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lamda = 0??...

Andiguys 85
October 04 2025
0 0

@dharr I have solved it using KKT Condition, but lamda is zero. Is it right???

> 

restart
> 

with(Optimization); with(plots); with(LinearAlgebra)
> 

_local(Pi)

Pi

 (1)
> 

NULL
> 

Pi := proc (Pn, Pr, w) options operator, arrow; (Pn-Cn)*(1-(Pn-Pr)/(1-delta))+(Pr-w-Crm)*alpha*(1/2+(1/2)*(((-2*delta+2)*tau^2+((-Pn+Pr-delta+1)*Cr+(2*delta-2)*(w+i2))*tau+Cr*i2*(Pn-Pr+delta-1))/((4*delta-4)*tau+Cr*(Pn-Pr+delta-1))-i2)/tau)*(1-(Pn-Pr)/(1-delta))-Ce*rho0*(1-(Pn-Pr)/(1-delta)) end proc

proc (Pn, Pr, w) options operator, arrow; (Pn-Cn)*(1-(Pn-Pr)/(1-delta))+(Pr-w-Crm)*alpha*(1/2+(1/2)*(((-2*delta+2)*tau^2+((-Pn+Pr-delta+1)*Cr+(2*delta-2)*(w+i2))*tau+Cr*i2*(Pn-Pr+delta-1))/((4*delta-4)*tau+Cr*(Pn-Pr+delta-1))-i2)/tau)*(1-(Pn-Pr)/(1-delta))-Ce*rho0*(1-(Pn-Pr)/(1-delta)) end proc

 (2)
> 

NULL
> 

NULL
> 

C1 := (((Cr+4*tau)*delta-4*tau+(Pn-Pr-1)*Cr)*rho0+(delta-1)*(i2-tau))/(delta-1) <= w

(((Cr+4*tau)*delta-4*tau+(Pn-Pr-1)*Cr)*rho0+(delta-1)*(i2-tau))/(delta-1) <= w

 (3)
> 

L := Pi(Pn, Pr, w)-lambda*((((Cr+4*tau)*delta-4*tau+(Pn-Pr-1)*Cr)*rho0+(delta-1)*(i2-tau))/(delta-1)-w)

(Pn-Cn)*(1-(Pn-Pr)/(1-delta))+(Pr-w-Crm)*alpha*(1/2+(1/2)*(((-2*delta+2)*tau^2+((-Pn+Pr-delta+1)*Cr+(2*delta-2)*(w+i2))*tau+Cr*i2*(Pn-Pr+delta-1))/((4*delta-4)*tau+Cr*(Pn-Pr+delta-1))-i2)/tau)*(1-(Pn-Pr)/(1-delta))-Ce*rho0*(1-(Pn-Pr)/(1-delta))-lambda*((((Cr+4*tau)*delta-4*tau+(Pn-Pr-1)*Cr)*rho0+(delta-1)*(i2-tau))/(delta-1)-w)

 (4)
> 

K1 := diff(L, Pn) = 0; K2 := diff(L, Pr) = 0; K3 := diff(L, w) = 0; K4 := diff(lambda*((((Cr+4*tau)*delta-4*tau+(Pn-Pr-1)*Cr)*rho0+(delta-1)*(i2-tau))/(delta-1)-w), lambda) = 0

1-(Pn-Pr)/(1-delta)-(Pn-Cn)/(1-delta)+(1/2)*(Pr-w-Crm)*alpha*((Cr*i2-Cr*tau)/((4*delta-4)*tau+Cr*(Pn-Pr+delta-1))-((-2*delta+2)*tau^2+((-Pn+Pr-delta+1)*Cr+(2*delta-2)*(w+i2))*tau+Cr*i2*(Pn-Pr+delta-1))*Cr/((4*delta-4)*tau+Cr*(Pn-Pr+delta-1))^2)*(1-(Pn-Pr)/(1-delta))/tau-(Pr-w-Crm)*alpha*(1/2+(1/2)*(((-2*delta+2)*tau^2+((-Pn+Pr-delta+1)*Cr+(2*delta-2)*(w+i2))*tau+Cr*i2*(Pn-Pr+delta-1))/((4*delta-4)*tau+Cr*(Pn-Pr+delta-1))-i2)/tau)/(1-delta)+Ce*rho0/(1-delta)-lambda*Cr*rho0/(delta-1) = 0

