The current ticket price at a local theatre is $4, and the theatre attracts an average of 250 customers per show. every $.20 increase in ticket price reduces average attendance by 10 customers, while every $.20 decrease results in 10 extra customers.

a) let x respresent the change in ticket price. show that revenue R from ticket sales depends of x according to $\displaystyle R(x) = (4+x)(250-50x)$.

I differentiated this and solved for x and found x=1/2 and because the rate of change is not zero the sales depend on the value of x. Correct?

b) if the seating capacity is 400, show that $\displaystyle -3 \le x \le 5$

I'm not sure what to do, I rewrote the function as $\displaystyle (4+.2x)(400-10x$ but I just ended up with a x value of 10 when I differentiated.

c) find the ticket price that will maximize revenue.

so I wrote

$\displaystyle (4+.2x)(250-10x)$ then I differentiated and came to a value of x = 2.5

then solve R(x) for x = -3, x = 2.5, x=5 and found the optimum price at x=2.5 for a price of $4.50

solving $\displaystyle (4-.2x)(250+10x)$ gave me the same result but x = -2.5

Are A and C correct, and what am I doing incorrectly for B?