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

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