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A chemical company is designing a plant for producing two types of polymers, P1\mathrm{P}_{1} and P2.\mathrm{P}_{2}. The plant must be capable of producing at least 520 units of P1\mathrm{P}_{1} and 330 units of P2\mathrm{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 $300,000 and is capable of producing 40 units of$ P\mathrm{P}{1}and20unitsofand 20 units of P\mathrm{P}{2}perday;typeBisamoreexpensivedesign,costing$400,000,andiscapableofproducing40unitsofper day; type B is a more expensive design, costing \$400,000, and is capable of producing 40 units of P\mathrm{P}{1}and30unitsofand 30 units of P\mathrm{P}{2}$ per day. Because of operating costs, it is necessary to have at least five chambers of each type in the plant. How many chambers of each type should be included to minimize the cost of construction and still meet the required production schedule?

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We have a plant in a chemical company is used for producing 520520 of polymer P1P_{1} and 330330 of polymer P2P_{2} per day at least. This plant contains two types of chambers, AA is capable for producing 4040 of p1p_{1} and 2020 units of P2P_{2} per day, and BB is capable for producing 4040 of p1p_{1} and 3030 units of P2P_{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

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

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

40A+40B52020A+30B330A5B5\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

40A+40B=52020A+30B=330A=5B=5\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 40A+40B52040 A + 40 B \geq 520, we have to substitute with the the point (0 , 0)(0 \ , \ 0) (as an example) as

40×0+40×05200520(is not true)\begin{gather*} 40 \times 0 + 40 \times 0 \geq 520 \\ 0 \geq 520 \,\,\, \text{(is not true)} \end{gather*}

Then the point (0 , 0)(0 \ , \ 0) is not a solution for this inequality. Since the point (0 , 0)(0 \ , \ 0) is located below the line 40A+40B=52040 A + 40 B = 520, then its solution's region is located above it.

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