Mixing
Optimal time for three people to travel
$begingroup$
Three men need to travel 60 km. They have a motorbike that can travel at 50 km/h but only two people can fit on it. They can walk at a speed of 5km/h. Can they get to their destination in 3 hours?
I found that they could do it in less than 3 hours but what is theoretically the best time they can do it in?
If anyone wants the explantion for how I found that they can do it in three hours, please comment and I will provide it.
applications
asked Jan 30 at 14:15
Linus DerehedLinus Derehed
12
$endgroup$
add a comment |
$begingroup$
Three men need to travel 60 km. They have a motorbike that can travel at 50 km/h but only two people can fit on it. They can walk at a speed of 5km/h. Can they get to their destination in 3 hours?
I found that they could do it in less than 3 hours but what is theoretically the best time they can do it in?
If anyone wants the explantion for how I found that they can do it in three hours, please comment and I will provide it.
applications
asked Jan 30 at 14:15
Linus DerehedLinus Derehed
12
$endgroup$
$begingroup$
I would start by putting person A and B to the bike, and C can start walking. Person A drives B closer to the destonation, but not all the way, and leaves him at a distance so that B will arrive walking exactly on time to the destination. Then he drives back to C ...
$endgroup$
– Matti P.
Jan 30 at 14:30
add a comment |
$begingroup$
Three men need to travel 60 km. They have a motorbike that can travel at 50 km/h but only two people can fit on it. They can walk at a speed of 5km/h. Can they get to their destination in 3 hours?
I found that they could do it in less than 3 hours but what is theoretically the best time they can do it in?
If anyone wants the explantion for how I found that they can do it in three hours, please comment and I will provide it.
applications
asked Jan 30 at 14:15
Linus DerehedLinus Derehed
12
$endgroup$
Three men need to travel 60 km. They have a motorbike that can travel at 50 km/h but only two people can fit on it. They can walk at a speed of 5km/h. Can they get to their destination in 3 hours?
I found that they could do it in less than 3 hours but what is theoretically the best time they can do it in?
If anyone wants the explantion for how I found that they can do it in three hours, please comment and I will provide it.
applications
applications
asked Jan 30 at 14:15
Linus DerehedLinus Derehed
12
asked Jan 30 at 14:15
Linus DerehedLinus Derehed
12
asked Jan 30 at 14:15
Linus DerehedLinus Derehed
12
asked Jan 30 at 14:15
Linus DerehedLinus Derehed
12
asked Jan 30 at 14:15
Linus DerehedLinus Derehed
12
12
$begingroup$
I would start by putting person A and B to the bike, and C can start walking. Person A drives B closer to the destonation, but not all the way, and leaves him at a distance so that B will arrive walking exactly on time to the destination. Then he drives back to C ...
$endgroup$
– Matti P.
Jan 30 at 14:30
add a comment |
$begingroup$
I would start by putting person A and B to the bike, and C can start walking. Person A drives B closer to the destonation, but not all the way, and leaves him at a distance so that B will arrive walking exactly on time to the destination. Then he drives back to C ...
$endgroup$
– Matti P.
Jan 30 at 14:30
$begingroup$
I would start by putting person A and B to the bike, and C can start walking. Person A drives B closer to the destonation, but not all the way, and leaves him at a distance so that B will arrive walking exactly on time to the destination. Then he drives back to C ...
$endgroup$
– Matti P.
Jan 30 at 14:30
$begingroup$
I would start by putting person A and B to the bike, and C can start walking. Person A drives B closer to the destonation, but not all the way, and leaves him at a distance so that B will arrive walking exactly on time to the destination. Then he drives back to C ...
$endgroup$
– Matti P.
Jan 30 at 14:30
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Matti P has given a good approach in a comment. A and B travel most of the way, drop off B to walk the rest, C starts walking, A goes back to get C and they all arrive together.
Once you have a route that is described like this, you can parameterize it to find the minimum time it takes. Here you let $t$ be the time that A drops off B and $u$ the time that A picks up C. B arrives at $t+(60-50t)/5$. C arrives at $u+(60-5u)/60$. A arrives at $t+(50t-5u)/50+(60-5u)/50$. You want all of these to be the same, which lets you solve for $t,u$ and find the total time.
