I want to maximize a function with respect to tau1, tau2, and alpha, and it is like the following:

V + β1 τ1 - α τ1^2 - β2 τ1^2 + \τ2 - 2 τ2^2 + α τ2^2

I know I can try solve for the partial derivatives with respect to the 3 variables and set them equal to 0, but alpha is linear in this function, and when I tried the following code, it returns a weird solution.

Simplify[Solve[D[π1 , τ1] == 0 && D[π1, τ2] == 0, D[π1, α] == 0, {τ1, τ2, α }]]

enter image description here

Since this function involves symbolic variable, V, and beta, I know I cannot go to the numerical route to find maximum, so what should I do then?

  • $\begingroup$ There are syntax errors in your code. You put a comma in front of the last equation instead of &&. $\endgroup$ – Sjoerd Smit May 30 at 14:23
  • $\begingroup$ i see thanks! So, it still makes sense to solve for alpha == 0 even thought the partial derivative of the pi1 is linear in alpha? $\endgroup$ – lll May 30 at 14:35
  • $\begingroup$ are all variables ($\alpha, \tau_1, \tau_2$) and parameters $\beta_1$ and $\beta_2$ positive or can they take negative values? $\endgroup$ – kglr May 30 at 15:49
  • $\begingroup$ yes, all these variables are positive $\endgroup$ – lll May 30 at 15:53

Looking at and thinking about, one finds

ClearAll[\[Alpha], \[Tau]1, \[Tau]2]; Maximize[V + \[Beta]1 \[Tau]1 - \[Alpha] \[Tau]1^2 - 
\[Beta]2 \[Tau]1^2 + \\[Tau]2 -2 \[Tau]2^2 + \[Alpha] \[Tau]2^2 /. {\[Tau]1 -> 
0, \[Alpha] -> \[Tau]2}, \[Tau]2]


Addition. Taking into account the comments of kglr and III done after my answer (French call such behavior l'esprit d'escalier), the same result is obtained by

ClearAll[\[Alpha], \[Tau]1, \[Tau]2]; Maximize[{V + \[Beta]1 \[Tau]1 \- \[Alpha] \[Tau]1^2 - \[Beta]2 \[Tau]1^2 + \[Tau]2 - 
2 \[Tau]2^2 + \[Alpha] \[Tau]2^2, \[Beta]1 >= 0 && \[Beta]2 >= 0 &&
 V >= 0} /. {\[Tau]1 -> 1/100, \[Alpha] -> \[Tau]2}, \[Tau]2]

$$\left\{ \begin{array}{cc} \{ & \begin{array}{cc} -\infty & \neg (\text{$\beta $2}\geq 0\land \text{$\beta $1}\geq 0\land V\geq 0) \\ \infty & \text{True} \\ \end{array} \\ \end{array} ,\{\text{$\tau $2}\to \text{Indeterminate}\}\right\} $$


Your Answer

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

Not the answer you're looking for? Browse other questions tagged or ask your own question.