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  • Refer to Exercise 11. How many circuits of Type A and of Type B, should be produced by the manufacturer so as to maximise his profit? Determine the maximum profit.

Refer to Exercise 11. How many circuits of Type A and of Type B, should be produced by the manufacturer so as to maximise his profit? Determine the maximum profit.

Answers (1)

If we refer to Exercise 11, we get the following information.

The manufacturer produces X units of type A circuits and y units of type B circuits. We make the following table from the given data:

\begin{array}{|l|l|l|l|} \hline & \text { Type A (X) } & \text { Type B (Y) } & \text { Maximum Stock } \\ \hline \text { Resistors } & 20 & 10 & 200 \\ \hline \text { Transistors } & 10 & 12 & 120 \\ \hline \text { Capacitors } & 10 & 30 & 150 \\ \hline \text { Profit } & \text { Rs. 50 } & \text { Rs. 60 } & \\ \hline \end{array}

Looking at the table, we can see that profit becomes Z=50x+60y.

When we maximize the profit, i.e. maximize Z=50x+60y.

If we see at the constraints that we have obtained, then the subject to constraints, 

20x+10y$ \leq $ 200 [this is resistor constraint]

Dividing it throughout by 10, we get the answer as:
2x+y$ \leq $ 20$ \ldots $ $ \ldots $ $ \ldots $ $ \ldots $ ..(i)\\ \\

 And 10x+20y$ \leq $ 120 [this is transistor constraint]

 Dividing it by 10 throughout we get 
x+2y$ \leq $ 12$ \ldots $ $ \ldots $ $ \ldots $ $ \ldots $ ..(ii)\\ \\

And 10x+30y$ \leq $ 150 [this is capacitor constraint]

Then divide it by 10, we get 
 x+3y$ \leq $ 15$ \ldots $ $ \ldots $ $ \ldots $ $ \ldots $ ..(iii)\\ \\

And x$ \geq $ 0, y$ \geq $ 0 [non-negative constraint]
So, when we look at the maximize profit it is Z=50x+60y, subject to

2 x+y$ \leq $ 20\\ \\ 

\\ x+2y$ \leq $ 12\\ \\ x+3y$ \leq $ 15\\ \\ x$ \geq $ 0, y$ \geq $ 0\\ \\

When we convert it to equation, we get the following equation

\\ 2x+y$ \leq $ 20\\ \\ 2x+y= 20\\ \\ x+2y$ \leq $ 12\\ \\ x+2y= 12\\ \\ x+3y$ \leq $ 15\\ \\ x+3y= 15\\ \\ x $ \geq $ 0\\ \\ x=0\\ \\ y $ \geq $ 0\\ \\ y=0\\ \\

The region that is represented by 2x+y\leq 20:

When the line 2x+y=20 meets the coordinate axes (10,0) and (0,20) respectively. Once after the lines are joined, we obtain the line 2x+y=20. It is clear that (0,0) satisfies the inequation 2x+y\leq 20. So the region then represents the set of solutions.

The region represented by x+2y\leq 12:

Once the line x+2y=12 meets the coordinate axes (12,0) and (0,6) respectively, then the points are joined to get the final result that is x+2y=12. It is then clear that it satisfies the inequation and the region does not contain the origin.

The region represented by x+3y\leq15:

The line x+3y=15 then meets the coordinate axes (15,0) and (0,5) respectively. Once we join these points we obtain the line x+3y=15. It is clear that (0,0) satisfies the inequation x+3y≤ 15. So the region that contains the origin represents the solution set of the inequation x+3y\leq15:.

Region represented by x≥0 and y≥0 is first quadrant, since every point in the first quadrant satisfies these inequations

The graph of the same is given below:

The shaded region OABCD 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), A(0,5) , B(6,3), C(9.3,1.3) and D(10,0)
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=50 x+60 y \\ \hline 0(0,0) & Z=50(0)+60(0)=0+0=0 \\ A(0,5) & Z=50(0)+60(5)=0+300=300 \\ B(6,3) & Z=50(6)+60(3)-300+180=480 \\ C(9.3,1.3) & Z=50(9.3)+60(1.3)=465+78=543 \rightarrow \max \\ D(10,0) & Z=50(10)+60(0)=500+0=500 \\ \hline \end{array}

So from the above table the maximum value of Z is at point (9.3,1.3), but as the manufacturer is required to produce two type of circuits, so the parts of resistors, transistors and capacitors cannot be decimals. So we will consider the next maximum number.

Hence, the maximum value of Z is 480 at the point (6,3) i.e., the manufacturer should produce 6 circuits of type A and 3 circuits of type B so as to maximize his profit.

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