The challenge is often to find all the reasonable approaches. Here one might reason that the bike carries two people, so each person should walk $frac {50}3$ km and ride $frac {100}3$ km. That fails because the walk takes more than $3$ hours. If you didn't think of the other approach you might say the problem cannot be solved.
answered Jan 30 at 14:54
Ross MillikanRoss Millikan
301k24200375
$endgroup$
1
$begingroup$
What you describe is how to compute a strategy which I also believe to be optimal. However, this is not really a proof that this is the best one can do.
$endgroup$
– quarague
Jan 31 at 9:39
$begingroup$
I agree with you. I can't offer a general approach to assure you have found the best one. Sometimes there is some parameter you can identify that is minimized/maximized by your solution and you can say yours is optimal. I don't see that here.
$endgroup$
– Ross Millikan
Jan 31 at 15:15
add a comment |
$begingroup$
Suppose B rides the motorbike and A and C are passenger/pedestrians. Let the journey start at O and end at D. It is clear that B has to do separate journeys for the others, who can also spend time walking, and that they don't necessarily need to arrive at D all at the same time, as long as they are all there within 3 hours. We assume all drop-offs and changeovers are instantaneous.
We can break the whole episode into five time intervals:
- B takes A to a point P so that OP $=x$ while C starts to walk.
- B returns to pick up C while A is left to continue walking.
- B and C go all the way to D.
- B returns to pick up A.
- A and B go to D, and the process is complete.
Accordingly,
- $t_1=frac{x}{50}$
- In this time, C has walked $5t_1=frac{x}{10}$. B and C are approaching each other at speed $55$, so B reaches C in time $t_2=frac{9x}{550}$. In this time, both A and C have walked a distance $frac{9x}{110}$.
- B and C go all the way to D which is a distance $60-x-frac{9x}{110}=60-frac{2x}{11}$. This takes time $$t_3=frac{1}{50}[60-frac{2x}{11}]=frac 65-frac{2x}{550}.$$
- B returns to pick up A who meanwhile in time $t_3$ has walked a distance $5t_{3}=6-frac{2x}{110}$. Therefore, when B sets off from D to pick up A, they are separated by a distance $$60-x-frac{9x}{550}-6+frac{2x}{550}=54-frac{557x}{550}.$$
They are approaching each other at speed $55$ so they meet in time $$t_4=frac{1}{55}[54-frac{557x}{550}].$$
- While B was coming back to pick up A, A had walked an extra distance $5t_4=frac{1}{11}[54-frac{557x}{550}]$, so the remaining distance between A and B is $$54-frac{557x}{550}-frac{1}{11}[54-frac{557x}{550}]=frac{540}{11}-frac{557x}{605}.$$ They drive this remaining distance in time $$t_5=frac{1}{50}[frac{540}{11}-frac{557x}{605}].$$
Putting all this together, we get a total time of $$frac{174}{55}-frac{62x}{15125}$$
This is less than $3$ provided $$x>39frac{57}{62}$$
So the optimal time is when $x=60$, that is, B drives CA all the way to D and comes back to pick up C. The journey time is 2 hours 55 minutes 3.67 sec
answered Jan 31 at 9:48
David QuinnDavid Quinn
24.1k21141
$endgroup$
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
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votes
active
oldest
votes
$begingroup$
Matti P has given a good approach in a comment. A and B travel most of the way, drop off B to walk the rest, C starts walking, A goes back to get C and they all arrive together.
Once you have a route that is described like this, you can parameterize it to find the minimum time it takes. Here you let $t$ be the time that A drops off B and $u$ the time that A picks up C. B arrives at $t+(60-50t)/5$. C arrives at $u+(60-5u)/60$. A arrives at $t+(50t-5u)/50+(60-5u)/50$. You want all of these to be the same, which lets you solve for $t,u$ and find the total time.
