A chemical company is designing a plant for producing two types of polymers, $P_1$ and $P_2$ . The plant must be capable of producing at least 100 units of $P_1$ and 420 units of $P_2$ each day.

There are two possible designs for the basic reaction chambers that are to be included in the plant. Each chamber of type A costs $600.000 and is capable of producing 10 units of$ P_1 $and 20 units of$ P_2$per day; type B is of cheaper design, costing$300.000, and is capable of producing 4 units of $P_1$ and 30 units of $P_2$ per day.

Because of operating costs, it is necessary to have at least four chambers of each type in the pant. How many chambers of each type should be included to minimize the cost of construction and still meet the required production schedule? (Assume that a minimum cost exists.)

Question

A chemical company is designing a plant for producing two types of polymers, $P_1$ and $P_2$ . The plant must be capable of producing at least 100 units of $P_1$ and 420 units of $P_2$ each day.

There are two possible designs for the basic reaction chambers that are to be included in the plant. Each chamber of type A costs $600.000 and is capable of producing 10 units of$ P_1 $and 20 units of$ P_2$per day; type B is of cheaper design, costing$300.000, and is capable of producing 4 units of $P_1$ and 30 units of $P_2$ per day.

Because of operating costs, it is necessary to have at least four chambers of each type in the pant. How many chambers of each type should be included to minimize the cost of construction and still meet the required production schedule? (Assume that a minimum cost exists.)

Quizlet Admin · Accepted Answer

We have a plant in a chemical company is used for producing $520$ of polymer $P_{1}$ and $330$ of polymer $P_{2}$ per day at least. This plant contains two types of chambers,
$A$ is capable for producing $40$ of $p_{1}$ and $20$ units of $P_{2}$ per day, and $B$ is capable for producing $40$ of $p_{1}$ and $30$ units of $P_{2}$ per day, 
This plant is necessary to be had five chambers of each type at least and the cost of this plant is being obtained from the objective function

$$
\begin{gather*}
C = 300,000 \ A + 400,000 \ B 	ag{1}
\end{gather*}
$$

and we can set its system of inequality that describes the possible combinations as the following :

$$
\begin{gather*}
40 A + 40 B \geq 520 \
20 A + 30 B \geq 330 \
\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, A \geq 5 \
\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, B \geq 5
\end{gather*}
$$

and we have to minimize this objective function as the following technique :

For all constraints, We have to replace the inequality symbol, $(\geq)$, by the equality sign only, $(=)$, then we have

$$
\begin{gather*}
40 A + 40 B = 520 \
20 A + 30 B = 330 \
\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, A = 5 \
\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, B = 5 
\end{gather*}
$$

After that, we have to graph these constraints as shown in the figure below.

After that, we have to define the solution's region for each constraint as the following :

For the first constraint $40 A + 40 B \geq 520$, we have to substitute with the the point $(0 \ , \ 0)$ (as an example) as

$$
\begin{gather*}
40 	imes 0 + 40 	imes 0 \geq 520 \
0 \geq 520 \,\,\, 	ext{(is not true)}
\end{gather*}
$$

Then the point $(0 \ , \ 0)$ is not a solution for this inequality. Since the point $(0 \ , \ 0)$ is located below the line $40 A + 40 B = 520$, 
then its solution's region is located above it.

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