 Hello, I am welcome to the session. I am Deepika and I am going to help you to solve the following question. The question says, If a young man rides his motorcycle at 25 km per hour, he had to spend rupees 2 per kilometre on petrol. If he rides at a faster speed of 40 km per hour, the petrol cost increases at rupees 5 per kilometre. He has rupees 100 to spend on petrol and wishes to find what is the maximum distance he can travel within one hour. Express this as an LVP and solve it graphically. So, let's start the solution. Now, according to the given question, if a young man rides his motorcycle at 25 km per hour, he had to spend rupees 2 per kilometre on petrol and if he rides at a faster speed of 40 km per hour, the petrol cost increases at rupees 5 per kilometre. So, let X km be the distance covered by young man with the speed at 25 km per hour and Y km be the distance covered with the speed at 40 km per hour. Now, again according to the given question, if he rides his motorcycle at 25 km per hour, he had to spend rupees 2 per kilometre on petrol and if he rides at a faster speed of 40 km per hour, the petrol cost increases at rupees 5 per kilometre. He has rupees 100 to spend on petrol. So, according to the given question, we have X5Y less than equal to 100. So, this is a constraint related to expenditure on petrol. Again, according to the given question, the young man wishes to find what is the maximum distance he can travel within one hour. Now, we know that speed is equal to distance upon time and time is equal to distance upon speed. So, the time taken by the young man with the speed at 25 km per hour is equal to X over 25 hour because we have assumed that X km be the distance covered at 25 km per hour. So, time taken by young man with the speed at 25 km per hour is equal to distance covered over speed which is equal to X over 25 hour. Similarly, the time taken by young man with the speed at 40 km per hour is equal to Y over 40 hour. Therefore, according to the given question, X over 25 plus Y over 40 less than equal to 1. So, this is a constraint related to time. This can be written as 40X plus 25Y less than equal to 1000 or 8X plus 5Y less than equal to 200. So, this is a constraint related to time. Now, we have X greater than equal to 0 and Y greater than equal to 0. These are the non-negative constraints. Now, we have assumed that X km be the distance covered at 25 km per hour and Y km be the distance covered at 40 km per hour. So, the total distance covered is equal to Z is equal to X plus 5. So, the mathematical formulation of the given problem is maximize Z is equal to X plus Y subject to the constraints 5Y less than equal to 100. Let us give this as number 1 and 8X plus 5Y less than equal to 200. Let us give this as number 2 and X greater than equal to 0 and Y greater than equal to 0. Let us give this as number 3. So, now we will draw the graph and find the feasible region subject to these given constraints. So, we will first draw the line representing the equation 2X plus 5Y is equal to 100 corresponding to the inequality 2X plus 5Y less than equal to 100. Now, when X is equal to 0, Y is equal to 20 and when Y is equal to 0, X is equal to 50. So, the points 0, 20 and 50 0 satisfy the equation 2X plus 5Y is equal to 100. Therefore, the graph of this line can be drawn by plotting points 0, 20 and 50 0 and then joining them. So, let A be the point 0, 20 and B be the point 50 0. So, A B represents the equation of the line 2X plus 5Y is equal to 100. Now, this line A B divides the plane into two half planes. So, we will consider the half plane which will satisfy one. Clearly, the origin satisfies this inequality. So, the half plane containing the origin is the graph of one. Now, the equation corresponding to the inequality 2 is AX plus 5Y is equal to 200. Now, clearly the points 0, 40 and 25 0 satisfy the equation AX plus 5Y is equal to 200. So, the graph of this line can be drawn by plotting the points 0, 40 and 25 0 and then joining them. Now, let C be the point 0, 40 and D be the point 25 0. So, C D represents the equation of the line AX plus 5Y is equal to 200. Now, again the line C D divides the plane into two half planes. So, we will consider the half plane which will satisfy two. Now, clearly origin satisfies this inequality. So, the half plane containing the origin is the graph of two. Now, X greater than equal to 0 and Y greater than equal to 0 implies that the graph lies in the first quadrant only. So, the shaded region in this graph is the physical region satisfying all the given constraints. Now, here the shaded region is bounded. So, we will see the coordinates of its corner points. So, the coordinates of its corner points are A with coordinates 0, 20, O with coordinates 0, 0, D with coordinates 25, 0. Let us take this point as the point F. So, the coordinates of this F point are 50 over 3 and 40 over 3. Now, the coordinates of this corner point can be find by either by just observing the graph or by solving the equations 8X plus 5Y is equal to 200 and 2X plus 5Y is equal to 100. So, on solving these two equations, we get the coordinates of the point F as 50 over 3 and 40 over 3. Hence, we can write here the physical region is bounded with coordinates of its corner points as A with coordinates 0, 20, O with coordinates 0, 0, D with coordinates 25, 0 and F has coordinates 50 over 3 and 40 over 3. Now, we will evaluate Z which is equal to X plus Y at each of these corner points. So, at the point A whose coordinates are 0, 20, Z is equal to 0 plus 20 which is equal to 20 and at the point O with coordinates 0, 0, Z is equal to 0 and at D whose coordinates are 25, 0, Z is equal to 25 plus 0 which is equal to 25 and at the point F whose coordinates are 50 over 3 and 40 over 3, Z is equal to 50 over 3 plus 40 over 3 which is equal to 90 over 3 and this is again equal to 30. So, maximum value of Z is equal to 30 when X is equal to 50 over 3 and Y is equal to 40 over 3. So, to cover the maximum distance within one hour the young men should travel 50 over 3 km with the speed at 25 km per hour and 40 over 3 km at 40 km per hour. So, the answer for the above question is 50 over 3 km at 25 km per hour and 40 over 3 km at 40 km per hour and then the maximum distance is equal to 30 km. So, this is the answer for the above question. This completes our session. I hope the solution is clear to you. Bye and have a nice day.