  

(Pn-Cn)/(1-delta)+alpha*(1/2+(1/2)*(((-2*delta+2)*tau^2+((-Pn+Pr-delta+1)*Cr+(2*delta-2)*(w+i2))*tau+Cr*i2*(Pn-Pr+delta-1))/((4*delta-4)*tau+Cr*(Pn-Pr+delta-1))-i2)/tau)*(1-(Pn-Pr)/(1-delta))+(1/2)*(Pr-w-Crm)*alpha*((-Cr*i2+Cr*tau)/((4*delta-4)*tau+Cr*(Pn-Pr+delta-1))+((-2*delta+2)*tau^2+((-Pn+Pr-delta+1)*Cr+(2*delta-2)*(w+i2))*tau+Cr*i2*(Pn-Pr+delta-1))*Cr/((4*delta-4)*tau+Cr*(Pn-Pr+delta-1))^2)*(1-(Pn-Pr)/(1-delta))/tau+(Pr-w-Crm)*alpha*(1/2+(1/2)*(((-2*delta+2)*tau^2+((-Pn+Pr-delta+1)*Cr+(2*delta-2)*(w+i2))*tau+Cr*i2*(Pn-Pr+delta-1))/((4*delta-4)*tau+Cr*(Pn-Pr+delta-1))-i2)/tau)/(1-delta)-Ce*rho0/(1-delta)+lambda*Cr*rho0/(delta-1) = 0

  

-alpha*(1/2+(1/2)*(((-2*delta+2)*tau^2+((-Pn+Pr-delta+1)*Cr+(2*delta-2)*(w+i2))*tau+Cr*i2*(Pn-Pr+delta-1))/((4*delta-4)*tau+Cr*(Pn-Pr+delta-1))-i2)/tau)*(1-(Pn-Pr)/(1-delta))+(1/2)*(Pr-w-Crm)*alpha*(2*delta-2)*(1-(Pn-Pr)/(1-delta))/((4*delta-4)*tau+Cr*(Pn-Pr+delta-1))+lambda = 0

  

(((Cr+4*tau)*delta-4*tau+(Pn-Pr-1)*Cr)*rho0+(delta-1)*(i2-tau))/(delta-1)-w = 0

 (5)
> 

sol := solve({K1, K2, K3, K4}, {Pn, Pr, lambda, w})

{Pn = (4*alpha*rho0^2*tau+Crm*alpha*rho0-alpha*delta*rho0+alpha*i2*rho0-alpha*rho0*tau+Ce*rho0+alpha*rho0+Cn)/(alpha*rho0+1), Pr = (4*alpha*rho0^2*tau+Crm*alpha*rho0+alpha*i2*rho0-alpha*rho0*tau+Ce*rho0+Cn+delta-1)/(alpha*rho0+1), lambda = 0, w = 4*rho0*tau+i2-tau}

 (6)
 

``

Download Q_1.mw

Optimum Pn, w and Pr...

Andiguys 85
October 04 2025
0 0

@dharr 

The above solution was obtained when I solved the following objective function with a constraint, but it appears to be incorrect.

Could you please help me solve the objective function under the condition that the constraint is binding? I would also like to know the KKT condition under which the objective function is maximized and the optimal values of Pn, Pr, and w are obtained.

