Using Markov Process for this question assume that the space is finite.

$\displaystyle X_t\epsilon\begin{Bmatrix} 1,2,3,4,...,m \end{Bmatrix}$

The stochastic process which determines the future of the process from a probabilistic standpoint is defined as follows.

Let 

$\displaystyle T_i=\text{time spent in state }i$

$\displaystyle Pr{\begin{Bmatrix} T_i>t+s|T_i>s \end{Bmatrix}}=Pr\begin{Bmatrix} T_i>t \end{Bmatrix}$

Recall that the exponential distribution, and thus the $\displaystyle T_i$ could have the density function,

$\displaystyle F_i(t)=\lambda_ie^{-\lambda_it}$

Now formulate the Markov Process specifically for this problem.

$\displaystyle X_t\epsilon\begin{Bmatrix} 0,1,2,..., 32\end{Bmatrix}$

The transitions of computers in and out of the shop are,

$\displaystyle X_t=i$ to $\displaystyle X_t=i+1$ or $\displaystyle X_t=i-1$

Now calculate the rates up and down.

To calculate $\displaystyle \lambda$ take the total number of computers fixed last year and divide it by the total months in the year. To calculate $\displaystyle \mu$ assume there are 22 working days in a month and it takes on average, two days to fix a computer.

$\displaystyle \\\lambda=5.9 \\\mu=7.3$

Its key to know that at zero we cannot move down a state and like wise when at 29, we cannot move up a state.

$\displaystyle [\text{Rate In}]=[\text{Rate Out}]$

$\displaystyle \\\lambda P_0=\mu P_1 \\(\mu +\lambda)P_1=\lambda P_0+\mu P_1 \\(\mu +\lambda)P_2=\lambda P_2+\mu P_3\\ \begin{matrix} .\\ .\\ . \end{matrix} \\\mu P_{32}=\lambda P_{31}$

Solving with 

$\displaystyle \sum P_i=1$

$\displaystyle EX_t=\sum iP_i$

Continuing in this fashion results in the following.

$\displaystyle \\P_n=(\lambda / \mu)P_{n-1} \\P_n=(\lambda / \mu)^nP_0 \\\\\sum_{n=0}^{32}P_n=P_0\sum_{n=0}^{32}\left(\frac{\lambda}{\mu} \right )^n=1 \\\\P_0=\frac{1-\rho }{1-\rho^{33}}$

Now, writing the sum of a finite geometric series is

$\displaystyle P_n=\rho^nP_0=\frac{\rho^n(1-\rho)}{1-\rho^{30}}$

Now 

$\displaystyle \rho=\frac{\lambda}{\mu}=\frac{5.9}{7.3}=0.808$

therefore,

$\displaystyle 1-\rho^{33}=1-0.808^{33}\approx 0.9991198131$

Next, 

$\displaystyle Pr\begin{Bmatrix} X_t>0 \end{Bmatrix}=1-P_0=\rho\approx 0.808$

$\displaystyle \\EX_t=\sum_{n=0}^{32}nP_n \\EX_t=\sum_{n=0}^{32}n\rho^n(1-\rho) \\EX_t\approx 4.176$

The objective of this particular problem is to maximize the profit.

Maximize the function:

$\displaystyle y=f(x_1,...,x_n)=c_1x_1+...+c_nx_n$

over the region defined by the constraints.

First identify the variables and assumptions of the problem.

$\displaystyle \\x_1=\text{acres of soy planted} \\x_2=\text{acres of corn planted} \\x_3=\text{acres of oats planted} \\l=\text{labor required} \\w=\text{water required} \\t=\text{total acreage planted} \\y=\text{total profit}$

$\displaystyle \\w=1.2x_1+3.4x_2+1.9x_3 \\l=0.1x_1+0.8x_2+0.3x_3 \\t=x_1+x_2+x_3 \\y=300x_1+550x_2+200x_3 \\w\leq 1200 \\l\leq 750 \\t\leq 575 \\x_1\geq0, x_2\geq0, x_3\geq 0$ 

Now using the constraints formulate the linear programming model.

$\displaystyle \\1.2x_1+3.4x_2+1.9x_3\leq1200 \\0.1x_1+0.8x_2+0.3x_3\leq750 \\x_1+x_2+x_3\leq575 \\x_1\geq0, x_2\geq0, x_3\geq 0$

From here use computer implementation to solve using the simplex method.

$\displaystyle \\x_1=343.182 \\x_2=231.818 \\x_3=0$

Therefore, the optimal solution will occur when there are 343.182 acres of soy planted and 231.818 acres of corn planted. 

The maximum profit at this solution will be, $230,454.50.