# Can we get closed form solution for such a problem?

\begin{align}\min&\quad\sum_{i=1}^N\frac{A_i}{x_i}\\\text{s.t.}&\quad\sum x_i \le X\\&\quad x_i \ge 0\end{align}

wherein $$A_i>0, (i\in\{1,\dots,N\})$$ is constant, $$x_i, (i\in\{1,\dots,N\})$$ is a continuous optimization variable.

If some $$A_i$$ is negative the problem is unbounded: we can make the objective arbitrarily small by making $$x_i$$ arbitrarily close to 0.

Assuming $$A_i \geq 0$$, the optimality is obtained when all $$\frac{A_i}{x_i^2}$$ are equal (KKT optimality conditions), or equivalently all $$\frac{x_i^2}{A_i}$$ are equal, and $$\sum x_i = X$$ (otherwise you can increase some $$x_i$$ and reduce the objective).

Stating $$x_i = \sqrt{A_i} y$$, you can deduce $$y = \frac{X}{\sum \sqrt{A_i}}$$ from the equality.

Therefore $$x_i = \frac{\sqrt{A_i}}{\sum \sqrt{A_j}}X$$

• This does not work if some $A_i=0$. Jul 20 '20 at 16:53
• Yes. I was too lazy to spell it out, but you need an $A_i \gt 0$ for the equality to hold. Thanks Jul 20 '20 at 17:02
• (As far as I can tell you just need one, though. And if all are zero the problem is trivial) Jul 20 '20 at 17:05
• In this case all $x_i$ where $A_i=0$ must be 0, just like the general formula Jul 20 '20 at 17:10
• No, I mean if even one $A_i=0$ your argument that all $A_i/x_i^2$ are equal fails. Furthermore, you cannot have $x_i=0$ ever because it appears in the denominator of a summand in the objective. Jul 20 '20 at 17:53