Question

The average ticket price for a concert at the opera house was $50. The average attendance was 4000. When the ticket price was raised to $52, attendance declined to an average of 3800 persons per performance. What should the ticket price be to maximize the revenue for the opera house?

Answer 1

Answer

given,

opera house ticket = $50

attendance = 4000 persons

now,

opera house ticket = $52

attendance = 3800 person

assuming these are the points on the demand curve

(x, p) = (4000,50) and (x,p) = (3800,52)

using point slope formula

$p-50 = \dfrac{50-52}{4000-3800}(x - 4000)$

$p-50 = \dfrac{-2}{200}(x - 4000)$

$p-50 = \dfrac{-x}{100}+ 40$

$p = \dfrac{-x}{100}+ 90$

R(x) = x . p

$R(x) = x (\dfrac{-x}{100}+ 90)$

$R(x) = \dfrac{-x^2}{100}+ 90x)$

$\dfrac{d}{dx}(R(x)) = \dfrac{d}{dx}(\dfrac{-x^2}{100}+ 90x))$

$\dfrac{d}{dx}(R(x)) = (\dfrac{-2x}{100})+90)$

at $\dfrac{d}{dx}(R(x)) = 0$

$\dfrac{-2x}{100}= -90$

x = 4500

$\dfrac{d^2}{d^2x}(R(x)) = -ve$

hence at x =4500 the revenue is maximum

for maximum revenue ticket price will be

$p = \dfrac{-4500}{100}+ 90$

p = $45