Ans: ticket price should be= $ 350
soln:
current revenue=$ 200*40,000=$ 8,000,000
now, let they suppose increase the price by=$ 10*x, x is a natural number
no ticket loss per week will be= 800*x
now, revenue =f(revenue)=(200+10*x)*(40,000-800*x)= -8,000*x2+240,000*x+8,000,000
for maximum or minimum--differentiating f(revenue)=0
=>-16000*x+240,000 = 0
=>x=15
by double differentiating f(revenue) we get <0
=>at x=15 it will be maximum value.
Ans:ticket should be= $200 + $ 10*15 =$ 350.
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