The challenge is often to find all the reasonable approaches. Here one might reason that the bike carries two people, so each person should walk $frac {50}3$ km and ride $frac {100}3$ km. That fails because the walk takes more than $3$ hours. If you didn't think of the other approach you might say the problem cannot be solved.
answered Jan 30 at 14:54
Ross MillikanRoss Millikan
301k24200375
$endgroup$
1
$begingroup$
What you describe is how to compute a strategy which I also believe to be optimal. However, this is not really a proof that this is the best one can do.
$endgroup$
– quarague
Jan 31 at 9:39
$begingroup$
I agree with you. I can't offer a general approach to assure you have found the best one. Sometimes there is some parameter you can identify that is minimized/maximized by your solution and you can say yours is optimal. I don't see that here.
$endgroup$
– Ross Millikan
Jan 31 at 15:15
add a comment |
$begingroup$
Matti P has given a good approach in a comment. A and B travel most of the way, drop off B to walk the rest, C starts walking, A goes back to get C and they all arrive together.
Once you have a route that is described like this, you can parameterize it to find the minimum time it takes. Here you let $t$ be the time that A drops off B and $u$ the time that A picks up C. B arrives at $t+(60-50t)/5$. C arrives at $u+(60-5u)/60$. A arrives at $t+(50t-5u)/50+(60-5u)/50$. You want all of these to be the same, which lets you solve for $t,u$ and find the total time.
The challenge is often to find all the reasonable approaches. Here one might reason that the bike carries two people, so each person should walk $frac {50}3$ km and ride $frac {100}3$ km. That fails because the walk takes more than $3$ hours. If you didn't think of the other approach you might say the problem cannot be solved.
answered Jan 30 at 14:54
Ross MillikanRoss Millikan
301k24200375
$endgroup$
1
$begingroup$
What you describe is how to compute a strategy which I also believe to be optimal. However, this is not really a proof that this is the best one can do.
$endgroup$
– quarague
Jan 31 at 9:39
$begingroup$
I agree with you. I can't offer a general approach to assure you have found the best one. Sometimes there is some parameter you can identify that is minimized/maximized by your solution and you can say yours is optimal. I don't see that here.
$endgroup$
– Ross Millikan
Jan 31 at 15:15
add a comment |
$begingroup$
Matti P has given a good approach in a comment. A and B travel most of the way, drop off B to walk the rest, C starts walking, A goes back to get C and they all arrive together.
Once you have a route that is described like this, you can parameterize it to find the minimum time it takes. Here you let $t$ be the time that A drops off B and $u$ the time that A picks up C. B arrives at $t+(60-50t)/5$. C arrives at $u+(60-5u)/60$. A arrives at $t+(50t-5u)/50+(60-5u)/50$. You want all of these to be the same, which lets you solve for $t,u$ and find the total time.
The challenge is often to find all the reasonable approaches. Here one might reason that the bike carries two people, so each person should walk $frac {50}3$ km and ride $frac {100}3$ km. That fails because the walk takes more than $3$ hours. If you didn't think of the other approach you might say the problem cannot be solved.
answered Jan 30 at 14:54
Ross MillikanRoss Millikan
301k24200375
$endgroup$
Matti P has given a good approach in a comment. A and B travel most of the way, drop off B to walk the rest, C starts walking, A goes back to get C and they all arrive together.
Once you have a route that is described like this, you can parameterize it to find the minimum time it takes. Here you let $t$ be the time that A drops off B and $u$ the time that A picks up C. B arrives at $t+(60-50t)/5$. C arrives at $u+(60-5u)/60$. A arrives at $t+(50t-5u)/50+(60-5u)/50$. You want all of these to be the same, which lets you solve for $t,u$ and find the total time.
The challenge is often to find all the reasonable approaches. Here one might reason that the bike carries two people, so each person should walk $frac {50}3$ km and ride $frac {100}3$ km. That fails because the walk takes more than $3$ hours. If you didn't think of the other approach you might say the problem cannot be solved.
answered Jan 30 at 14:54
Ross MillikanRoss Millikan
301k24200375
answered Jan 30 at 14:54
Ross MillikanRoss Millikan
301k24200375
answered Jan 30 at 14:54
Ross MillikanRoss Millikan
301k24200375
answered Jan 30 at 14:54
Ross MillikanRoss Millikan
301k24200375
301k24200375
1
$begingroup$
What you describe is how to compute a strategy which I also believe to be optimal. However, this is not really a proof that this is the best one can do.
