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Mathematik-Online lexicon:

Farmer


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A farmer employs 25 workers and wishes to optimize the cattle breading on his 120 hectare of meadland. He assumes that 10 sheep require 1 hectare and 10 cows 3 hectare of land. Moreover, he calculates 1 worker per 40 sheep and per 20 cows, respectively. His profit is 100 EUR per sheep and 250 EUR per cow. How should the farmer partition his meadowland, if he takes into account that the state pays a premium of 500 EUR per hectare for land set-aside?

The mathematical formulation of this problem is

$\displaystyle S$ $\displaystyle \geq$ 0  
$\displaystyle K$ $\displaystyle \geq$ 0  
$\displaystyle \frac{1}{10} S + \frac{3}{10} K$ $\displaystyle \leq$ $\displaystyle 120$  
$\displaystyle \frac{1}{40} S + \frac{1}{20} K$ $\displaystyle \leq$ $\displaystyle 25$  
$\displaystyle \frac{1}{10} S + \frac{3}{10} K + B$ $\displaystyle =$ $\displaystyle 120$  
$\displaystyle 100 S + 250 K + 500 B$ $\displaystyle \to$ $\displaystyle \max$  

Eliminating $ B$ with the aid of the second equation and rewriting the inequalities as well as the target function yields
$\displaystyle - S - 3 K$ $\displaystyle \geq$ $\displaystyle - 1200$  
$\displaystyle S$ $\displaystyle \geq$ 0  
$\displaystyle K$ $\displaystyle \geq$ 0  
$\displaystyle - S - 2 K$ $\displaystyle \geq$ $\displaystyle - 1000$  
$\displaystyle 50 S + 100 K + 60\,000$ $\displaystyle \to$ $\displaystyle \max
\,,$  

i.e., a linear program $ c^{\text{t}} x \to \min, Ax \geq b$ with

$\displaystyle c = \begin{pmatrix}- 50 \\ - 100 \end{pmatrix},\quad
A = \begin{...
...{pmatrix},\quad
b = \begin{pmatrix}0\\ 0\\ -1200 \\ -100 \end{pmatrix}
\,.
$

\includegraphics[width=0.5\linewidth]{bsp_lineares_programm1}

In the figure, the admissible set is bounded by

$\displaystyle S = 0,\quad K =0,\quad - S - 3 K = - 1200,\quad -S - 2K = - 1000
\,,
$

and the target function $ c^{\text{t}} x$ increases in the direction of the gradient $ c$. Since $ c$ is perpendicular to the line through $ (600,200)$ and $ (1000,0)$, the entire segment from $ (600,200)$ to $ (1000,0)$ is optimal. Hence, the farmer maximizes his profit, if all workers are needed and if he keeps at least 600 sheep.

see also:


  automatisch erstellt am 26.  1. 2017