Three men travelling

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Three men want to travel from town $P$ to town $Q$ (distance between towns: $a$). They have a motorbike (vehicle speed: $b$), but only two of them can be on the motorbike at the same time. Their walking speed is $c$ ($b$ > $c$). How much time do they need to travel? I believe that the best solution goes like this:

1. Two people ($A$ and $B$) travel on motorbike, the third one ($C$) walks.


2. $A$ drops $B$ at point $X$ and goes back to pick up $C$. $B$ walks to town $Q$.


3. After picking up $C$, motorbike goes to town $Q$. All three men arrive at the same time.

However, I don't know, how to prove that this is the best solution.

algebra-precalculus

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edited Jul 27 at 10:20

N. F. Taussig

38.1k93053

asked Jul 27 at 10:13

Rakaho

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up vote
5
down vote

favorite

1

Three men want to travel from town $P$ to town $Q$ (distance between towns: $a$). They have a motorbike (vehicle speed: $b$), but only two of them can be on the motorbike at the same time. Their walking speed is $c$ ($b$ > $c$). How much time do they need to travel? I believe that the best solution goes like this:

1. Two people ($A$ and $B$) travel on motorbike, the third one ($C$) walks.


2. $A$ drops $B$ at point $X$ and goes back to pick up $C$. $B$ walks to town $Q$.


3. After picking up $C$, motorbike goes to town $Q$. All three men arrive at the same time.

However, I don't know, how to prove that this is the best solution.

algebra-precalculus

share|cite|improve this question

edited Jul 27 at 10:20

N. F. Taussig

38.1k93053

asked Jul 27 at 10:13

Rakaho

283

add a comment | 

up vote
5
down vote

favorite

1

up vote
5
down vote

favorite

1

1

Three men want to travel from town $P$ to town $Q$ (distance between towns: $a$). They have a motorbike (vehicle speed: $b$), but only two of them can be on the motorbike at the same time. Their walking speed is $c$ ($b$ > $c$). How much time do they need to travel? I believe that the best solution goes like this:

1. Two people ($A$ and $B$) travel on motorbike, the third one ($C$) walks.


2. $A$ drops $B$ at point $X$ and goes back to pick up $C$. $B$ walks to town $Q$.


3. After picking up $C$, motorbike goes to town $Q$. All three men arrive at the same time.

However, I don't know, how to prove that this is the best solution.

algebra-precalculus

share|cite|improve this question

edited Jul 27 at 10:20

N. F. Taussig

38.1k93053

asked Jul 27 at 10:13

Rakaho

283

Three men want to travel from town $P$ to town $Q$ (distance between towns: $a$). They have a motorbike (vehicle speed: $b$), but only two of them can be on the motorbike at the same time. Their walking speed is $c$ ($b$ > $c$). How much time do they need to travel? I believe that the best solution goes like this:

1. Two people ($A$ and $B$) travel on motorbike, the third one ($C$) walks.


2. $A$ drops $B$ at point $X$ and goes back to pick up $C$. $B$ walks to town $Q$.


3. After picking up $C$, motorbike goes to town $Q$. All three men arrive at the same time.

However, I don't know, how to prove that this is the best solution.

