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Mathematics-Online problems:

Solution to the problem of the (previous) week

Problem:

The following table shows the freight charges per unit of crude oil deliveries.

Kuwait Caracas Anchorage
Houston $ 4$ $ 1$ $ 2$
Rotterdam $ 2$ $ 3$ $ 4$

According to the table below calculate the delivery quantities $ q$ based on $ s$ units delivered (Kuwait $ \longrightarrow$ Houston) and $ t$ units delivered (Anchorage $ \longrightarrow$ Rotterdam). The quantities produced are in the top row and the quantities ordered are in the extreme left column of the table. Supply and demand are supposed to match, i.e. the numerical values on top of the columns or at the left of the rows show the sums of the columns or the rows, respectively.

Kuwait: $ 6$ Caracas: $ 4$ Anchorage: $ 5$
Houston: $ 7$ $ s$ $ q_{CH}$ $ q_{AH}$
Rotterdam: $ 8$ $ q_{KR}$ $ q_{CR}$ $ t$


Also determine the total freight charges $ c$ depending on the units $ s$ and $ t$ delivered. For which choice is $ c$ minimal? Indicate all solutions.


Answer:

$ q_{CH}$ = $ +$ $ s$ $ +$ $ t$
$ q_{AH}$ = $ +$ $ s$ $ +$ $ t$
$ q_{KR}$ = $ +$ $ s$ $ +$ $ t$
$ q_{CR}$ = $ +$ $ s$ $ +$ $ t$
$ c$ = $ +$ $ s$ $ +$ $ t$

$ c$ is minimal for $ \leq s \leq$ and $ \leq t \leq$

Solution:

Considering the quantities ordered and produced, we obtain:

Kuwait: $ 6$ Caracas: $ 4$ Anchorage: $ 5$
Houston: $ 7$ $ s$ $ 2-s+t$ $ 5-t$
Rotterdam: $ 8$ $ 6-s$ $ 2+s-t$ $ t$

Freight charges:

$\displaystyle c$ $\displaystyle =$ $\displaystyle 4s + 1(2-s+t) + 2(5-t) + 2(6-s) + 3(2+s-t) + 4t$  
  $\displaystyle =$ $\displaystyle 30 + 4s$  

Positivity: $ \Longrightarrow \quad s,t \ge 0, \quad t\le 2+s $

$ c$ is minimal for:     $ s=0,\quad t \in [0,2]$

[problem of the week]