If the price of a candy bar is p(x) cents then x thousand candy bars are sold. The price p(x) = 124-(x/16). How many candy bars must be sold to maximize revenue?
p(x) = 124-x/16
P(x) is the price of a candy bar. Then if we assume we sold x thousand of candy bars the revenue R(x) = p(x)*x
So R(x) = (124-x/16)*x = 124x-x^2/16
To R(x) to be a maximum or minimum R'(x) = 0
But if R(x) is a maximum then R''(x) <0 or negative.
R'(x) = 124-2x/16
If R(x) is a maximum or minimum R'(x)= 0
Then 124-2x/16 = 0
Lets consider R''(x)
R''(x) = 0-2/16 = -1/8<0 negative value.
This means R(x) is a maximum. R(x) is maximum when x=992 thousand.
So to get maximum revenue you should sell 992000 candy bars.
The price of a candy bar is given as a function of the number sold as `p(x) = 124-(x/16)` where x is in 1000s
If the price of the candy bar is p(x), the revenue function `R(x) = p(x)*x = 124*x - x^2/16`
To maximize R(x), find the solution of R'(x) = 0
`R'(x) = 124 - x/8`
`124 - x/8 = 0`
=> `x = 992`
As x is in 1000s, the number of candy bars that must be sold to maximize revenue is 992000