A man rides his motorcycle at the speed of 50km/h.He has to spend Rs 2 per km in petrol.If he rides it at a faster speed of 80km/h,the petrol cost increases to Rs 3 per km.He has atmost Rs 120 to spend on petrol and one hour's time.Using LPP find the maximum distance he can travel.

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Solution

Let the distance travelled at the rate of 50 km/h be x km and that at the rate of 80 km/h be y km.
To Maximize:D=x+y
Subject to constraints: $2 x + 3 y \leq 120, \frac{x}{50} + \frac{y}{80} \leq 1 o r 8 x + 5 y \leq 400, x \geq 0, y \geq 0$
Vertices are (0,40),(300/7,80/7),(50,0)
Maximum D is at (300/7,80/7)
Maximum D= $\frac{300}{7} k m = 54 \frac{2}{7} k m .$

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A man rides his motorcycle at the speed of 50km/h.He has to spend Rs 2 per km in petrol.If he rides it at a faster speed of 80km/h,the petrol cost increases to Rs 3 per km.He has atmost Rs 120 to spend on petrol and one hour's time.Using LPP find the maximum distance he can travel.

Let the distance travelled at the rate of 50 km/h be x km and that at the rate of 80 km/h be y km. To Maximize:D=x+y Subject to constraints:2x+3y≤120,x50+y80≤1 or 8x+5y≤400,x≥0,y≥0 Vertices are (0,40),(300/7,80/7),(50,0) Maximum D is at (300/7,80/7) Maximum D=3007km=5427km.