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
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 50 y 20 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 x 0 25 y 40 0 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
\[\therefore \] 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. 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) \[Z=0+20=20\]
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