$endgroup$
– quarague
Jan 31 at 9:39
$begingroup$
I agree with you. I can't offer a general approach to assure you have found the best one. Sometimes there is some parameter you can identify that is minimized/maximized by your solution and you can say yours is optimal. I don't see that here.
$endgroup$
– Ross Millikan
Jan 31 at 15:15
add a comment |
1
$begingroup$
What you describe is how to compute a strategy which I also believe to be optimal. However, this is not really a proof that this is the best one can do.
$endgroup$
– quarague
Jan 31 at 9:39
$begingroup$
I agree with you. I can't offer a general approach to assure you have found the best one. Sometimes there is some parameter you can identify that is minimized/maximized by your solution and you can say yours is optimal. I don't see that here.
$endgroup$
– Ross Millikan
Jan 31 at 15:15
1
1
$begingroup$
What you describe is how to compute a strategy which I also believe to be optimal. However, this is not really a proof that this is the best one can do.
$endgroup$
– quarague
Jan 31 at 9:39
$begingroup$
What you describe is how to compute a strategy which I also believe to be optimal. However, this is not really a proof that this is the best one can do.
$endgroup$
– quarague
Jan 31 at 9:39
$begingroup$
I agree with you. I can't offer a general approach to assure you have found the best one. Sometimes there is some parameter you can identify that is minimized/maximized by your solution and you can say yours is optimal. I don't see that here.
$endgroup$
– Ross Millikan
Jan 31 at 15:15
$begingroup$
I agree with you. I can't offer a general approach to assure you have found the best one. Sometimes there is some parameter you can identify that is minimized/maximized by your solution and you can say yours is optimal. I don't see that here.
$endgroup$
– Ross Millikan
Jan 31 at 15:15
add a comment |
$begingroup$
Suppose B rides the motorbike and A and C are passenger/pedestrians. Let the journey start at O and end at D. It is clear that B has to do separate journeys for the others, who can also spend time walking, and that they don't necessarily need to arrive at D all at the same time, as long as they are all there within 3 hours. We assume all drop-offs and changeovers are instantaneous.
We can break the whole episode into five time intervals:
- B takes A to a point P so that OP $=x$ while C starts to walk.
- B returns to pick up C while A is left to continue walking.
- B and C go all the way to D.
- B returns to pick up A.
- A and B go to D, and the process is complete.
Accordingly,
- $t_1=frac{x}{50}$
- In this time, C has walked $5t_1=frac{x}{10}$. B and C are approaching each other at speed $55$, so B reaches C in time $t_2=frac{9x}{550}$. In this time, both A and C have walked a distance $frac{9x}{110}$.
- B and C go all the way to D which is a distance $60-x-frac{9x}{110}=60-frac{2x}{11}$. This takes time $$t_3=frac{1}{50}[60-frac{2x}{11}]=frac 65-frac{2x}{550}.$$
- B returns to pick up A who meanwhile in time $t_3$ has walked a distance $5t_{3}=6-frac{2x}{110}$. Therefore, when B sets off from D to pick up A, they are separated by a distance $$60-x-frac{9x}{550}-6+frac{2x}{550}=54-frac{557x}{550}.$$
They are approaching each other at speed $55$ so they meet in time $$t_4=frac{1}{55}[54-frac{557x}{550}].$$
- While B was coming back to pick up A, A had walked an extra distance $5t_4=frac{1}{11}[54-frac{557x}{550}]$, so the remaining distance between A and B is $$54-frac{557x}{550}-frac{1}{11}[54-frac{557x}{550}]=frac{540}{11}-frac{557x}{605}.$$ They drive this remaining distance in time $$t_5=frac{1}{50}[frac{540}{11}-frac{557x}{605}].$$
Putting all this together, we get a total time of $$frac{174}{55}-frac{62x}{15125}$$
This is less than $3$ provided $$x>39frac{57}{62}$$
So the optimal time is when $x=60$, that is, B drives CA all the way to D and comes back to pick up C. The journey time is 2 hours 55 minutes 3.67 sec
answered Jan 31 at 9:48
David QuinnDavid Quinn
24.1k21141
$endgroup$
add a comment |
$begingroup$
Suppose B rides the motorbike and A and C are passenger/pedestrians. Let the journey start at O and end at D. It is clear that B has to do separate journeys for the others, who can also spend time walking, and that they don't necessarily need to arrive at D all at the same time, as long as they are all there within 3 hours. We assume all drop-offs and changeovers are instantaneous.
