Question

Demand The demand function for a product is given by

p = 8000(1 - 5/5 + e^-0.002x)
where p is the price per unit (in dollars) and x is the number of units sold. Find the numbers of units sold for prices of (a) p = $200 and (b) p = $800.

Answer 1

`x= -(ln [(5P)/(8000-P)])/(0.002)`

a)
`x= -(ln [(5*200)/(8000-200)])/(0.002) =1027.062 \approx 1027`

b)
`x= -(ln [(5*800)/(8000-800)])/(0.002) =293.893 \approx 294`

Explanation:

For this case we have the following function:

`P= 8000 (1- (5)/(5 +e^(-0.002 x)))`

We can solve for x like this. First we can reorder the expression like this:

`(P)/(8000) = 1- (5)/(5+e^(-0.002x))`

`(5)/(5+e^(-0.002x)) = 1 -(P)/(8000) = (8000-P)/(8000)`

`(40000)/(8000-P) = 5 + e^(-0.002x)`

Now we can apply natura log on both sids and we got:

`ln[(40000)/(8000-P) -5] = ln e^(-0.002x)`

`ln [(5P)/(8000-P)] = -0.002x`

And if we solve for x we got:

`x= -(ln [(5P)/(8000-P)])/(0.002)`

Part a

For this case we can replace P = 200 and see what we got for x like this:

`x= -(ln [(5*200)/(8000-200)])/(0.002) =1027.062 \approx 1027`

Part b

For this case we can replace P = 800 and see what we got for x like this:

`x= -(ln [(5*800)/(8000-800)])/(0.002) =293.893 \approx 294`