algebra-precalculus

share|cite|improve this question

edited Jul 27 at 10:20

N. F. Taussig

38.1k93053

asked Jul 27 at 10:13

Rakaho

283

share|cite|improve this question

share|cite|improve this question

edited Jul 27 at 10:20

N. F. Taussig

38.1k93053

edited Jul 27 at 10:20

N. F. Taussig

38.1k93053

edited Jul 27 at 10:20

N. F. Taussig

38.1k93053

38.1k93053

asked Jul 27 at 10:13

Rakaho

283

asked Jul 27 at 10:13

Rakaho

283

asked Jul 27 at 10:13

Rakaho

283

283

add a comment | 

add a comment | 

1 Answer
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We may assume $a=1$, $b=1$ and $c=1over w$, whereby $w>1$ denotes the time needed to walk the full stretch. In order to simplify matters I'm assuming that $A$ is always staying with his bike. Given some protocol $cal P$ (maybe more complicated than the one described in the question), denote by $t_A$, $t_B$, $t_C$ the time needed by the three people in order to arrive at $Q$, when this protocol is in force. We want a universal lower bound for the quantity
$$T_cal P:=maxt_A,t_B,t_C .$$
Let $rho_B$ be the fraction of the way $B$ is riding forward on the bike together with $A$, and define $rho_C$ similarly. Then
$$t_Bgeq rho_B+(1-rho_B)w=w-(w-1)rho_B,qquad t_Cgeq w-(w-1)rho_C .$$
It follows that
$$maxt_B,t_Cgeq w-(w-1)minrho_B,rho_Cgeq w-(w-1)rho ,qquad rho:=rho_B+rho_Cover2 .tag1$$
In order to get a bound for $t_A$ we have to distinguish the cases (i): $rho_B+rho_Cleq1$ and (ii): $rho_B+rho_C>1$. In case (i) all we can say is $t_Ageq1$. In case (ii) the parts of the road where $B$ as well as $C$ are riding with $A$ overlap by at least $rho_B+rho_C-1$. It follows that $A$ has to drive back at least this amount. In this way we obtain in case (ii) the bound
$$t_Ageqrho_B+rho_C+(rho_B+rho_C-1)=4rho-1 .$$
It is therefore safe to say that
$$t_Ageqmax1,4rho-1 .tag2$$
Taking $(1)$ and $(2)$ together we see that
$$T_cal Pgeqmaxw-(w-1)rho, 1, 4rho-1 .tag3$$
An analysis of $(3)$ for $0leqrholeq1$ then shows that the RHS is minimal at $rho=w+1over w+3$. This leads to the universal estimate
$$T_cal Pgeq3w+1over w+3 .$$
I'd say that the protocol you describe realizes this bound; hence you cannot do better.

share|cite|improve this answer

edited Jul 29 at 8:20

answered Jul 28 at 9:39

Christian Blatter

163k7107306

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1 Answer
1

active

oldest

votes

1 Answer
1

active

oldest

votes

active

oldest

votes

active

oldest

votes

up vote
4
down vote



accepted

We may assume $a=1$, $b=1$ and $c=1over w$, whereby $w>1$ denotes the time needed to walk the full stretch. In order to simplify matters I'm assuming that $A$ is always staying with his bike. Given some protocol $cal P$ (maybe more complicated than the one described in the question), denote by $t_A$, $t_B$, $t_C$ the time needed by the three people in order to arrive at $Q$, when this protocol is in force. We want a universal lower bound for the quantity
$$T_cal P:=maxt_A,t_B,t_C .$$
Let $rho_B$ be the fraction of the way $B$ is riding forward on the bike together with $A$, and define $rho_C$ similarly. Then
$$t_Bgeq rho_B+(1-rho_B)w=w-(w-1)rho_B,qquad t_Cgeq w-(w-1)rho_C .$$
It follows that
$$maxt_B,t_Cgeq w-(w-1)minrho_B,rho_Cgeq w-(w-1)rho ,qquad rho:=rho_B+rho_Cover2 .tag1$$
In order to get a bound for $t_A$ we have to distinguish the cases (i): $rho_B+rho_Cleq1$ and (ii): $rho_B+rho_C>1$. In case (i) all we can say is $t_Ageq1$. In case (ii) the parts of the road where $B$ as well as $C$ are riding with $A$ overlap by at least $rho_B+rho_C-1$. It follows that $A$ has to drive back at least this amount. In this way we obtain in case (ii) the bound
$$t_Ageqrho_B+rho_C+(rho_B+rho_C-1)=4rho-1 .$$
It is therefore safe to say that
$$t_Ageqmax1,4rho-1 .tag2$$
Taking $(1)$ and $(2)$ together we see that
$$T_cal Pgeqmaxw-(w-1)rho, 1, 4rho-1 .tag3$$
An analysis of $(3)$ for $0leqrholeq1$ then shows that the RHS is minimal at $rho=w+1over w+3$. This leads to the universal estimate
$$T_cal Pgeq3w+1over w+3 .$$
I'd say that the protocol you describe realizes this bound; hence you cannot do better.

