where I use a FOR loop to iterate over different values of varepsilon. Within each iteration, I perform the following steps:

1. I optimize the function R_out and obtain a result, denoted as Pc

2. I then substitute Pc into another function, L_out and optimize it to find the values of p1,p2, and the corresponding function value.

I follow a similar procedure for a second case as well.

However, I'm encountering an issue: I'm unable to successfully substitute the result from R_out into L_out within the loop. This is causing an error in execution.

Finally, I intend to generate a plot with varepsilon  on the x-axis and certain result variables on the y-axis in a single plot (as indicated at the end of the sheet).

Sheet: Question_New.mw

Can someone help me:

1. Fix the substitution error in the loop, and

2. Provide the correct syntax for generating the desired plot?

I'm optimizing a function with constraints, meaning the decision variable has both a lower and an upper bound. When I use the  (MAX) syntax in Maple, it returns the lower bound as the optimal value (Pc) along with the corresponding function value. However, when I plot the graph, it shows that the function actually reaches its maximum at the upper bound. What could be causing this discrepancy

Sheet:Q_result_1.mw