# 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?

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

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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

x= 992

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.

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