1. ## optimization in price

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?

2. ## Re: optimization in price

1. a) You are not supposed to differentiate this. You are asked to GENERATE the equation you are given.

3. ## Re: optimization in price

okay is this correct then,

Total revenue is equal to the amount of customers times the price, if R is the revenue, p is equal the price, a is equal to the increase in price, q is equal to the amount of customers and b is to the amount of customers lost and x is equal to the amount of the number of times the price is increased. For every increase in x value there is a increase in price but a decrease in customers

R = (p+ax)(q-bx)

and when the price is 4+the increase times x, and the average customers is 250, and loss of customers is 50 times the value of x

R(x) = (4+x)(250-50x)

and so R(x) is dependent on x.