0
$\begingroup$

I have define the following functions:

w[k_] := k*0.8 + (1 - k)*0.3

b[k_] := 0.3*Sqrt[k]

and I want to solve the following system of equations

optimal = {0.5*x^(-1/2) == w[k]/(p + sP), D[b[k], k]/D[w[k], k] == x*(1 - r)/r}

by calling the Solve function

Solve[optimal, {x, k}, PositiveReals]

However, $p>0$, $sP>0$ and $r\in[0,1]$. So, my question is how to tell mathematica to account for these constraints on the value of coefficients?

$\endgroup$
  • $\begingroup$ It does not answer your question but if your are using Solve, I would suggest adding \\ Rationalize after the definition of optimal. $\endgroup$ – anderstood Oct 28 at 14:47
  • $\begingroup$ Also, the condition can be simplified. If $p$ and $sP$ are positive, since $x^(-1/2)>0$, w[k] has to be positive too, Reduce[w[k] > 0 // Rationalize] returns k > -(3/5). $\endgroup$ – anderstood Oct 28 at 14:51
  • $\begingroup$ w[k] is always positive, since $k\in[0,1]$ $\endgroup$ – Yorgos Oct 28 at 14:54
  • 2
    $\begingroup$ w[k_] = k*0.8 + (1 - k) 0.3 // Rationalize; b[k_] = Rationalize@0.3*Sqrt[k] // Rationalize; optimal = {0.5*x^(-1/2) == w[k]/(p + sP), D[b[k], k]/D[w[k], k] == x*(1 - r)/r, x > 0, k > 0, p > 0, sp > 0, 0 < r < 1} // Rationalize; Reduce[optimal, {x, k}, Reals] $\endgroup$ – cvgmt Oct 28 at 15:10

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.