Question

The manager of a CD store has found that if the price of a CD is P(X)=82-(X/6), then x CDs will be sold. Find an expression for the total revenue from the sale of x CDs (hint: revenue= demand x price) Use your expression. What is the maximum revenue?

Answer 1

Maximum revenue of `r=10086` at `x=246.`

Explanation:

It's usually crazy to maximize revenue instead of profit. Is this some sort of stock scam?

The price `p` is given by

`p = 82 - x/6`

The revenue `r` is

`r = px = 82 x- x^2/6`

Let's maximizing `r` by completing the square:

`r = -1 /6 ( x^2- (6)82 x) = -1/6( x^2 - 6(82) x + (6^2 41^2 ) ) + 6(41)^2`

`r = -1/6 (x - 3(82) )^2 + 6(41)^2`

`r` is maximized when `x=3(82)=246` so `r =6(41)^2=10086`

Answer 2

Maximum revenue is 10,086

Explanation:

We know a function for the price of the CD, which is a function of demand. We also know that Revenue is demand x price. Let's write a function that determines Revenue:

`R(x)=x xx P(x)`

`R(x)=x xx (82-x/6)`

`R(x)=(-x^2)/6+82x`

Now, take the derivative of the revenue function `R(x)`, and set it to zero. This will determine the maximum revenue. We know it will be a maximum because the function opens downwards (negative `x^2` term tells us that).

`(dR(x))/(dx)=2xx(-x)/6+82`

`(dR(x))/(dx)=(-x)/3+82`

`0=(-x)/3+82`

`x/3=82`

`color(purple)(x=246)`

Now, we know the demand required to optimize revenue. Plugging that back into `R(x)`:

`R(246)=(-246^2)/6+82(246)`

`R(246)=(-60516)/6+20172`

`R(246)=-10086+20172`

`color(green)(R(246)=10086)`