Answer

Solution

1. 1

   Each consumer at $x$ gets net surplus $4-d(x)-2=2-d(x)$, served only if $d(x)\le 2$. Since $[0,1]$ has length $1$, $d(x)\le 1/2$ always, so every consumer is served and contributes positive surplus.

2. 2

   Total welfare:

3. 3
   $$W=500\int_0^1 \bigl(2-d(x)\bigr)\,dx \;-\; 2\cdot 300=1000-500\int_0^1 d(x)\,dx -600.$$
4. 4

   Minimise $\int_0^1 d(x)\,dx$ over locations $a,b$. With two facilities on $[0,1]$, the welfare-minimising travel placement is the classical $1$-median for two facilities on a uniform line: $a=1/4,b=3/4$ with cut-point $1/2$:

5. 5
   $$\int_0^{1/2}|x-\tfrac14|dx+\int_{1/2}^{1}|x-\tfrac34|dx=2\cdot \tfrac{1}{2}\cdot\tfrac{1}{2}\cdot\tfrac{1}{4}\cdot 2=\tfrac{1}{8}.$$
6. 6

   (Each segment $[0,1/2]$ contributes $\int_0^{1/2}|x-1/4|dx=2\cdot \frac{1}{2}\cdot \frac{1}{4}\cdot \frac{1}{4}\cdot 2 = \frac{1}{16}$, total $\frac{1}{8}$.)

7. 7

   For $(1/3,2/3)$:

8. 8
   $$\int_0^{1/2}|x-\tfrac13|dx+\int_{1/2}^{1}|x-\tfrac23|dx = \tfrac{5}{36}+\tfrac{5}{36}=\tfrac{5}{18}>\tfrac{1}{8}.$$
9. 9

   For both at $1/2$: $\int_0^1|x-1/2|dx=1/4>1/8$.

10. 10

    So $(1/4,3/4)$ minimises mean travel and maximises welfare. Check welfare positive:

11. 11
    $$W=1000-500\cdot\tfrac{1}{8}-600=1000-62.5-600=337.5>0.$$

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