Question 9.12: Given the supply and demand functions for two related goods,......
Given the supply and demand functions for two related goods, A and B,
Good A : \begin{matrix} \{Q_{da} = 30 − 8P_a + 2P_b \\ \{Q_{sa} = −15 + 7P_a \end{matrix} Good B : \begin{matrix} \{Q_{db} = 28 + 4P_a − 6P_b\\ \{Q_{sb} = 12 + 2P_b \end{matrix}
(a) Write down the equilibrium condition for each good. Hence, deduce two equations in P_a and P_b .
(b) Use Cramer’s rule to find the equilibrium prices and quantities for goods A and B.
(a) The equilibrium condition for each good is that Q_s = Q_d .
For good A:
−15 + 7P_a = 30 − 8P_a + 2P_a
15P_a − 2P_b = 45
For good B:
12 + 2P_b = 28 + 4P_a − 6P_b
−4P_a + 8P_b = 16
Therefore, the simultaneous equations are
(1) 15P_a – 2P_b = 45
(2) –4P_a + 8P_b = 16
(b) Applying Cramer’s rule:
P_{a}={\frac{\Delta_{P_{a}}}{\Delta}}= \frac {\begin{vmatrix} 45 & -2 \\ 16 & 8 \end{vmatrix} }{\begin{vmatrix} 15 & -2 \\ -4 & 8 \end{vmatrix} } = \frac {(45)(8) − (16)(−2)}{(15)(8) − (−4)(−2)} = \frac {360 − (−32)}{120 − 8} = \frac {392}{112} = 3.5
P_{b}={\frac{\Delta_{P_{b}}}{\Delta}}= \frac {\begin{vmatrix} 15 & 45 \\ -4 & 16 \end{vmatrix} }{\begin{vmatrix} 15 & -2 \\ -4 & 8 \end{vmatrix} } = \frac {(15)(16) − (−4)(45)}{(15)(8) − (−4)(−2)} = \frac {240 − (−180)}{120 − 8} = \frac {420}{112} = 3.75
Substituting these values of P_a and P_b into any of the original equations, solve for Q_a and Q_b .
Q_a = 9.5, Q_b = 19.5.