Pembahasan Ujian PAI: A30 – No. 6 – November 2018

Masukkan P=0 dan Q=0

• Ketika P = 0
  \(P = 90 – 0,5{Q_d}\) \(0 = 90 – 0,5{Q_d}\) \(0,5{Q_d} = 90\) \({Q_d} = 180\)

• Ketika Q = 0
  \(P = 90 – 0,5{Q_d}\) \(P = 90 – 0,5(0)\) \(P = 90\)
• Sehingga \(\left( {P,Q} \right) = \left( {90\,,\,\,180} \right)\)

Consumer Surplus ada 2

1. \(Cs = \int\limits_0^{{Q_d}} {f\left( Q \right)\,dQ} – \left( {{Q_d} \cdot {P_d}} \right)\)
2. \(Cs = \int\limits_{{P_d}}^P {f\left( P \right)\,dP} \)

Dengan rumus 1
\(Cs = \int\limits_0^{{Q_d}} {f\left( Q \right)\,dQ} – \left( {{Q_d} \cdot {P_d}} \right)\) \(= \int\limits_0^{50} {\left( {90 – 0,5Q} \right)dQ} – \left( {50 \cdot 65} \right)\) \(= \left[ {\left. {90Q – \frac{{0,5{Q^2}}}{2}\,} \right|_0^{50}} \right] – 3250\) \(= \left[ {\left( {90 \cdot 50} \right) – \frac{{0,5\left( {{{50}^2}} \right)}}{2}\,\, – \,0} \right] – 3250\) \(= 3875 – 3250 = 625\)

Dengan rumus 2
\(Cs = \int\limits_{{P_d}}^P {f\left( P \right)\,dP} \) \(= \int\limits_{65}^{90} {180 – 2P\,\,dP} \) \(= \left. {180P – \frac{{2{P^2}}}{2}\,\,} \right|_{65}^{90}\) \(= \left[ {180\left( {90} \right) – \frac{{2\left( {{{90}^2}} \right)}}{2}} \right] – \left[ {180\left( {65} \right) – \frac{{2\left( {{{65}^2}} \right)}}{2}} \right]\) \(= 8100 – 7475 = 625\)