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Suppose $f:\mathbb{R}_+ \to \mathbb{R}_+$ is a strictly decreasing continuous function such that. Let $n$ be a natural number. I want to solve the following maximization problem $$ S_n = \max_{x_1 \leq \dots \leq x_n} x_1 f(x_1) + (x_2-x_1) f(x_2) + \dots + (x_n-x_{n-1}) f(x_n). $$ See the plot below for graphical illustration when $n=3$. I need to find two things:

  1. Characterization of optimal $(x_1,\dots,x_n)$
  2. Lower bound for the area covered.

This is a standard problem for example in economics (price discrimination) and it looks like someone probably has solved the problem, but I haven't been able to find it.

enter image description here

TomH
  • 695

1 Answers1

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It seems to me that you have already done most of the work, writing the problem as an optimization problem for multiple variable. In general calling $\theta$ the actual value of the area as a function of the different points chosen.

$\theta(x_1,x_2,...,x_N) = x_1\cdot f(x_1) + \sum_{I=2}^N{f(x_I)\cdot(x_I-x_{I-1})}$

And we want to find the maximum of theta, so we search the point with all partial derivatives equal to zero

$\begin{cases}\frac{\partial \theta }{\partial x_1} = f(x_1) + x_1 \cdot f^{'}(x_1) - 0 \cdot f^{'}(x_1) - f(x_2) \\ \frac{\partial \theta }{\partial x_2} = f(x_2) + x_2 \cdot f^{'}(x_2) - x_1 \cdot f^{'}(x_2) - f(x_3) \\.... \\ \frac{\partial \theta }{\partial x_N} = f(x_N) + x_N \cdot f^{'}(x_N) - x_{N-1} \cdot f^{'}(x_N) - 0 \end{cases}$

This in addition to the conditions

$ \begin{cases} x_J>x_I & \text{for $J>I$} \\ x_I > 0 & \forall I \end{cases} $

will give you a system of equation that you can solve for the value of $x_I$. The system may not always be solvable, depending on the definition of $f$. Also you should check that the chosen point is a maxima and not a minima.

Some examples: For $f= 1-x$ the solution is always to split in equal part. For $f= 1-x^2$ and $N=2$ the two points are $ \begin{cases} x_1 = \sqrt{\frac{1}{9-2\sqrt3}} \approx 0.425\\ x_2 = \sqrt{\frac{9}{9-2\sqrt3}} \approx 0.736 \end{cases}$

Maurizio De Leo
  • 143