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

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 $-3 \le x \le 5$

I'm not sure what to do, I rewrote the function as $(4+.2x)(400-10x$ 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

$(4+.2x)(250-10x)$ 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 $(4-.2x)(250+10x)$ 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.