Dear all,

This is a question about efficient coding. In the code below, I write demands as integrals of piecewise functions (three pieces). I then need to find the price for which the derivative is null. In spite of my numerous trials, I didn't manage to make the code efficient enough for Mathematica to find the answer.

Do you see a way to make it more efficient, or do you think the problem in itself is too intricate to be solved?

Code:

Clear["Global`*"]
a = 1/2;
t = 4/10;

da = Integrate[Piecewise[{{1, pa + t*x < 1/4},
     {((3/4) - (pa + t*x))/((3/4) - (1/4)) + ((pa + t*x - 1/4) (t*
          x))/(2*((3/4) - (1/4))) + ((pa + t*x - 1/4) (1 - pb - 
          t*x))/((3/4) - (1/4)), 1/4 < pa + t*x && pb + t*x < 1},
     {((3/4) - (pa + t*x))/((3/4) - (1/4)) + (1 - pb)^2/(
       2*((3/4) - (1/4))), 1/4 < pa + t*x && pb + t*x > 1}}], {x, 0, 
    1}];

db = Integrate[Piecewise[{{(1 - pb), pa + t*x < 1/4},
     {((1 - pb) ((3/4) - (pa + t*x)))/((3/4) - (1/4)) + ((pa + t*x - 
          1/4) (t*x))/(
       2*((3/4) - (1/4))) + ((pa + t*x - 1/4) (1 - pb - t*x))/((3/
          4) - (1/4)), 1/4 < pa + t*x && pb + t*x < 1},
     {((1 - pb) ((3/4) - (pa + t*x)))/((3/4) - (1/4)) + (1 - pb)^2/(
       2*((3/4) - (1/4))), 1/4 < pa + t*x && pb + t*x > 1}}], {x, 0, 
    1}];

pi = da*pa + pb*db;

dbdev = Integrate[Piecewise[{{1 - pbdev, pa + t*x < 1/4},
     {((1 - pbdev) ((3/4) - (pa + pb + t*x - pbdev)))/((3/4) - (1/
          4)) + ((pb + t*x - pbdev) (pa + pb + t*x - pbdev - 1/4))/(
       2*((3/4) - (1/4))) + ((pa + pb + t*x - pbdev - 1/4) (1 - pb - 
          t*x))/((3/4) - (1/4)), 1/4 < pa + t*x && pb + t*x < 1},
     {((3/4 - (pa + pb + t*x - pbdev)) (1 - pbdev))/((3/4) - (1/
          4)) + (1 - pbdev)^2/(2*((3/4) - (1/4))), 
      1/4 < pa + t*x && pb + t*x > 1}}], {x, 0, 1}];

pidev = da*pa + pbdev*dbdev;

Solve[D[pidev, pbdev] == 0 && D[D[pidev, pbdev], pbdev] < 0, pbdev, Reals]

Thank you so much for your help !