I have a question on solving constrained maximization problem when Lagrangian method fails because of the linear parts of objective function or and constraints. I would really appreciate any thought!

The problem I had in mind:

Maximizing by choosing $c_t$, $i_t$ $$ max \sum_{t=1}^{\infty} \beta^{t-1}log(c_t) $$ such that $$\sum_{t=1}^{\infty} \frac{1}{1+r}^{t-1}c_t = \sum_{t=1}^{\infty} \frac{1}{1+r}^{t-1}\pi_t $$ $$\pi_t = zk_t - i_t$$ $$k_t = k_{t-1} + i_{t-1}$$ $$0 <= i_t <= zk_t$$

where $k_1 > 0$ is given

I noticed that, the constraint is linear in $i_t$, and the Lagrangian method fails since $i_t$ does not appear in the objective function. Hence, the gradients of objective and constraint will never be parallel.

In short, my question is:

Is there any general approach to solving maximization problems involving this issue. (i.e. the issue that gradients for objective and constraint are never equal because one of them is linear.)


  • $\begingroup$ why can't the gradient of the constraints be parallel to 0? $\endgroup$ – LinAlg Nov 9 '16 at 17:18
  • $\begingroup$ I was thinking that the gradient for objective is 0 for $i_t$ but non-zero for $c_t$, and the gradient for constraint is constant in both $i_t$ and $c_t$. Then Lagrangian multiplier can not be zero. $\endgroup$ – Matcha Nov 9 '16 at 21:16
  • $\begingroup$ each constraint has its own multiplier, and multipliers can be 0 $\endgroup$ – LinAlg Nov 9 '16 at 21:53
  • $\begingroup$ Thanks LinAlg for replies. I was referring to the summation constraint linear in $i_t$ and $c_t$. I'm sure if I'm allowed to separate it into two constraints. $\endgroup$ – Matcha Nov 10 '16 at 0:16

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.