> 

restart
> 

with(Optimization); with(plots); with(LinearAlgebra)
> 

_local(Pi)

Pi

 (1)
> 

``
> 

Pi := proc (Pn, Pr, w) options operator, arrow; (Pn-Cn)*(1-(Pn-Pr)/(1-delta))+(Pr-w-Crm)*alpha*(1/2+(1/2)*(((-2*delta+2)*tau^2+((-Pn+Pr-delta+1)*Cr+(2*delta-2)*(w+i2))*tau+Cr*i2*(Pn-Pr+delta-1))/((4*delta-4)*tau+Cr*(Pn-Pr+delta-1))-i2)/tau)*(1-(Pn-Pr)/(1-delta))-Ce*rho0*(1-(Pn-Pr)/(1-delta)) end proc

proc (Pn, Pr, w) options operator, arrow; (Pn-Cn)*(1-(Pn-Pr)/(1-delta))+(Pr-w-Crm)*alpha*(1/2+(1/2)*(((-2*delta+2)*tau^2+((-Pn+Pr-delta+1)*Cr+(2*delta-2)*(w+i2))*tau+Cr*i2*(Pn-Pr+delta-1))/((4*delta-4)*tau+Cr*(Pn-Pr+delta-1))-i2)/tau)*(1-(Pn-Pr)/(1-delta))-Ce*rho0*(1-(Pn-Pr)/(1-delta)) end proc

 (2)
> 

``
> 

NULL
> 

C1 := (((Cr+4*tau)*delta-4*tau+(Pn-Pr-1)*Cr)*rho0+(delta-1)*(i2-tau))/(delta-1) <= w

(((Cr+4*tau)*delta-4*tau+(Pn-Pr-1)*Cr)*rho0+(delta-1)*(i2-tau))/(delta-1) <= w

 (3)
 

``

Download Q_1.mw

solve 3 equations? showing error...

Andiguys 85
October 03 2025
0 0

@dharr 

How can we simultaneously solve three equations of the form:
w=f(Pn,Pr)
Pn=g(w,Pr)
Pr=h(Pn,w)

so that we obtain standalone expressions for w, Pn and Pr?

> 

restart
> 

with(Optimization); with(plots); with(LinearAlgebra)
> 

_local(D)

D

 (1)
 
> 

D := 1-(Pn-Pr)/(1-delta)

1-(Pn-Pr)/(1-delta)

 (2)
> 

N := 2*delta*tau*(i2-tau+w)-2*tau*(i2-tau+w)+Cr*(i2-tau)*(Pn-Pr+delta-1); M := (4*delta-4)*tau+Cr*(Pn-Pr+delta-1)

2*delta*tau*(i2-tau+w)-2*tau*(i2-tau+w)+Cr*(i2-tau)*(Pn-Pr+delta-1)

  

(4*delta-4)*tau+Cr*(Pn-Pr+delta-1)

 (3)
> 

A := Cr*(M*(i2-tau)-N)/(tau*M^2); B := (delta-1)/M; lambda := alpha*D*(rho0*M/(1-delta)-Pr+w+Crm)

Cr*(((4*delta-4)*tau+Cr*(Pn-Pr+delta-1))*(i2-tau)-2*delta*tau*(i2-tau+w)+2*tau*(i2-tau+w)-Cr*(i2-tau)*(Pn-Pr+delta-1))/(tau*((4*delta-4)*tau+Cr*(Pn-Pr+delta-1))^2)

  

(delta-1)/((4*delta-4)*tau+Cr*(Pn-Pr+delta-1))

  

alpha*(1-(Pn-Pr)/(1-delta))*(rho0*((4*delta-4)*tau+Cr*(Pn-Pr+delta-1))/(1-delta)-Pr+w+Crm)

 (4)
> 

w := Pr-Crm-rho0*M/(1-delta)+lambda/(alpha*D); Pn := Cn+(1-delta)*(1-alpha*rho0-(Pr-w-Crm)*alpha*A+lambda*A)-(Pr-w-Crm)*alpha*rho0-Ce*rho0; Pr := Pn-(1-delta)*(1-alpha*rho0-(Pr-w-Crm)*alpha*A-lambda*A)-(Pr-w-Crm)*alpha*rho0-Ce*rho0

w

  

Error, recursive assignment

  

Error, recursive assignment

  
 

``

Download Simultaneously_solve.mw

Got it. Thankyou...

Andiguys 85
October 03 2025
0 0

@sand15 Thank you for the detailed clarification. I now understand that the FOC applies regardless of whether the maximum lies on the boundary or not, and that in constrained problems the maximum may occur at the boundary even if the FOC is not satisfied there. Your explanation cleared up my confusion.

