Question #146512

The Mohan Bagan football team plays in a stadium with a seating capacity of 1,80,000. How
ever, during the past season, attendance averaged only 1,50,000. The average ticket price was 30. If price elasticity is -4, what price would the team have to charge in order to fill
27 and the average attendance increased the stadium? If the price were to be decreased to
to 1,60,000, what is the price elasticity?

Expert's answer

P1=30,P2=?,Q1=150000,Q2=180000P_{1} = 30, P_{2} = ?, Q_{1} = 150000, Q_{2} = 180000

-4 = ΔQ(Q1+Q2)/2÷ΔP(P1+P2)/2=(180000−150000)(150000+180000)÷(P2−30)(30+P2=30000330000×(30+P2)P2−30\frac{\Delta Q}{(Q_{1}+Q_{2})/2} \div \frac{\Delta P}{(P_{1}+P_{2})/2} = \frac{(180000-150000)}{(150000+180000)} \div \frac{(P_{2}-30)}{(30+P_{2}} = \frac{30000}{330000}\times \frac{(30+P_{2})}{P_{2}-30}

−4-4 =0.091×30+P2P2−30=0.091\times \frac{30+P_{2}}{P_{2}-30} =−43.956=30+P2P2−30=-43.956=\frac{30+P_{2}}{P_{2}-30}

−43.956P2+1318.68=30+P2-43.956P_{2}+1318.68=30+P_{2}

1288.68=44.956P21288.68=44.956P_{2}

P2=28.665P_{2} = 28.665

Price elasticity of demand =Q2−Q1Q1×100%÷P2−P1P1×100%=160000−180000180000×100÷27−28.66528.665×100= \frac{Q_{2}-Q_{1}}{Q_{1}}\times 100\% \div \frac{P_{2}-P_{1}}{P_{1}}\times 100\%=\frac{160000-180000}{180000}\times 100 \div \frac{27-28.665}{28.665}\times 100

=−11.11−5.81=1.912=\frac{-11.11}{-5.81}=1.912


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