f the price charged for a candy bar is p (x) cents, where p (x) equals 162 minus StartFraction x Over 10 EndFraction , then x thousand candy bars will be sold in a certain city.

a. 1620-x^2

b. x=810

c. Maximum value revenue=$656,100

Step-by-step explanation:

(a) Total revenue from sale of x thousand candy bars

P (x) = 162 - x/10

Price of a candy bar=p (x) / 100 in dollars

1000 candy bars will be sold for

=1000*p (x) / 100

=10*p (x)

x thousand candy bars will be

Revenue=price * quantity

=10p (x) * x

=10 (162-x/10) * x

=10 (1620-x/10) * x

=1620-x * x

=1620x-x^2

R (x) = 1620x-x^2

(b) Value of x that leads to maximum revenue

R (x) = 1620x-x^2

R' (x) = 1620-2x

If R' (x) = 0

Then,

1620-2x=0

1620=2x

Divide both sides by 2

810=x

x=810

(C) find the maximum revenue

R (x) = 1620x-x^2

R (810) = 1620x-x^2

=1620 (810) - 810^2

=1,312,200-656,100

=$656,100