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Find the function for a decision variable

Shafa Hananta
Posted 2 years ago

Hi, I'm currently trying to find the equation for a decision variable from a total profit equation using Mathematica. From what I've read, I can use the solve function to do this. Here's a example i found:

(*Define the total cost function*)
TC[Q_] := (D S/Q) + (H Q/2)

(*Compute the first derivative*)
dTCdQ = D[TC[Q], Q]

(*Solve for Q*)
solution = Solve[dTCdQ == 0, Q]

(*Simplify the solution*)
simplifiedSolution = Simplify[solution]

However, when I try to apply this to my actual equation, the program runs for hours without completing (it stuck). Here is my specific code:

TP[x_] = pr/x ((a (1 - b) (x^2))/(2 L) ((y g)/(1 + y g)))^(1/(
   1 - b)) - ((c + 1) (Kr/x + Kp/(n x) + Kf/(
      n x) + (hr + g)/
       x (2 x (-((a (-1 + b) g x^2 y)/(L + g L y)))^(1/(1 - b))) + (
       hp + g)/(2 R x) ((a (1 - b) (x^2))/(2 L) ((y g)/(1 + y g)))^(
       2/(1 - b)) + ((hp + g) (n - 1))/(
       2 x) ((a (1 - b) (x^2))/(2 L) ((y g)/(1 + y g)))^(2/(
       1 - b)) (x ((a (1 - b) (x^2))/(2 L) ((y g)/(1 + y g)))^(1/(
          1 - b)) - 1/R) + 
      1/x^2 S + (v/x + vp/(n x)) ((a (1 - b) (x^2))/(
         2 L) ((y g)/(1 + y g)))^(1/(1 - b)) + 
      1/(n x)^2 Sv + (Pv w)/( 
       x f ww) ((a (1 - b) (x^2))/(2 L) ((y g)/(1 + y g)))^(1/(
       1 - b)) + (((cf f) + (m bk))/( 
        x f ww ) {k n x + 
          k/q (Log[1 + z E^(-q n x)] - Log[1 + z])}) )) + (c ((pp/x + 
        pf/x) ((a (1 - b) (x^2))/(2 L) ((y g)/(1 + y g)))^(1/(
      1 - b)))) + (pr/x l (
     2 a g (x^2)
       y (((-((a (-1 + b) g x^2 y)/(L + g L y)))^(1/(1 - b)))^b) )/(
     L + g L y) - ((a (1 - b) (x^2))/(2 L) ((y g)/(1 + y g)))^(1/(
     1 - b)) (M - x))
(*Compute the first derivative*)
dTPdx = D[TP[x], x]

(*Solve for p*)
solution = Solve[dTPdx == 0, x]

(*Simplify the solution*)
simplifiedSolution = Simplify[solution]

Any suggestions on how to resolve this issue? Are there more efficient methods or best practices I should consider for solving complex equations in Mathematica?

Attachments:
coba2.nb
POSTED BY: Shafa Hananta
2 Replies
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Depending on the parameters, your equation has a variable number of solutions:

TP[x_] = 
 pr/x  ((a  (1 - b)  (x^2))/(2  L)  ((y  g)/(1 + y  g)))^(1/(1 - 
         b)) - ((c + 1)  (Kr/x + Kp/(n  x) + 
       Kf/(n  x) + (hr + g)/
         x  (2  x  (-((a  (-1 + b)  g  x^2  y)/(L + g  L  y)))^(1/(1 -
                b))) + (hp + 
           g)/(2  R  x)  ((a  (1 - b)  (x^2))/(2  L)  ((y  g)/(1 + 
               y  g)))^(2/(1 - b)) + ((hp + g)  (n - 
             1))/(2  x)  ((a  (1 - b)  (x^2))/(2  L)  ((y  g)/(1 + 
               y  g)))^(2/(1 - 
             b))  (x  ((a  (1 - b)  (x^2))/(2  L)  ((y  g)/(1 + 
                  y  g)))^(1/(1 - b)) - 1/R) + 
       1/x^2  S + (v/x + 
          vp/(n  x))  ((a  (1 - b)  (x^2))/(2  L)  ((y  g)/(1 + 
               y  g)))^(1/(1 - b)) + 
       1/(n  x)^2  Sv + (Pv  w)/(x  f  ww)  ((a  (1 - 
                b)  (x^2))/(2  L)  ((y  g)/(1 + y  g)))^(1/(1 - 
             b)) + (((cf  f) + (m  bk))/(x  f  ww)  (k  n  x + 
           k/q  (Log[1 + z  E^(-q  n  x)] - 
              Log[1 + z]))))) + (c  ((pp/x + 
         pf/x)  ((a  (1 - b)  (x^2))/(2  L)  ((y  g)/(1 + 
              y  g)))^(1/(1 - b)))) + (pr/
       x  l  (2  a  g  (x^2)  y  (((-((a  (-1 + b)  g  x^2  y)/(L + 
                   g  L  y)))^(1/(1 - b)))^b))/(L + 
         g  L  y) - ((a  (1 - b)  (x^2))/(2  L)  ((y  g)/(1 + 
             y  g)))^(1/(1 - b))  (M - x)) // Simplify
(*Compute the first derivative*)
dTPdx = D[TP[x], x] // Simplify;
effe = Block[{a = 1, b = 2, bk = 1, c = 1, cf = 4, f = 1, g = 1, 
   hp = 1, hr = 1, k = 1, Kf = 1, Kp = 1, Kr = 1, l = 1, L = 1, m = 1,
    M = 1, n = 1, pf = 1, pp = 1, pr = 1, Pv = 1, q = 1, R = 1, S = 1,
    Sv = 1, v = 1, vp = 1, w = 1, ww = 1, y = 1, z = 1},
  dTPdx]
Plot[effe, {x, -10, 10}, PlotRange -> {-1, 1}]
Solve[effe == 0, x, Reals]
gi = Block[{a = 1, b = 0, bk = 1, c = 1, cf = 4, f = 1, g = 1, hp = 1,
    hr = 1, k = 1, Kf = 1, Kp = 1, Kr = 1, l = 1, L = 1, m = 1, M = 1,
    n = 1, pf = 1, pp = 1, pr = 1, Pv = 1, q = 1, R = 1, S = 1, 
   Sv = 1, v = 1, vp = 1, w = 1, ww = 1, y = 1, z = 1},
  dTPdx]
Plot[gi, {x, -10, 10}, PlotRange -> {-1, 1}]
Solve[gi == 0, x, Reals]
acca = Block[{a = 4, b = 2, bk = 1, c = 1, cf = 4, f = 3, g = 2, 
   hp = 1, hr = 1, k = 1, Kf = 2, Kp = 1, Kr = 5, l = 3, L = 1, m = 1,
    M = 1, n = 1, pf = 4, pp = 1, pr = 1, Pv = 1, q = 1, R = 10, 
   S = 1, Sv = 1, v = 1, vp = 1, w = 1, ww = 1, y = 1, z = 2},
  dTPdx]
Plot[acca, {x, -10, 10}]
Solve[acca == 0, x, Reals]

You cannot expect a simple analytical solution for the general case.
POSTED BY: Gianluca Gorni
POSTED BY: Mariusz Iwaniuk
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