A baseball team plays in a stadium that holds 55,000 spectators

(a) Assume the demand function is p(x)=ax+bp(x)=ax+b for some constants a,ba,b. We are given that p(38000)=10p(38000)=10 and p(42000)=8p(42000)=8. Thus

10=38000a+b...→18=42000a+b...→210=38000a+b...→18=42000a+b...→2

Subtracting 11 from 22,

we have −2=4000a−2=4000a or a=−12000a=−12000.

Substituting a=−12000a=−12000 into 11,

we have 10=38000⋅(−12000)+b10=38000⋅(−12000)+b

or

b=10+19=29b=10+19=29.

Therefore the demand function is p(x)=−12000x+29p(x)=−12000x+29

(b)The revenue function is r(x)=xp(x)=−12000x2+29xr(x)=xp(x)=−12000x2+29x with the boundary condition 0≤x≤550000≤x≤55000.

We have r(x)=−11000x+29r(x)=−11000x+29. To find the critical points of r(x)r(x), we set r(x)=0r(x)=0 to get11000x=29x=2900011000x=29x=29000.

Nowr(29000)=−12000⋅290002+29⋅29000=420050r(29000)=−12000⋅290002+29⋅29000=420050.

Thus, (29000,420050)(29000,420050) is the critical point.

At the boundaries, we have r(0)=−12000⋅02+29⋅0=0r(0)=−12000⋅02+29⋅0=0

r(55000)=−12000⋅550002+29⋅55000=0r(55000)=−12000⋅550002+29⋅55000=0

Therefore (29000,420050)(29000,420050) is the absolute maximum in the interval 0≤x≤550000≤x≤55000.

Sincep(29000)=−12000⋅29000+29=14.5p(29000)=−12000⋅29000+29=14.5,

So the ticket price should be set to $14.50$14.50 to maximize revenue.