Is there a solution when the constraint ...

Andiguys 85
October 02 2025
0 0

@sand15 

I agree that there is no interior maximum.

However, after rearranging the constraint by isolating w on one side, I obtain the following form of the constraint.


C1:=  -w < -(((Cr + 4*tau)*rho0 + i2 - tau)*delta + (-4*tau + (-1 + Pn - Pr)*Cr)*rho0 - i2 + tau)/(-1 + delta)

 In this case, when the constraint is binding, what would be the optimal values of Pn,Pr, and w that maximize the objective function? could you please indicate the analytical equations?

Just to be sure I’m following correctly...

Andiguys 85
October 02 2025
0 0

@sand15 
 

Apologies, i1 = (2*rho0 - 1)*tau + i2,   i1 is between [0,1]

If I understand correctly, the maximum occurs on the boundary. In that case, what are the analytical first-order/optimality equations for Pn, Pr​, and w?

                                                        OR 

In the unconstrained case, the Hessian has a zero in one of its principal minors and Determinant zero

When the constraint is binding, the Lagrangian Hessian also has a zero in one of the minors, and its determinant is zero.

So, does this imply that in both cases there is no maximum solution?

 

Maximizing Objective...

Andiguys 85
October 01 2025
0 0

@sand15 
I am working on maximizing an objective function with three decision variables: Pn, Pr and w. I want to understand the difference between cases where the constraint is binding versus not binding. Specifically, when the constraint is binding, what are the analytical optimal values of Pn​, Pr and w? Also, under what conditions do we get optimal values, and what will be the value of the KKT multiplier μ?

 Data = [C0 = 0.02, Ce = 0.05, Cr = 0.05, alpha=0.9, delta = 0.7, i2 = 0.1, rho0 = 0.5, tau = 0.05, Cn=0.1, Crm=0.05] Any missing DATA you can assume.

Note: Pn, Pr and w should be between 0 and 1

Apologies and Updated Worksheet with Obj...

Andiguys 85
October 01 2025
0 0

@sand15 

Thank you for pointing that out. I understand your concern and I apologize for mixing different problems together earlier.
Below is a clean worksheet with only the objective function and constraints so it’s clearer.

Also, I’m sorry if my way of following up seemed inconsiderate. I truly appreciate the time and effort you put into answering my queries, and I’ll make sure to acknowledge the solutions properly before moving on to new questions.

The last syntax you shared worked well and its correct.

> 

restart
> 

_local(Pi)

Pi

 (1)
> 

``
> 

NULL
> 

Pi := proc (Pn, Pr, w) options operator, arrow; (Pn-Cn)*(1-(Pn-Pr)/(1-delta))+(Pr-w-Crm)*alpha*rho0*(1-(Pn-Pr)/(1-delta))-Ce*rho0*(1-(Pn-Pr)/(1-delta)) end proc
> 

C1 := (2*(-(1-(Pn-Pr)/(1-delta))/(2*tau)-Cr*(1-(Pn-Pr)/(1-delta))^2/(8*tau^2)))*((2*rho0-1)*tau+i2)+(w/(2*tau)-1/2+i2/(2*tau))*(1-(Pn-Pr)/(1-delta))-Cr*(1/2-i2/(2*tau))*(1-(Pn-Pr)/(1-delta))^2/(2*tau) > 0
 

 

Download Proposal_follow_up.mw

Solving optimization with three decision...

Andiguys 85
October 01 2025
0 0

@sand15 Thankyou for the syntax. Just one more question.

If I have three decision variables and one constraint, can I still solve it using the same syntax? How would that work?

