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I have the following optimization problem:

minimize $f(x)=\dfrac{a}{x} + \dfrac{b}{d-c x} + f$

s.t $x\leq k$

Where $a,b,c,d,f>0$

Here is my approach using Lagrange multiplier: $L=f(x)+\lambda (x-k)$

(1)$\dfrac{\partial{L}}{\partial{x}}=-\dfrac{a}{x^2}+\dfrac{bc}{(d-cx)^2} + \lambda=0$

(2)$\dfrac{\partial{L}}{\partial{\lambda}}=x-k=0$

(3)$\lambda=\dfrac{a}{k^2}-\dfrac{b}{(d-ck)^2}$

I am stuck here. Replacing (3) in (1) yields an equation with degree 4. Is there any easier way to solve this. Need help!

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    $\begingroup$ How do you know a solution exists? If ${d \over c} \le k$, then $f$ is unbounded below on the feasible set. $\endgroup$ – copper.hat Dec 29 '16 at 6:09
  • $\begingroup$ $f(x)$ is convex within $(0,d/c)$. So there will be a solution $sol$ within $(0,d/c)$. Now can I say the solution is $min(sol,k)$ provided $k>0$? $\endgroup$ – usr109876787 Dec 29 '16 at 6:59
  • $\begingroup$ but from (2) we get $$x=k$$? $\endgroup$ – Dr. Sonnhard Graubner Dec 29 '16 at 7:58
  • $\begingroup$ Then will it be the suboptimal solution? $\endgroup$ – usr109876787 Dec 29 '16 at 8:15

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