A man rides his motorcycle at the speed of 50 km/hour. He has to spend Rs 2 per km on petrol. If he rides it at a faster speed of 80 km/hour, the petrol cost increases to Rs 3 per km. He has atmost Rs 120 to spend on petrol and one hour’s time. He wishes to find the maximum distance that he can travel. Express this problem as a linear programming problem.

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Turning Maths, Physics & CS Struggles into Strengths for 100+ Months | B.Tech Grad.

Time constraint: Time = Distance/Speed. Time taken at 50 km/hr is x50\dfrac{x}{50} , and time taken at 80 km/hr is y80\dfrac{y}{80} . Total time available is at most 1 hour: x50+y80≤1\frac{x}{50} + \frac{y}{80} \leq 1 Multiplying through by 400 (LCM of 50 and 80) to clear denominators: 8x+5y≤4008x...
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Time constraint: Time = Distance/Speed. Time taken at 50 km/hr is x50\dfrac{x}{50} , and time taken at 80 km/hr is y80\dfrac{y}{80} . Total time available is at most 1 hour: x50+y80≤1\frac{x}{50} + \frac{y}{80} \leq 1 Multiplying through by 400 (LCM of 50 and 80) to clear denominators: 8x+5y≤4008x + 5y \leq 400 Petrol cost is Rs 2 per km at 50 km/hr, and Rs 3 per km at 80 km/hr. Total cost is at most Rs 120: 2x+3y≤1202x + 3y \leq 1208x+5y≤400 ---------- (i)2x + 3y \leq 1202x+3y≤120----------- (ii) read less
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C Language Faculty (Online Classes ) Having 10 Years Experience

The problem can be formulated as a Linear Programming Problem (LPP) to maximize the total distance \(Z = x + y\), subject to cost and time constraints.Here is the step-by-step mathematical formulation: 1. Decision Variables Let \(x\) be the distance traveled at the speed of 50 km/hour (in km).Let \(y\)...
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The problem can be formulated as a Linear Programming Problem (LPP) to maximize the total distance \(Z = x + y\), subject to cost and time constraints.Here is the step-by-step mathematical formulation: 1. Decision Variables Let \(x\) be the distance traveled at the speed of 50 km/hour (in km).Let \(y\) be the distance traveled at the speed of 80 km/hour (in km). 2. Objective Function The goal is to maximize the total distance traveled (\(Z\)).\(\text{Maximize\ }Z=x+y\) 3. Constraints • Petrol Cost Constraint: The cost at 50 km/h is ₹2/km, and at 80 km/h is ₹3/km. The total budget is at most ₹120.\(2x+3y\le 120\)• Time Constraint: \(\text{Time} = \frac{\text{Distance}}{\text{Speed}}\). The total available time is at most 1 hour.\(\frac{x}{50}+\frac{y}{80}\le 1\)Multiplying by 400 to simplify:\(8x+5y\le 400\)• Non-negativity Constraints: Distance cannot be negative.\(x\ge 0,\quad y\ge 0\) Final LPP Formulation \(\text{Maximize\ }Z=x+y\)Subject to the constraints:• \(2x + 3y \le 120\) (Cost constraint)• \(8x + 5y \le 400\) (Time constraint)• \(x, y \ge 0\) (Non-negativity constraint) read less
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M.Sc. (Mathematics) | Former Faculty at Vedantu (5 Years) & Doon International School (6 Years)

Let xx be the distance (in km) travelled at 50 km/hour and yy be the distance (in km) travelled at 80 km/hour. 1. Objective function The total distance travelled is x+yx+y. Since he wants to travel the maximum distance, the objective function...
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Let xx be the distance (in km) travelled at 50 km/hour and yy be the distance (in km) travelled at 80 km/hour. 1. Objective function The total distance travelled is x+yx+y. Since he wants to travel the maximum distance, the objective function is MaximizeZ=x+yMaximizeZ=x+y 2. Constraints Petrol cost constraint: At 50 km/hour, the cost is Rs 2 per km, and at 80 km/hour, it is Rs 3 per km. The total petrol budget is at most Rs 120. 2x+3y≤1202x+3y≤120 Time constraint: Time taken at 50 km/hour is x5050x hours, and at 80 km/hour is y8080y hours. The total time available is at most 1 hour. x50+y80≤150x+80y≤1 Multiplying by 400 gives 8x+5y≤4008x+5y≤400 Non-negativity constraints: x≥0,y≥0x≥0,y≥0 Final answer: Linear Programming Problem (LPP) Maximize Z=x+yZ=x+y Subject to the constraints 2x+3y≤1208x+5y≤400x,y≥02x+3y8x+5yx,y≤120≤400≥0 read less
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Turning Maths, Physics & CS Struggles into Strengths for 100+ Months | B.Tech Grad.

Let x = km at 50 km/h, y = km at 80 km/h. Maximise Z = x + y subject to: 2x + 3y ≤ 120 (petrol) 8x + 5y ≤ 400 (time: x/50 + y/80 ≤ 1) x ≥ 0, y ≥ 0 (Maximum Z = 380/7 ≈ 54.29 km, at x = 300/7, y = 80/7.)
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Let x = km at 50 km/h, y = km at 80 km/h. Maximise Z = x + y subject to: 2x + 3y ≤ 120 (petrol) 8x + 5y ≤ 400 (time: x/50 + y/80 ≤ 1) x ≥ 0, y ≥ 0 (Maximum Z = 380/7 ≈ 54.29 km, at x = 300/7, y = 80/7.) read less
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Turning Maths, Physics & CS Struggles into Strengths for 100+ Months | B.Tech Grad.

Let x = distance (km) travelled at 50 km/h and y = distance (km) travelled at 80 km/h. Maximise Z = x + y Subject to: Petrol cost: 2x + 3y ≤ 120 Time: x/50 + y/80 ≤ 1, which simplifies to 8x + 5y ≤ 400 x ≥ 0, y ≥ 0 (Multiplying x/50 + y/80 ≤ 1 by 400 gives 8x + 5y ≤...
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Let x = distance (km) travelled at 50 km/h and y = distance (km) travelled at 80 km/h. Maximise Z = x + y Subject to: Petrol cost: 2x + 3y ≤ 120 Time: x/50 + y/80 ≤ 1, which simplifies to 8x + 5y ≤ 400 x ≥ 0, y ≥ 0 (Multiplying x/50 + y/80 ≤ 1 by 400 gives 8x + 5y ≤ 400.) read less
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