Answer :
We have a linear optimization problem.
We have two plants with different capacity for each type of fertilizer. Each hour, the plant can produce a mix of the products.
Each plant has an associated cost per hour of operation.
Let:
J: the number of operation hours of the Jonesboro plant
M: the number of operation hours of the Macon plant
We can write the total production for the LP fertilizer as:
[tex]\begin{gathered} 1\cdot J+1\cdot M\ge100 \\ J+M\ge100 \end{gathered}[/tex]Each plant produces one ton per hour, and the total production of both plants togheter has to be 100 tons or more.
For the MP fertilizer, we can similarly write:
[tex]5J+2M\ge260[/tex]For the HP fertilizer, we can write:
[tex]J+3M\ge180[/tex]We have the equations of the restrictions:
[tex]\begin{gathered} J+M\ge100 \\ 5J+2M\ge260 \\ J+3M\ge180 \end{gathered}[/tex]The objective function is the cost function, that have to be minimized:
[tex]1000J+600M[/tex]We can graph the restrictions as:
(J is written as x and M as y)
Now we can graph the objective function, and minimize satisfying all the restrictions:
We can see that the minimization happens when the MP and HP restrictions meet (both restrictions are "saturated").
Then, we can write:
[tex]\begin{gathered} 5J+2M=260 \\ J+3M=180\longrightarrow J=180-3M \\ \\ 5(180-3M)+2M=260 \\ 900-15M+2M=260 \\ 900-13M=260 \\ 13M=900-260 \\ 13M=640 \\ M=\frac{640}{13}\approx49.231 \\ \\ J=180-3M=180-3(49.231)=180-147.693=32.307 \end{gathered}[/tex]The cost is:
[tex]1000J+600M=1000\cdot32.307+600\cdot49.231=32,307+29,538.60=61,845.60[/tex]The combination that minimize the cost is:
Hours of Jonesboro plant: 32.307 hours
Hours of Macon plant: 49.231 hours.
Total cost: $ 61,845.60
