The marketing research department of a computer company used a large city to test market the firm's new laptop. The department found the relationship between price p (dollars per unit) and the demand x (units per week) was given approximately by the following equation. p= 1275 = 0.17x^2 0 < x < 80

So, weekly revenue can be approximated by the following equation.

R(x)= rp = 1275x- 0.17x^3 0 < x <80

Required:
a. Find the local extrema for the revenue function. What is/are the local maximum/a?
b. On which intervals is the graph of the revenue function concave upward?
c. On which intervals is the graph of the revenue function concave downward?

a. Local maximum = 50 units per week.

b. The graph is never concave upward.

c. (0, 80)

Step-by-step explanation:

a. The revenue function is:

The derivate of the revenue function for which R'(x) = 0 gives us the local extrema:

The second derivate of the revenue function determines if x =50 is local maximum or minimum:

Since the second derivate yields a negative value, x = 50 units per week is a local maximum.

b. Since there are no local minimums in the range of 0 < x < 80, the graph is never concave upward.

c. Since there is only one local maximum  in the range of 0 < x < 80, the graph is concave downward from x>0 to x<80 or (0, 80)

i think it would be 43 movies that would have to be watched.

step-by-step explanation:

think about it. for the second plan, every month is 8 dollars. 8 x 12= 96. so thats the number for the second plan. now for the first plan, its annual meaning you have to pay 10 dollars every year. so now you have to get an equivalent amount of 86 since 96-10=86. 86 divided by 2 since every movie costs 2$ then you would have to watch 43 movies a year to get the same as the second plan. i really hoped this . have a wonderful day as well!