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  • Refer to Exercise 14. How many sweaters of each type should the company make in a day to get a maximum profit? What is the maximum profit?

Refer to Exercise 14. How many sweaters of each type should the company make in a day to get a maximum profit? What is the maximum profit?

Answers (1)

When we refer to the exercise, we get the following information:

Let’s take an assumption that the company manufactures x number of type A sweaters and y number of type B. We have made a table for your understanding which is given below:

\begin{array}{|l|l|l|l|} \hline & \begin{array}{l} \text { Type A Sweaters } \\ (\mathrm{X}) \end{array} & \begin{array}{l} \text { Type B Sweaters } \\ (\mathrm{Y}) \end{array} & \\ \hline \text { cost per day } & \text { Rs. } 360 & \text { Rs. } 120 & \text { Rs. } 72000 \\ \hline \begin{array}{l} \text { Number of } \\ \text { Sweaters } \end{array} & 1 & 1 & 300 \\ \hline \text { Profit } & \text { Rs.200 } & \text { Rs.120 } & \\ \hline \end{array}

Thus according to the table, the profit becomes, Z=200x+120y

If we have to maximize the profit, i.e., maximize Z=200x+120y

The subject to constraints is:

The company spends at most Rs 72000 a day

. $\therefore 360 x+120 y \leq 72000$
Divide throughout by 120, we get

$\Rightarrow 3 x+y \leq 600 \ldots$...(i)

Also, company can make at most 300 sweaters.:$x+y \leq 300 \ldots$.. (ii)

Also, the number of sweaters of type $B$ cannot exceed the number of sweaters of type A by more than 100
i.e., $y-x \leq 100$
$y \leq 100+x \ldots \ldots \ldots$(iii)
And $x \geq 0, y \geq 0$  [non-negative constraint]

\begin{aligned} &\text { So, to maximize profit we have to maximize, } Z=200 x+120 y, \text { subject to }\\ &3 x+y \leq 600\\ &x+y \leq 300\\ &y \leq 100+x\\ &x \geq 0, y \geq 0 \end{aligned}

So, when we convert the inequalities into the equation we get the following answer:

\\ \mathrm{3x}+\mathrm{y} \leq 600 \\ \Rightarrow 3 \mathrm{x}+\mathrm{y}=600 \\ \mathrm{x}+\mathrm{y} \leq 300 \\ \Rightarrow \mathrm{x}+\mathrm{y}=300 \\ \mathrm{y} \leq 100+\mathrm{x} \\ \Rightarrow \mathrm{y}=100+\mathrm{x} \\ \mathrm{x} \geq 0 \\ \\ \Rightarrow \mathrm{x}=0

\begin{aligned} &y \geq 0\\ &\Rightarrow y=0\\ &\text { The region represented by } 3 \mathrm{x}+\mathrm{y} \leq 600 \text { : } \end{aligned}

We can say that the line 3x+y=600 meets the coordinate axes (200,0) and (0,600) respectively. Once the points are joined, we get the answer that is 3x+y=600. It is justified that 3x+y≤ 600. So the region that contain the origin represents the solution set of the inequation 3x+y≤ 600

The region represented by x+y≤ 300:

The line when it meets the coordinate axes respectively that is y= 100+x meets (-100,0) and (0,100) respectively and when the lines are joined we get the line y= 100+x. Then it is clarified that it satisfies the inequation. Therefore, the region contains the origin that represents the solution of sets of inequation.

The graph of the same is given:

The shaded region OBCDE is the feasible region is bounded, so, minimum value will occur at a corner point of the feasible region.
Corner Points are O(0,0), B(0,100), C(100,200) , D(150,150) and E(200,0)
Now we will substitute these values in Z at each of these corner points, we get

\begin{aligned} &\begin{array}{|l|l|} \hline \text { Corner Point } & \text { Value of } Z=200 x+120 y \\ \hline O(0,0) & Z=200(0)+120(0)=0+0=0 \\ B(0,100) & Z=200(0)+120(100)=0+12000=12000 \\ C(100,200) & Z=200(100)+120(200)=20000+24000=44000 \\ D(150,150) & Z=200(150)+120(150)=30000+18000=48000 \rightarrow \\ E(200,0) & Z=200(200)+120(0)=40000+0=40000 \\ \hline \end{array}\\ &\text { So from the above table the maximum value of } Z \text { is at point }(150,150) \text { . } \end{aligned}

Hence, the final answer is the maximum profit to the manufacturer is Rs. 48,000, for making 150 sweaters each of type A and type B.

 

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