share|cite|improve this answer

edited Jul 29 at 8:20

answered Jul 28 at 9:39

Christian Blatter

163k7107306

add a comment | 

up vote
4
down vote



accepted

We may assume $a=1$, $b=1$ and $c=1over w$, whereby $w>1$ denotes the time needed to walk the full stretch. In order to simplify matters I'm assuming that $A$ is always staying with his bike. Given some protocol $cal P$ (maybe more complicated than the one described in the question), denote by $t_A$, $t_B$, $t_C$ the time needed by the three people in order to arrive at $Q$, when this protocol is in force. We want a universal lower bound for the quantity
$$T_cal P:=maxt_A,t_B,t_C .$$
Let $rho_B$ be the fraction of the way $B$ is riding forward on the bike together with $A$, and define $rho_C$ similarly. Then
$$t_Bgeq rho_B+(1-rho_B)w=w-(w-1)rho_B,qquad t_Cgeq w-(w-1)rho_C .$$
It follows that
$$maxt_B,t_Cgeq w-(w-1)minrho_B,rho_Cgeq w-(w-1)rho ,qquad rho:=rho_B+rho_Cover2 .tag1$$
In order to get a bound for $t_A$ we have to distinguish the cases (i): $rho_B+rho_Cleq1$ and (ii): $rho_B+rho_C>1$. In case (i) all we can say is $t_Ageq1$. In case (ii) the parts of the road where $B$ as well as $C$ are riding with $A$ overlap by at least $rho_B+rho_C-1$. It follows that $A$ has to drive back at least this amount. In this way we obtain in case (ii) the bound
$$t_Ageqrho_B+rho_C+(rho_B+rho_C-1)=4rho-1 .$$
It is therefore safe to say that
$$t_Ageqmax1,4rho-1 .tag2$$
Taking $(1)$ and $(2)$ together we see that
$$T_cal Pgeqmaxw-(w-1)rho, 1, 4rho-1 .tag3$$
An analysis of $(3)$ for $0leqrholeq1$ then shows that the RHS is minimal at $rho=w+1over w+3$. This leads to the universal estimate
$$T_cal Pgeq3w+1over w+3 .$$
I'd say that the protocol you describe realizes this bound; hence you cannot do better.

