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

Solution:

According to the solution of exercise 12, we have

The objective function for minimum cost is \mathrm{Z}=400 \mathrm{x}+200 \mathrm{y}

NCERT Exemplar Solutions Class 12 Mathematics Chapter 12 - 13

Subject to the constrains;

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

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

x \leq y \ldots (iii)

and x \geq 0, y \geq 0 (non-negative constraints)

Now, let’s construct a constrain table for the above Table for (i)

    \[\begin{tabular}{|l|l|l|} \hline x & 0 & 6 \\ \hline y & 15 & 0 \\ \hline Table for (ii) \\ \hline x & 0 & $7.5$ \\ \hline y & 15 & 0 \\ \hline Table & for (iii) \\ \hline x & 1 & 0 \\ \hline y & 1 & 0 \\ \hline \end{tabular}\]

Next, solving (i) and (iii), we obtain

x=30 / 7 and y=30 / 7, so the corner point is A (30 / 7,30 / 7)

On solving (ii) and (iii), we obtain

x=5 and y=5, so the corner point is B(5,5)

Here, \mathrm{ABC} is the shaded feasible region whose corner points are \mathrm{A}(30 / 7,30 / 7), \mathrm{B}(5,5) and \mathrm{C}(0,15) On evaluating the value of Z, we get

    \[\begin{tabular}{|l|l|} \hline Corner point & Value of $\mathrm{Z}=400 \mathrm{x}+200 \mathrm{y}$ \\ \hline $\mathrm{A}(30 / 7,30 / 7)$ & $\mathrm{Z}=400(30 / 7)+200(30 / 7)=18000 / 7=2571.4$ \\ \hline $\mathrm{B}(5,5)$ & $\mathrm{Z}=400(5)+200(5)=3000$ \\ \hline $\mathrm{C}(0,15)$ & $\mathrm{Z}=400(0)+200(15)=3000$ \\ \hline \end{tabular}\]

It is seen from the table that the minimum value is 2571.4

So, the required minimum cost is Rs 2571.4 at (30 / 7,30 / 7)