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  • 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.

Answers (1)

When we refer to the Exercise, we get the following information:

If the man rides his motorcycle for a distance of x km at a speed of 50km/hr then he has to spend Rs. 2/km on petrol.

And if he rides his motorcycle for a distance of y km at a speed of 80 km/hr then he has to spend Rs. 3/km on petrol.

He has Rs.120 to spend on petrol for covering the total distance so the constraint becomes

2x+3y≤120…………(i)

Given that he has only one hour’s time to cover the total distance, then the constraint becomes

\frac{x}{50}+\frac{y}{80} \leq 1 {as distance speed x time}
Now taking the LCM as 400, we get
$\Rightarrow 8 x+5 y \leq 400 \ldots \ldots \ldots \ldots$(ii)
And $\mathrm{x} \geq 0, \mathrm{y} \geq 0$ [non-negative constraint]
He want to find out the maximum distance travelled, here total distance, $\mathrm{Z}=\mathrm{x}+\mathrm{y}$

Now, we have to maximize the distance, i.e., maximize $\mathrm{Z}=\mathrm{x}+\mathrm{y}$
So, to maximize distance we have to maximize, $\mathrm{Z}=\mathrm{x}+\mathrm{y}$,$ subject to
\\$2 x+3 y \leq 120$ \\$8 x+5 y \leq 400$ \\$x \geq 0, y=0$

When we convert it into equation we get

\\ 2x+3y$ \leq $ 120$ \Rightarrow $ 2x+3y=120\\ \\ 8x+5y$ \leq $ 400 $ \Rightarrow $ 8x+5y=400\\ \\ x $ \geq $ 0 $ \Rightarrow $ x=0\\ \\ y $ \geq $ 0 $ \Rightarrow $ y=0\\ \\

The region that is represented by 2x+3y≤120:

The line that is 2x+3y=120 further meets the coordinate axes (60,0) and (0,40) respectively and then when we will join these points we obtain the line 2x+3y=120. It is clear that (0,0) justifies the inequation 2x+3y≤120. So the region that contain the origin represents the solution set of the inequation 2x+3y≤120

The region represented by 8x+5y≤400:

We can see that the line 8x+5y=400 meets the coordinate axes (50,0) and (0,80) respectively. When we join these points we obtain the line 8x+5y=400. Therefore, it is clear that (0,0) satisfies the inequation 8x+5y≤400. So the region that contain the origin represents the solution set of the inequation 8x+5y≤400

The first quadrant of the region represented is x≥0 and y≥0. The graph is as follows:

The shaded region OBCD is the feasible region is bounded, and maximum value will occur at a corner point of the feasible region.
Corner Points are \mathrm{O}(0,0), \mathrm{B}(0,40), \mathrm{C}\left(\frac{300}{7}, \frac{90}{7}\right)$ and $\mathrm{D}(50,0)$
When we substitute the values in Z, we get the following answer:

\begin{array}{|l|l|} \hline \text { Corner Point } & \text { Value of } Z=x+y \\ \hline O(0,0) & Z=0+0=0 \\ B(0,40) & Z=0+40=40 \\ C\left(\frac{300}{7}, \frac{00}{7}\right) & z=\frac{300}{7}+\frac{80}{7}=\frac{380}{7}=54 \frac{2}{7} \rightarrow \max \\ D(50,0) & Z=50+0=50 \\ \hline \end{array}

So from the above table the maximum value of Z is at point \left(\frac{300}{7}, \frac{80}{7}\right)$
Hence, the maximum distance the man can travel is 54 \frac{2}{7} \mathrm{~km}$ or  54.3 km

 

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