share|cite|improve this answer

edited Jul 29 at 8:20

answered Jul 28 at 9:39

Christian Blatter

163k7107306

add a comment | 

up vote
4
down vote



accepted

up vote
4
down vote



accepted

We may assume $a=1$, $b=1$ and $c=1over w$, whereby $w>1$ denotes the time needed to walk the full stretch. In order to simplify matters I'm assuming that $A$ is always staying with his bike. Given some protocol $cal P$ (maybe more complicated than the one described in the question), denote by $t_A$, $t_B$, $t_C$ the time needed by the three people in order to arrive at $Q$, when this protocol is in force. We want a universal lower bound for the quantity
$$T_cal P:=maxt_A,t_B,t_C .$$
Let $rho_B$ be the fraction of the way $B$ is riding forward on the bike together with $A$, and define $rho_C$ similarly. Then
$$t_Bgeq rho_B+(1-rho_B)w=w-(w-1)rho_B,qquad t_Cgeq w-(w-1)rho_C .$$
It follows that
$$maxt_B,t_Cgeq w-(w-1)minrho_B,rho_Cgeq w-(w-1)rho ,qquad rho:=rho_B+rho_Cover2 .tag1$$
In order to get a bound for $t_A$ we have to distinguish the cases (i): $rho_B+rho_Cleq1$ and (ii): $rho_B+rho_C>1$. In case (i) all we can say is $t_Ageq1$. In case (ii) the parts of the road where $B$ as well as $C$ are riding with $A$ overlap by at least $rho_B+rho_C-1$. It follows that $A$ has to drive back at least this amount. In this way we obtain in case (ii) the bound
$$t_Ageqrho_B+rho_C+(rho_B+rho_C-1)=4rho-1 .$$
It is therefore safe to say that
$$t_Ageqmax1,4rho-1 .tag2$$
Taking $(1)$ and $(2)$ together we see that
$$T_cal Pgeqmaxw-(w-1)rho, 1, 4rho-1 .tag3$$
An analysis of $(3)$ for $0leqrholeq1$ then shows that the RHS is minimal at $rho=w+1over w+3$. This leads to the universal estimate
$$T_cal Pgeq3w+1over w+3 .$$
I'd say that the protocol you describe realizes this bound; hence you cannot do better.

share|cite|improve this answer

edited Jul 29 at 8:20

answered Jul 28 at 9:39

Christian Blatter

163k7107306

We may assume $a=1$, $b=1$ and $c=1over w$, whereby $w>1$ denotes the time needed to walk the full stretch. In order to simplify matters I'm assuming that $A$ is always staying with his bike. Given some protocol $cal P$ (maybe more complicated than the one described in the question), denote by $t_A$, $t_B$, $t_C$ the time needed by the three people in order to arrive at $Q$, when this protocol is in force. We want a universal lower bound for the quantity
$$T_cal P:=maxt_A,t_B,t_C .$$
Let $rho_B$ be the fraction of the way $B$ is riding forward on the bike together with $A$, and define $rho_C$ similarly. Then
$$t_Bgeq rho_B+(1-rho_B)w=w-(w-1)rho_B,qquad t_Cgeq w-(w-1)rho_C .$$
It follows that
$$maxt_B,t_Cgeq w-(w-1)minrho_B,rho_Cgeq w-(w-1)rho ,qquad rho:=rho_B+rho_Cover2 .tag1$$
In order to get a bound for $t_A$ we have to distinguish the cases (i): $rho_B+rho_Cleq1$ and (ii): $rho_B+rho_C>1$. In case (i) all we can say is $t_Ageq1$. In case (ii) the parts of the road where $B$ as well as $C$ are riding with $A$ overlap by at least $rho_B+rho_C-1$. It follows that $A$ has to drive back at least this amount. In this way we obtain in case (ii) the bound
$$t_Ageqrho_B+rho_C+(rho_B+rho_C-1)=4rho-1 .$$
It is therefore safe to say that
$$t_Ageqmax1,4rho-1 .tag2$$
Taking $(1)$ and $(2)$ together we see that
$$T_cal Pgeqmaxw-(w-1)rho, 1, 4rho-1 .tag3$$
An analysis of $(3)$ for $0leqrholeq1$ then shows that the RHS is minimal at $rho=w+1over w+3$. This leads to the universal estimate
$$T_cal Pgeq3w+1over w+3 .$$
I'd say that the protocol you describe realizes this bound; hence you cannot do better.

share|cite|improve this answer

edited Jul 29 at 8:20

answered Jul 28 at 9:39

Christian Blatter

163k7107306

share|cite|improve this answer

share|cite|improve this answer

edited Jul 29 at 8:20

edited Jul 29 at 8:20

edited Jul 29 at 8:20

answered Jul 28 at 9:39

Christian Blatter

163k7107306

answered Jul 28 at 9:39

Christian Blatter

163k7107306

answered Jul 28 at 9:39

Christian Blatter

163k7107306

163k7107306

add a comment | 

add a comment | 

 

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