We can break the whole episode into five time intervals:
- B takes A to a point P so that OP $=x$ while C starts to walk.
- B returns to pick up C while A is left to continue walking.
- B and C go all the way to D.
- B returns to pick up A.
- A and B go to D, and the process is complete.
Accordingly,
- $t_1=frac{x}{50}$
- In this time, C has walked $5t_1=frac{x}{10}$. B and C are approaching each other at speed $55$, so B reaches C in time $t_2=frac{9x}{550}$. In this time, both A and C have walked a distance $frac{9x}{110}$.
- B and C go all the way to D which is a distance $60-x-frac{9x}{110}=60-frac{2x}{11}$. This takes time $$t_3=frac{1}{50}[60-frac{2x}{11}]=frac 65-frac{2x}{550}.$$
- B returns to pick up A who meanwhile in time $t_3$ has walked a distance $5t_{3}=6-frac{2x}{110}$. Therefore, when B sets off from D to pick up A, they are separated by a distance $$60-x-frac{9x}{550}-6+frac{2x}{550}=54-frac{557x}{550}.$$
They are approaching each other at speed $55$ so they meet in time $$t_4=frac{1}{55}[54-frac{557x}{550}].$$
- While B was coming back to pick up A, A had walked an extra distance $5t_4=frac{1}{11}[54-frac{557x}{550}]$, so the remaining distance between A and B is $$54-frac{557x}{550}-frac{1}{11}[54-frac{557x}{550}]=frac{540}{11}-frac{557x}{605}.$$ They drive this remaining distance in time $$t_5=frac{1}{50}[frac{540}{11}-frac{557x}{605}].$$
Putting all this together, we get a total time of $$frac{174}{55}-frac{62x}{15125}$$
This is less than $3$ provided $$x>39frac{57}{62}$$
So the optimal time is when $x=60$, that is, B drives CA all the way to D and comes back to pick up C. The journey time is 2 hours 55 minutes 3.67 sec
answered Jan 31 at 9:48
David QuinnDavid Quinn
24.1k21141
$endgroup$
add a comment |
$begingroup$
Suppose B rides the motorbike and A and C are passenger/pedestrians. Let the journey start at O and end at D. It is clear that B has to do separate journeys for the others, who can also spend time walking, and that they don't necessarily need to arrive at D all at the same time, as long as they are all there within 3 hours. We assume all drop-offs and changeovers are instantaneous.
We can break the whole episode into five time intervals:
- B takes A to a point P so that OP $=x$ while C starts to walk.
- B returns to pick up C while A is left to continue walking.
- B and C go all the way to D.
- B returns to pick up A.
- A and B go to D, and the process is complete.
Accordingly,
- $t_1=frac{x}{50}$
- In this time, C has walked $5t_1=frac{x}{10}$. B and C are approaching each other at speed $55$, so B reaches C in time $t_2=frac{9x}{550}$. In this time, both A and C have walked a distance $frac{9x}{110}$.