> 

restart
> 

with(Optimization); with(plots); with(LinearAlgebra)
> 

_local(Pi)

Pi

 (1)
> 

NULL
> 

NULL
> 

> 

SC_CS_sols := solve(
  {
    diff(?????) - mu[1] = 0,
    mu[1]*(-2*(-(1 - (Pn - Pr)/(1 - delta))/(2*tau) - Cr*(1 - (Pn - Pr)/(1 - delta))^2/(8*tau^2))*((2*rho0 - 1)*tau + i2) + (w/(2*tau) - 1/2 + i2/(2*tau))*(1 - (Pn - Pr)/(1 - delta)) - Cr*(1/2 - i2/(2*tau))*(1 - (Pn - Pr)/(1 - delta))^2/(2*tau)) = 0
   
  }
  ,
  {i1, mu[1]}
  , 'parametric'='full', 'parameters'={P, Q, A, B}
)

SC_CS_sols := piecewise(P = 0, piecewise(Q <> 0, [], Q = 0, [{i1 = i1, mu[1] = 0}]), P <> 0, [{i1 = -Q/(2*P), mu[1] = 0}])

 (2)
> 

pi__m := (Pn - Cn)*(1 - (Pn - Pr)/(1 - delta)) + (Pr - w - Crm)*alpha*(1/2 + (i1 - i2)/(2*tau))*(1 - (Pn - Pr)/(1 - delta)) - Ce*rho0*(1 - (Pn - Pr)/(1 - delta)):

collect(pi__m, [Pn,Pr,w]):
PQR := [R, Q, P] =~ [coeffs(%, i1)]:

print~(PQR):

R = w*(1/2-(1/2)*i2/tau)*(1-(Pn-Pr)/(1-delta))-C0-(1/2)*Cr*(1/2-(1/2)*i2/tau)^2*(1-(Pn-Pr)/(1-delta))^2+Ce*rho0*(1-(Pn-Pr)/(1-delta))

  

Q = ((1/2)*w/tau-1/2+(1/2)*i2/tau)*(1-(Pn-Pr)/(1-delta))-(1/2)*Cr*(1/2-(1/2)*i2/tau)*(1-(Pn-Pr)/(1-delta))^2/tau

  

P = -(1/2)*(1-(Pn-Pr)/(1-delta))/tau-(1/8)*Cr*(1-(Pn-Pr)/(1-delta))^2/tau^2

 (3)
> 

AB := [A = `2*`(-(1 - (Pn - Pr)/(1 - delta))/(2*tau) - Cr*(1 - (Pn - Pr)/(1 - delta))^2/(8*tau^2))*((2*rho0 - 1)*tau + i2) + (w/(2*tau) - 1/2 + i2/(2*tau))*(1 - (Pn - Pr)/(1 - delta)) - Cr*(1/2 - i2/(2*tau))*(1 - (Pn - Pr)/(1 - delta))^2/(2*tau) ]

[A = (2*rho0-1)*tau+i2, B = tau+i2]

 (4)
> 

ind  := indets(rhs~(PQR)) union indets(rhs~(AB));

{C0, Ce, Cr, Pn, Pr, delta, i2, rho0, tau, w}

 (5)
> 

randomize();
data:=[C0 = 0.02, Ce = 0.05, Cr = 0.05, Pn = 0.603882321, Pr = 0.45, delta = 0.7, i2 = 0.1, rho0 = 0.5, tau = 0.05, w = 0.243374];

PQRAB := { eval(PQR, data)[], eval(AB, data)[] };

all_solutions := allvalues~(eval(SC_CS_sols, PQRAB)):
all_pi        := map(s -> eval(eval(pi__m, data), s), all_solutions):
here          := ListTools:-Search(min(all_pi), all_pi):
SOLUTION      := [all_pi[here], all_solutions[here]];


plot(eval(pi__m, data), i1=eval(eval(A..B, AB), data), gridlines=true)

Download Proposal_follow_up.mw

Got it. Thankyou...

Andiguys 85
September 28 2025
0 0

@sand15 

Thank you very much for your detailed and helpful comments

You’re right , I’ll substitute the solved expression for K2 (the w solution) into {K3,K4} and solve it to simplify Maple’s task.

Point well taken, I will carefully check feasibility and not assume any solution is correct without verifying it myself.

Thanks again for the clear, constructive feedback.

Query on Selecting the Correct Solution ...

Andiguys 85
September 27 2025
0 0

 rcorless 750,  @sand15 and @dharr Thankyou for the reply.

