I am very new to optimization andI have to solve this equation

max U(k)=tlog(1+ yhpg/(pg+s))+mpge^(-ky)) st k>0

can anyybody give me idea where should I start

  • $\begingroup$ This is hard to parse. is it $U(k)=t\log (1+\frac {yhpg}{pg+s})+mpge^{-ky}$? If so...is $k$ the only parameter you can vary? If so, then most of the expression is noise and you are just trying to optimize $e^{-ky}$. $\endgroup$ – lulu Jul 18 '17 at 1:20
  • $\begingroup$ The usual approach is to check the first-order and the second-order conditions. That is check whether there are points such that $U'(k)=0$ (so-called first order condition). Call a solution to $U'(k)=0$, $k^{\ast}$, and check if $U''(k^{\ast})$<0 (second-order condition). If the second-order condition is satisfied then $k^{\ast}$ is a local maximum. If $U''<0$ for all k then $k^{\ast}$ is the global maximum. $\endgroup$ – Mdoc Jul 18 '17 at 1:21
  • $\begingroup$ Yes you are right with equation , Thankyou! I will try as you have said $\endgroup$ – user7341333 Jul 18 '17 at 2:23

Notice that $\exp(-ky)$ is a monotone function.

If $y > 0$, it is a decreasing function.

If $y < 0$, it is an increasing function.

In either case, the maximum might not exist.


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