- B and C go all the way to D which is a distance $60-x-frac{9x}{110}=60-frac{2x}{11}$. This takes time $$t_3=frac{1}{50}[60-frac{2x}{11}]=frac 65-frac{2x}{550}.$$
- B returns to pick up A who meanwhile in time $t_3$ has walked a distance $5t_{3}=6-frac{2x}{110}$. Therefore, when B sets off from D to pick up A, they are separated by a distance $$60-x-frac{9x}{550}-6+frac{2x}{550}=54-frac{557x}{550}.$$
They are approaching each other at speed $55$ so they meet in time $$t_4=frac{1}{55}[54-frac{557x}{550}].$$
- While B was coming back to pick up A, A had walked an extra distance $5t_4=frac{1}{11}[54-frac{557x}{550}]$, so the remaining distance between A and B is $$54-frac{557x}{550}-frac{1}{11}[54-frac{557x}{550}]=frac{540}{11}-frac{557x}{605}.$$ They drive this remaining distance in time $$t_5=frac{1}{50}[frac{540}{11}-frac{557x}{605}].$$
Putting all this together, we get a total time of $$frac{174}{55}-frac{62x}{15125}$$
This is less than $3$ provided $$x>39frac{57}{62}$$
So the optimal time is when $x=60$, that is, B drives CA all the way to D and comes back to pick up C. The journey time is 2 hours 55 minutes 3.67 sec
answered Jan 31 at 9:48
David QuinnDavid Quinn
24.1k21141
$endgroup$
Suppose B rides the motorbike and A and C are passenger/pedestrians. Let the journey start at O and end at D. It is clear that B has to do separate journeys for the others, who can also spend time walking, and that they don't necessarily need to arrive at D all at the same time, as long as they are all there within 3 hours. We assume all drop-offs and changeovers are instantaneous.
We can break the whole episode into five time intervals:
- B takes A to a point P so that OP $=x$ while C starts to walk.
- B returns to pick up C while A is left to continue walking.
- B and C go all the way to D.
- B returns to pick up A.
- A and B go to D, and the process is complete.
Accordingly,
- $t_1=frac{x}{50}$
- In this time, C has walked $5t_1=frac{x}{10}$. B and C are approaching each other at speed $55$, so B reaches C in time $t_2=frac{9x}{550}$. In this time, both A and C have walked a distance $frac{9x}{110}$.
- B and C go all the way to D which is a distance $60-x-frac{9x}{110}=60-frac{2x}{11}$. This takes time $$t_3=frac{1}{50}[60-frac{2x}{11}]=frac 65-frac{2x}{550}.$$
- B returns to pick up A who meanwhile in time $t_3$ has walked a distance $5t_{3}=6-frac{2x}{110}$. Therefore, when B sets off from D to pick up A, they are separated by a distance $$60-x-frac{9x}{550}-6+frac{2x}{550}=54-frac{557x}{550}.$$
They are approaching each other at speed $55$ so they meet in time $$t_4=frac{1}{55}[54-frac{557x}{550}].$$
- While B was coming back to pick up A, A had walked an extra distance $5t_4=frac{1}{11}[54-frac{557x}{550}]$, so the remaining distance between A and B is $$54-frac{557x}{550}-frac{1}{11}[54-frac{557x}{550}]=frac{540}{11}-frac{557x}{605}.$$ They drive this remaining distance in time $$t_5=frac{1}{50}[frac{540}{11}-frac{557x}{605}].$$
Putting all this together, we get a total time of $$frac{174}{55}-frac{62x}{15125}$$
This is less than $3$ provided $$x>39frac{57}{62}$$
So the optimal time is when $x=60$, that is, B drives CA all the way to D and comes back to pick up C. The journey time is 2 hours 55 minutes 3.67 sec
answered Jan 31 at 9:48
David QuinnDavid Quinn
24.1k21141
answered Jan 31 at 9:48
David QuinnDavid Quinn
24.1k21141
answered Jan 31 at 9:48
David QuinnDavid Quinn
24.1k21141
answered Jan 31 at 9:48
David QuinnDavid Quinn
24.1k21141
24.1k21141
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$begingroup$
I would start by putting person A and B to the bike, and C can start walking. Person A drives B closer to the destonation, but not all the way, and leaves him at a distance so that B will arrive walking exactly on time to the destination. Then he drives back to C ...
$endgroup$
– Matti P.
Jan 30 at 14:30