question_answer

  If a young man rides his motorcycle at 25 km /h, then he had to spend Rs. 2 per km on petrol. If he rides at a faster speed of 40 km /h, then the petrol cost increases at a Rs. 5 per km.                  He has Rs. 100 to spend on petrol and-wishes to find what the maximum distance is, he can travel within one hour. Express this as a linear programming problem and solve it graphically.

Answer:

Let the distance covered at the speed of 25 km/h = x km and the distance covered at the speed of 40 km/h = y km Maximum distance \[Z=x+y\] Subject to the constraints,             \[2x+5y\le 100\]             \[\frac{x}{25}+\frac{y}{40}\le 1\]          \[\left[ \because \,\,\,\text{time}=\frac{\text{distance}}{\text{speed}} \right]\] or         \[8x+5y\le 200\] and       \[x,\,\,y\ge 0\] Table for \[2x+5y=100\] is

x	0	50
y	20	0

So, the line \[2x+5y=100\] passes through the points (0, 20) and (25, 0). On putting (0, 0) in the inequality \[2x+5y\le 100,\] we get   \[2(0)+5(0)\le 100,\] which true. \[\therefore \] Shaded region is towards the origin. Table for \[\frac{x}{25}+\frac{y}{40}=1\] or \[8x+5y=200\] is

x	0	25
y	40	0

So, the line \[8x+5y=200\] passes through the points (0, 40) and (25, 0). On putting (0, 0) in the inequality \[\frac{x}{25}+\frac{y}{40}\le 1,\]we get \[0+0\le 1,\] which is true. \[\therefore \] Shaded region is towards the origin. The graphical representation of these lines is given below The shaded portion (OABCO) represents the feasible region which is bounded.                Clearly, intersection point of lines (i) and (ii) is\[B\left( \frac{50}{3},\,\,\frac{40}{3} \right).\]. Thus, the coordinates of corner points are O (0, 0), A (25, 0), \[B\left( \frac{50}{3},\,\,\frac{40}{3} \right)\] and C (0, 20), respectively. Now, the values of Z at each comer point are given below

Corner points	\[\mathbf{Z =x+ y}\]
O (0, 0)	\[Z=0+0=0\]
A(25, 0)	\[Z=25+0=25\]
\[B\left( \frac{50}{3},\,\,\frac{40}{3} \right)\]	\[Z=\frac{50}{3}+\frac{40}{3}=\frac{90}{3}=30\] (maximum)
C (0, 20)	\[Z=0+20=20\]

\[\therefore \] Maximum value of \[Z=30\] at the point\[B\left( \frac{50}{3},\,\,\frac{40}{3} \right).\]. Hence, he travel maximum distance of 30 km in one hour out of which \[\frac{50}{3}\] km covered at a speed of 25 km/h and \[\frac{40}{3}\]km covered at a speed of 40 km/h.