I have modified my objective function and, as before, I am simultaneously solving three equations with three variables and multiple parameters. This process is giving me three different solution sets, and I am unsure which one is correct or should be considered.

Note: All parameters are positive and greater than zero. All decision variable Pn, Pr and w should be positive equal to or greater than zero.

> 

restart
> 

with(Optimization); with(plots); with(LinearAlgebra)
> 

``
> 

"Pi1(Pn,Pr,w):=(Pn-Cn) (1-(Pn-Pr)/(1-delta))+(Pr-w-Crm) (1/2+(((-2 delta+2) tau^2+((-Pn+Pr-delta+1) Cr+(-2+2 delta) (w+i2)) tau+Cr i2 (-1+delta+Pn-Pr))/((4 delta-4) tau+Cr (-1+delta+Pn-Pr))-i2)/(2 tau)) (1-(Pn-Pr)/(1-delta))-Ce rho0 (1-(Pn-Pr)/(1-delta))"

 

 (1)
> 

K2 := diff(Pi1(Pn, Pr, w), w) = 0

 

 (2)
> 

solve(K2, w)

-(1/2)*Crm+(1/2)*Pr+(1/2)*i2-(1/2)*tau

 (3)
> 

NULL
> 

K3 := 1-(Pn-Pr)/(1-delta)-(Pn-Cn)/(1-delta)+(Pr-w-Crm)*((Cr*i2-Cr*tau)/((4*delta-4)*tau+Cr*(-1+delta+Pn-Pr))-((-2*delta+2)*tau^2+((-Pn+Pr-delta+1)*Cr+(-2+2*delta)*(w+i2))*tau+Cr*i2*(-1+delta+Pn-Pr))*Cr/((4*delta-4)*tau+Cr*(-1+delta+Pn-Pr))^2)*(1-(Pn-Pr)/(1-delta))/(2*tau)-(Pr-w-Crm)*(1/2+(((-2*delta+2)*tau^2+((-Pn+Pr-delta+1)*Cr+(-2+2*delta)*(w+i2))*tau+Cr*i2*(-1+delta+Pn-Pr))/((4*delta-4)*tau+Cr*(-1+delta+Pn-Pr))-i2)/(2*tau))/(1-delta)+Ce*rho0/(1-delta) = 0

 

 (4)
> 

K4 := (Pn-Cn)/(1-delta)+(1/2+(((-2*delta+2)*tau^2+((-Pn+Pr-delta+1)*Cr+(-2+2*delta)*(w+i2))*tau+Cr*i2*(-1+delta+Pn-Pr))/((4*delta-4)*tau+Cr*(-1+delta+Pn-Pr))-i2)/(2*tau))*(1-(Pn-Pr)/(1-delta))+(Pr-w-Crm)*((-Cr*i2+Cr*tau)/((4*delta-4)*tau+Cr*(-1+delta+Pn-Pr))+((-2*delta+2)*tau^2+((-Pn+Pr-delta+1)*Cr+(-2+2*delta)*(w+i2))*tau+Cr*i2*(-1+delta+Pn-Pr))*Cr/((4*delta-4)*tau+Cr*(-1+delta+Pn-Pr))^2)*(1-(Pn-Pr)/(1-delta))/(2*tau)+(Pr-w-Crm)*(1/2+(((-2*delta+2)*tau^2+((-Pn+Pr-delta+1)*Cr+(-2+2*delta)*(w+i2))*tau+Cr*i2*(-1+delta+Pn-Pr))/((4*delta-4)*tau+Cr*(-1+delta+Pn-Pr))-i2)/(2*tau))/(1-delta)-Ce*rho0/(1-delta) = 0

 

 (5)
> 

sol := solve({K2, K3, K4}, {Pn, Pr, w})

 

 (6)

Download Q_solve.mw

How to Simultaneously solve?...

Andiguys 85
September 22 2025
0 0

@acer I am optimizing a profit function with respect to three decision variables Pn,Pr, and w. When I solve the problem, I obtain conditions such as Pn(Pr,w), Pr​(Pn​,w), and w(Pn,Pr), which makes them appear dependent on each other. How can I simultaneuosly solve K2,K3 and K4 so that I can obtain the optimal values of Pn, Pr​, and w simultaneously as independent decision variables?

