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  • Refer to Exercise 13. Solve the linear programming problem and determine the maximum profit to the manufacturer.

Refer to Exercise 13. Solve the linear programming problem and determine the maximum profit to the manufacturer.

Answers (1)

Refer to the Exercise 13, we get the following data:

Let us assume that the company manufactures x boxes of type A screws and y boxes of type B screws. We make the following table from the given data:

  Type A screws (x boxes) Type B screws (y boxes) Max time available on each machine in a week
Time required for screws on threading machine 2 8 60 hrs = 60*60min = 3600min

Time required for screws on slotting machine

 3                         3

60 hrs = 60*60min = 3600min
Profit Rs 100 Rs170  

When we look at the table, the profit becomes, Z=100x+170y

Thus according to the table, the profit becomes, Z=100x+170y

The constraints that we have obtained that is subject to constraints:

2x+8y$ \leq $ 3600 [time constraints for threading machine]

Divide it throughout by 2, we get

x+4y$ \leq $ 1800$ \ldots $ $ \ldots $ $ \ldots $ $ \ldots $ ..(i)\\

And 3x+2y$ \leq $ 3600 [time constraints for slotting machine]

3x+2y$ \leq $ 3600…………..(ii)

And x≥0, y≥0 [non-negative constraint]

So, to maximize profit we have to maximize, Z=100x+170y, subject to
\\ \\ x+4y$ \leq $ 1800\\ \\ 3x+2y$ \leq $ 3600\\ \\ x$ \geq $ 0, y$ \geq $ 0\\

Now let us convert the given inequalities into equation.

We obtain the following equation

\\ x+4y$ \leq $ 1800\\ \\ x+4y=1800\\ \\ 3x+2y$ \leq $ 3600\\ \\ 3x+2y=3600\\ \\ x $ \geq $ 0\\ \\ x=0\\ \\ y $ \geq $ 0\\ \\ y=0\\ \\

The region that represents x+4y≤ 1800:

We can say that the line x+4y=1800 meets the coordinate axes (1800,0) and (0,450) respectively. When we join these points, we get the desired line that is x+4y=1800. We can say that it is clear that it satisfies the inequation and therefore the region that contains the origin further represents the set of solutions of the inequation.

The region represented by 3x+2y\leq3600:

The line 3x+2y=3600 further meets the axes (1200,0) and (0,1800) respectively. After joining these points, we get the result that is 3x+2y\leq3600. It is then clear that it satisfies the inequation 3x+2y\leq3600. So the region that contains the origin represents the solution set of the inequation 3x+2y\leq3600.

The graph is given below:

 

The shaded region OBCD is the feasible region is bounded, so, maximum value will occur at a corner point of the feasible region.
Corner Points are O(0,0), B(0,450), C(1080,180) and D(1200,0)
Now we will substitute these values in Z at each of these corner points, we get

\begin{array}{|l|l|} \hline \text { Corner Point } & \text { Value of } Z=100 x+170 y \\ \hline O(0,0) & Z=100(0)+170(0)=0+0=0 \\ B(0,450) & Z=100(0)+170(450)=0+76500=76500 \\ C(1080,180) & Z=100(1080)+170(180)=108000+30600 \\ & Z=138600 \rightarrow \max \\ D(1200,0) & Z=100(1200)+170(0)=120000+0=120000 \\ \hline \end{array}

So from the above table the maximum value of Z is at point (1080,180) .
Therefore, the final answer is maximum profit to the manufacturer is Rs.
1,38,600

 

Posted by

infoexpert22

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