Answer

Solution

1. 1

   From Q26:

2. 2
   $$d_a = \begin{cases} w/p_a, & p_a \le p_b, \\ 0, & p_a > p_b. \end{cases} \qquad d_b = \begin{cases} w/p_b, & p_b < p_a, \\ 0, & p_b \ge p_a. \end{cases}$$
3. 3

   At $p_a = p_b = p$, $d_a = w/p$ but the limit from $p_a > p_b$ gives $d_a = 0$, so $d_a$ jumps. Similarly $d_b$ jumps from $w/p$ (as $p_b \to p_a^-$) down to $0$ at equality. Hence both have a discontinuity along the diagonal $p_a = p_b$.

Answer structure / marking notes

Content note