 

> 

restart
> 

``
> 

Pi1 := proc (Pn, Pr, w) options operator, arrow; (Pn-Cn)*(1-(Pn-Pr)/(1-delta))+(Pr-w-Cr)*((Pn-Pr)/(1-delta)-(-beta*(-(1/2)*tau+(1/2)*i2+(1/2)*w)*upsilon+Pr)/delta)-Ce*rho0*(1-(Pn-Pr)/(1-delta)) end proc

proc (Pn, Pr, w) options operator, arrow; (Pn-Cn)*(1-(Pn-Pr)/(1-delta))+(Pr-w-Cr)*((Pn-Pr)/(1-delta)-(-beta*(-(1/2)*tau+(1/2)*i2+(1/2)*w)*upsilon+Pr)/delta)-Ce*rho0*(1-(Pn-Pr)/(1-delta)) end proc

 (1)
> 

K2 := diff(Pi1(Pn, Pr, w), w) = 0

-(Pn-Pr)/(1-delta)+(-beta*(-(1/2)*tau+(1/2)*i2+(1/2)*w)*upsilon+Pr)/delta+(1/2)*(Pr-w-Cr)*beta*upsilon/delta = 0

 (2)
> 

diff(Pi1(Pn, Pr, w), `$`(w, 2))

-beta*upsilon/delta

 (3)
> 

solve(K2, w)

-(1/2)*(Cr*beta*delta*upsilon-Pr*beta*delta*upsilon+beta*delta*i2*upsilon-beta*delta*tau*upsilon-Cr*beta*upsilon+Pr*beta*upsilon-beta*i2*upsilon+beta*tau*upsilon-2*Pn*delta+2*Pr)/((-1+delta)*beta*upsilon)

 (4)
> 

K3 := diff(Pi1(Pn, Pr, w), Pn) = 0

1-(Pn-Pr)/(1-delta)-(Pn-Cn)/(1-delta)+(Pr-w-Cr)/(1-delta)+Ce*rho0/(1-delta) = 0

 (5)
> 

diff(Pi1(Pn, Pr, w), `$`(Pn, 2))

-2/(1-delta)

 (6)
> 

solve(K3, Pn)

(1/2)*Ce*rho0+(1/2)*Cn-(1/2)*Cr+Pr-(1/2)*delta-(1/2)*w+1/2

 (7)
> 

K4 := diff(Pi1(Pn, Pr, w), Pr) = 0

(Pn-Cn)/(1-delta)+(Pn-Pr)/(1-delta)-(-beta*(-(1/2)*tau+(1/2)*i2+(1/2)*w)*upsilon+Pr)/delta+(Pr-w-Cr)*(-1/(1-delta)-1/delta)-Ce*rho0/(1-delta) = 0

 (8)
> 

diff(Pi1(Pn, Pr, w), `$`(Pr, 2))

-2/(1-delta)-2/delta

 (9)
> 

solve(K4, Pr)

-(1/4)*beta*delta*i2*upsilon+(1/4)*beta*delta*tau*upsilon-(1/4)*beta*delta*upsilon*w-(1/2)*Ce*rho0*delta+(1/4)*beta*upsilon*i2-(1/4)*beta*upsilon*tau+(1/4)*beta*upsilon*w-(1/2)*delta*Cn+delta*Pn+(1/2)*Cr+(1/2)*w

 (10)
> 

``
> 

``
> 

``
> 

> 

``


 

Download Simultaneously_solve.mw

Follow-up...

Andiguys 85
August 12 2025
0 0

@acer
I had shared the complete details of the substitutions and their explanations in my message last week. Could you please confirm if you were able to review that? I can resend the details if needed. Thankyou. 

Got it...

Andiguys 85
August 11 2025
0 0

@sand15 Thankyou so much.

Still showing the same...

Andiguys 85
August 11 2025
0 0

@sand15 The z-axis values are still shown only up to zero—how can I extend it to display positive values? Also, how can I display the z-value at point P?

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