Refer to Exercise 15. Determine the maximum distance that the man can travel.
Refer to Exercise 15. Determine the maximum distance that the man can travel.

Solution:

According to the solution of exercise 15, we have

Maximize Z=x+y, subject to the constraints

2 x+3 y \leq 120 \ldots (i)

8 x+5 y \leq 400 \ldots (ii)

x \geq 0, y \geq 0

Let’s construct a constrain table for the above

NCERT Exemplar Solutions Class 12 Mathematics Chapter 12 - 16

Table for (i)

    \[\begin{tabular}{|l|l|l|} \hline $\mathrm{x}$ & 0 & 60 \\ \hline $\mathrm{y}$ & 40 & 0 \\ \hline \end{tabular}\]

Table for (ii)

    \[\begin{tabular}{|l|l|l|} \hline $\mathrm{x}$ & 0 & 50 \\ \hline $\mathrm{y}$ & 80 & 0 \\ \hline \end{tabular}\]

Next, solving (i) and (iii) we obtain
x=300 / 7 and y=80 / 7
It can be seen that the feasible region whose corner points are O(0,0), A(50,0), B(300 / 7,80 / 7) and C(0,40).
Let’s evaluate the value of Z

    \[\begin{tabular}{|l|l|} \hline Corner points & Value of $Z=x+y$ \\ \hline$O(0,0)$ & $Z=0+0=0$ \\ \hline$A(50,0)$ & $Z=50+0=50$ \\ \hline$B(300 / 7,80 / 7)$ & $Z=300 / 7+80 / 7=380 / 7=54.3$ \\ \hline$C(0,40)$ & $Z=0-40=40$ \\ \hline \end{tabular}\]

From above table the maximum value of Z is 54.3
So, the maximum distance that the man can travel is 54.3 \mathrm{~km} at (300 / 7,80 / 7).