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The conditions are: $w$ is a known value, and $x_{11} >0, x_{12}>0, ..., x_{nn}>0;$

\begin{equation} x_{11} \leq w \end{equation}

\begin{equation}x_{12} + x_{22} \leq w\end{equation}

\begin{equation}x_{13} + x_{23} + x_{33} \leq w\end{equation}

\begin{equation} ...\end{equation}

\begin{equation}x_{1n} + x_{2n} +... + x_{nn} \leq w\end{equation}

How can I choose $x_{ij}$ to minimize {R} and make these expressions be true?

\begin{equation} \frac{1}{x_{11}} + \frac{1}{x_{12}} + \frac{1}{x_{13}} + ... \frac{1}{x_{1n}} = R;\end{equation}

\begin{equation} \frac{1}{x_{22}} + \frac{1}{x_{23}} + ... \frac{1}{x_{2n}} = R;\end{equation}

\begin{equation}...\end{equation}

\begin{equation} \frac{1}{x_{nn}} = R;\end{equation}

The question is from a wire width issue: In a rectangle (width is w) have n wires, length are 1, 2, 3, 4 ... n I want they have same resistance, and the resistance is minimum. Their width can be changed in each step. So I need choose each line's width (x11, x12, ... ). I think a number computation way maybe can solve the problem, but have no idea about how to do it. enter image description here

Thanks

user3007260
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  • Please take some time to explain more about the context, and what have you tried so far. What are you trying to achieve actually? – Umberto Jan 26 '14 at 16:30
  • I tried lingo, it works and give result, but I want to know how to resolve it. – user3007260 Jan 27 '14 at 16:27

1 Answers1

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Let's look at the longest constraint with its equality: $$\begin{align}x_{1n}+x_{2n}+\cdots+x_{nn}\le w\\ \frac{1}{x_{1n}}+\frac{1}{x_{2n}}+\cdots+\frac{1}{x_{nn}}=R\end{align}$$ where $x_{ij}>0$ for all $i,j$. The objective is to minimize $R$. Taking the partial derivative w.r.t. $x_{in}$ we have $$\frac{\partial}{\partial x_{in}}R=-x_{in}^{-2}.$$ This cannot equal 0, so there is no finite stationary point, and the minimum is achieved for the maximum possible value of $x_{in}$. In other words, the constraint will be active: $$x_{1n}+x_{2n}+\cdots+x_{nn}=w.$$ Assume we have some $x_{in}<x_{jn}$ in that equality. Then $$-x_{in}^{-2}>-x_{jn}^{-2},$$ which means by increasing $x_{in}$ the value of $R$ changes more rapidly per unit than by decreasing $x_{jn}$, and the lowest value of $R$ is achieved when $x_{1n}=x_{2n}=\cdots=x_{nn}$.

Thus $$x_{in}=\frac{w}{n}$$ and $$R=\frac{n^2}{w}.$$ You can argue similarly to above that $x_{ij}=x_{kj}$ for all $j$, hence all your $x$ values are now fixed now that you know $R$: $$x_{ij}=\frac{wj}{n^2}.$$

Jeff Snider
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  • Hi, Jeff Snider – user3007260 Jan 31 '14 at 09:43
  • Hi, Jeff Snider

    Thanks for your comments. I agree the conclustion: lower width will make R change rapid. But in this case constraint is a little complicate: the width constraint is in x direction, and resistor constraint is in y (other) direction. So when x1n+x2n+⋯+xnn≤w, other resistor constrain is 1/xjj+1/xj(j+1)+...+1/xjn=R.

     – user3007260 Jan 31 '14 at 10:06