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I am working with the optimization problem from the paper, eq.(5)

$$\max_{X=(x_1, x_2, \ldots, x_{n+1})} f(X)=(A-B\sum_{i=1}^n \frac{1}{x_i})\times x_{n+1}$$ subject to $$x_{n+1}=1-2k\sum_{i=1}^n x_i,$$ $$x_i \geq 0, \quad i = 1,2,\ldots, n+1.$$

Here $A \gg B > 0 \in \mathbb{R}$, $k \in \mathbb{Z}^+$, and $f(X)>1$.

From the paper I know the stationary point $$X^*=(\underbrace{\sqrt{\frac{B}{2kA}}, \ldots, \sqrt{\frac{B}{2kA}}}_{n\text{ times}}, 1 − 2 kn\sqrt{\frac{B}{2kA}} )$$ and the optimal value $$f(X^*)=(\sqrt{A} − n \sqrt{ 2 \cdot k \cdot B})^2.$$

Question. How to find the stationary point under constraints analytically? Is it possible for $n=3, k=10$ case?

Attempt.

I have tried to use the Lagrange multiplier:

$$F(X, \lambda) = x_{n+1}(A-B\sum \frac{1}{x_i}) + \lambda(x_{n+1} - 1 + 2k\sum x_i)=0$$ and found the partical derivatives and have the $(n+2)$ system with $n+2$ variables:

\begin{cases} F'_{x_i}(X, \lambda)= x_{n+1}\frac{B}{x_i^2} +2 \lambda k x_i=0, \quad i=1,2,..., n, \\ F'_{x_{n+1}}(X, \lambda)= A - B\sum \frac{1}{x_i}+\lambda=0, \\ F'_{\lambda}(X, \lambda)=x_{n+1} -1+2k\sum x_i =0. \end{cases}

My problem now is how to express $x_i$, $i=1,2,..., n$, and $x_{n+1}$ through $\lambda$ and find roots.

I know the stationary point and have found the A & Q. I will use the notation $\sum_{i=1}^n x_i := n \cdot x$ $\sum_{i=1}^n \frac{1}{x_i} := \frac{n} {x}$, and $x_{n+1}:=y$. Then the system will be

$$y\frac{B}{x^2}+2\lambda k x=0, \tag{2.1}$$ $$\lambda=B\frac{n}{x}-A, \tag{2.2}$$ $$y=1-2knx. \tag{2.3}$$

Put $(2.2)$ and $(2.3)$ in $(1.1)$:

$$( 1-2knx )\frac{B}{x^2}+2 (B\frac{n}{x}-A ) k x=0, \tag{3.1}$$ Multiple both sides $(3.1)$ on $x^2$:

$$( 1-2knx )B+2 (B\frac{n}{x}-A ) k x^3=0, \tag{4.1}$$ Open brackets and collect tems: $$2kAx^3-2nBkx^2+2nBkx-B=0, \tag{5.1}$$ divide both sides on $2kA$: $$x^3 - n\frac{B}{A}x^2 + n \frac{B}{A}x-\frac{1}{2k}\frac{B}{A}=0. \tag{6.1}$$

One can see the equation of power $3$, I am looking for a root $x \in \mathbb{R}$. I think the equation $(6.1)$ should has a simple real root and complex pair.

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  • $\begingroup$ @ChristianBlatter, I have added some details $A \gg B > 0 \in \mathbb{R}$, $k \in \mathbb{Z}^+$, and $f(X)>1$. $\endgroup$ – Nick yesterday

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