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  • Refer to Exercise 12. What will be the minimum cost?

Refer to Exercise 12. What will be the minimum cost?

Answers (1)

We get the following information referring to the exercise 12:

Let us consider that the firm has X number of large vans and Y number of small vans . We have made the following table:

\begin{aligned} &\text { - }\\ &\begin{array}{|l|l|l|l|} \hline & \text { Large Van (X) } & \text { Small Van (Y) } & \begin{array}{l} \text { Maximum/Minim } \\ \text { um } \end{array} \\ \hline \text { Packages } & 200 & 80 & 1200 \\ \hline \text { cost } & \text { Rs.400 } & \text { Rs.200 } & \text { Rs. 3000 } \\ \hline \end{array} \end{aligned}

According to the table, the cost becomes Z=400x+200y.

When we have to minimize the cost, i.e., minimize Z=400x+200y.

The subject to the constraints are as follows:

200x+80y$ \geq $ 1200\\

Then divide it by 40, we get

5x+2y$ \geq $ 30$ \ldots $ $ \ldots $ $ \ldots $ $ \ldots $ ..(i)\\

$ And 400x+200y$ \leq $ 3000\\

Then divide it by 200, we get

2x+y$ \leq $ 15$ \ldots $ $ \ldots $ $ \ldots $ $ \ldots $ ..(ii)\\

The number of the vans that are large exceed the number of small vans

\\ x$ \leq $ y$ \ldots $ $ \ldots $ $ \ldots $ $ \ldots $ $ \ldots $ ..(iii)\\ \\ $ And x$ \geq $ 0, y$ \geq $ 0 [non-negative constraint]\\

So, when we minimize cost we have to minimize, Z=400x+200y, subject to

\\ 5x+2y$ \geq $ 30\\ \\ 2x+y$ \leq $ 15\\ \\ x$ \leq $ y\\ \\ x$ \geq $ 0, y$ \geq $ 0\\ \\

When we convert the inequalities into equation, we get the following equation

\\ \\ 5x+2y$ \geq $ 30\\ \\ 5x+2y=30\\ \\ 2x+y$ \leq $ 15\\ \\ 2x+y=15\\ \\ x$ \leq $ y\\ \\ x=y\\ \\ x $ \geq $ 0\\ \\ x=0\\ \\ y $ \geq $ 0\\ \\ y=0\\ \\

We can see the region represented by 5x+2y\geq 30:

The line that is 5x+2y=30 when it meets the coordinate axes (6,0) and (0,15) respectively we get the desired answer. We will then join these points to obtain the line 5x+2y=30. It is clear that (0,0) does not satisfy the inequation 5x+2y\geq 30. So the region that does not contain the origin represents the solution set of the inequation 5x+2y\geq 30.

The region that represents 2x+y\leq15:

We can see that the line 2x+y=15 meets the coordinate axes (7.5,0) and (0,15) respectively. We will have to join these points to obtain the line 2x+y=15. It is clear that (0,0) satisfies the inequation 2x+y\leq15. So the region that contains the origin represents the solution set of the inequation 2x+y\leq15.

The region represented by x≤y:

Then when the line x=y is a line that passes through the origin and doesn’t touch any coordinate axes at any other point except (0,0). When we join the points we get line x=y.

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

The shaded region ABC represents the feasible region is bounded, and minimum value will occur at a corner point of the feasible region.

\\ $ Corner Points are $ \mathrm{A}\left(\frac{30}{7}, \frac{30}{7}\right), \mathrm{B}(0,15) and \mathrm{C}(5,5)

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=400 \times+200 \mathrm{y} \\ \hline A\left(\frac{30}{7}, \frac{30}{7}\right) & 2=400\left(\frac{30}{7}\right)+200\left(\frac{30}{7}\right) \\ & =\frac{12000}{7}+\frac{6000}{7}=\frac{18000}{7} \\ B(0,15) & =2571.43 \rightarrow \min \\ C(5,5) & Z=400(0)+200(15)=0+3000=3000 \\ & Z=400(5)+200(5)=2000+1000=3000 \\ \hline \end{array}

So from the above table the minimum value of Z is at point \left(\frac{30}{7}, \frac{30}{7}\right)$

Hence, the minimum cost of the firm is Rs. 2571.43

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infoexpert22

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