0
$\begingroup$

Given constants $b_1$, $b_2$ such that $b_1 − e^{-b_2} ≥ 0$, maximise $f(x) =2\tan^{-1} x_1 + x_2$ subject to $x_1 + x_2 ≤ b_1$, $− \ln{x_2} ≤ b_2$, $x_1 ≥ 0$, $x_2 ≥ 0$.

I am trying to learn optimisation alone and I can't figure out this problem. I want to minimize $-f(x)$. My Lagrangian is

$L(x,\lambda,z) = 2\tan^{-1} x_1 + x_2 - \lambda_1(z_1+x_1+x_2-b_1)-\lambda_2(-\ln{x_2} -b_2+z_2)$

I got that $\lambda_1 \leq 0$ and I think $\lambda_2$ should be $0$ but I'm not sure. I also calculated the gradient and equaled it to $0$ but I'm not sure how to do the different cases.

I would appreciate some help.

$\endgroup$
0
$\begingroup$

Hint.

Considering that $-\ln x_2 \le b_2\equiv x_2 \ge e^{-b_2}$ for $x_2 \ge 0$

the restrictions can be transformed to

$$ g_1(x,b,\epsilon) = x_2-b_2-\epsilon^2\\ g_2(x,b,\epsilon)=b_1-b_2-\epsilon^2\\ g_3(x,b,\epsilon)=x_1+x_2-b_1+\epsilon^2\\ g_4(x,b,\epsilon)=x_1-\epsilon^2\\ g_5(x,b,\epsilon)=x_2-\epsilon^2\\ g_6(x,b,\epsilon)=b_2-\epsilon^2\\ $$

(here $b_2$ represents $e^{-b_2}$. Now with

$$ f(x) = 2\arctan x_1+x_2 $$

we have the Lagrangian

$$ L(x,b,\epsilon,\lambda) = f(x)+\sum_{k=1}^6 g(x,b,\epsilon_k) $$

so the stationary points are the solutions for

$$ \left\{ \begin{array}{rcl} \lambda_3+\lambda_4+\frac{2}{x_1^2+1}&=&0 \\ \lambda_1+\lambda_3+\lambda_5+1&=&0 \\ \lambda_2-\lambda_3&=&0 \\ -\lambda_1-\lambda_2+\lambda_6&=&0 \\ -\epsilon_1^2-b_2+x_2&=&0 \\ -\epsilon_2^2+b_1-b_2&=&0 \\ \epsilon_3^2-b_1+x_1+x_2&=&0 \\ x_1-\epsilon_4^2&=&0 \\ x_2-\epsilon_5^2&=&0 \\ b_2-\epsilon_6^2& =& 0 \\ -2 \epsilon_k \lambda_k & =& 0 \end{array} \right. $$

for $k = 1,\cdots,6$

| cite | improve this answer | |
$\endgroup$

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.