How to represent this problem using general linear programming? I have to represent this problem using general linear programming, which is not a problem in itself, but I can't figure out what would the $x_1, x_2,...,x_n$ be. A company is trying to minimize their cost for storing merchandise, and they know exactly how much extra space they'll need over the next 5 months.
$$\begin{array}{|c|c|}
\hline 
 \text{Month} & \text{Additional space required (sqr. ft)} \\ \hline
\ 1 & \ 30,000 \\ \hline
\text{2} & \ 20,000\\ \hline
\text{3} & \ 40,000\\ \hline
\text{4} & \ 10,000\\ \hline
\text{5} & \ 50,000\\ \hline
\end{array}$$
The company can either rent space monthly, or for have a special rate for periods of 2+ months. The costs go as follows:
$$\begin{array}{|c|c|}
\hline 
 \text{Months} & \text{Cost (\$/sqr. ft)} \\ \hline
\ 1 & \ 65 \\ \hline
\text{2} & \ 100\\ \hline
\text{3} & \ 135\\ \hline
\text{4} & \ 160\\ \hline
\text{5} & \ 190\\ \hline
\end{array}$$
The company has to minimize their cost while ensuring they have the space required. 
I originally thought that I would have 5 $x_n$ variables ($x_1$ to $x_5$) and that each $x_n$ would be the amount of sqr. ft. for each rate period and that my mimizing function would look like:
$$ min \ Z = 65x_1 + 100x_2 + 135x_3 + 160x_4 + 190x_5$$
but that wouldn't make sense, because if space is rented using a 2 months rate, that space can't be rented the next month. So I'm lost. What am I missing? What would be my $x_n$ and my constraints?
 A: The options that are available can be enumerated as follows:
denote $x_{ni}$ as an additional area contract occurred at $n$th next month which is durable for $i$ periods. Then all of the available options are
$$\begin{align}&x_{11},x_{12},x_{13},x_{14},x_{15}\\
&x_{21},x_{22},x_{23},x_{24}\\
&x_{31},x_{32},x_{33}\\
&x_{41},x_{42}\\
&x_{51}
\end{align}$$
and the following is the equations that represents the extra space needed for the company.
$$\begin{align}
&x_{11} + x_{12} + x_{13} + x_{14} + x_{15} = 30000\\
&x_{21} + x_{22} + x_{23} + x_{24} + x_{12} + x_{13} + x_{14} + x_{15} = 20000\\
&x_{31} + x_{32} + x_{33} + x_{22} + x_{23} + x_{24}+ x_{13} + x_{14} + x_{15} = 40000\\
&x_{41} + x_{42} + x_{32} + x_{33} + x_{24} + x_{14} + x_{15} = 10000\\
&x_{51} + x_{42} + x_{33} + x_{24} + x_{15} = 50000
\end{align}$$
Therefore, the cost you are minimizing can be written as
$$\begin{align}
Z &= 65(x_{11} + x_{21} + x_{31} + x_{41} + x_{51}) \\
&+ 100(x_{12} + x_{22} + x_{32} + x_{42})\\
&+ 135(x_{13} + x_{23} + x_{33})\\
&+ 160(x_{14} + x_{24})\\
&+ 190x_{15}
\end{align}
$$
