Help !! I need help solving a demand function...Please help?

Question

This is the demand function p=D(x)=1000/x

I need to find the average price in ($) over the damand intereval [400,600].

I need to use the Average Value of a continuous function f over [a,b}

Answer 1

A lot of it depends on how you mean 'average'. If it's simply the arithemetic average of the endpoint values on the interval, it's (D(400) + D(600))/2 which is 2.08333....
But if you want the actual mean value, it's the integral of 1000/x between 400 and 600 divided by the length of the interval (200) which is (1000*(ln(600) - ln(400))/200 or about 2.027325


Doug

Answer 2

Very interesting. At D(x) = 400, the price is 2.5.

At D(x) = 600, the price is 1.67.

At D(x) = 500, the price is 2.

Divide the interval 400 - 600 into tiny pieces such that the first interval is 400 to 400+a, the second is 400+a to 400+2a, and in general the nth interval is 400+(n-1)a to 400+na, so na = 200.

Each interval is so small that we can compute the price corresponding to each interval by using the starting value of the interval.

So the price for the first interval is 1000/400.

The price for the second interval is 1000/(400+a)

The price for the nth interval is 1000/[400+(n-1)a].

To find the average of all the prices, add 'em up and divide by n.

I guess this is why integral calculus was invented.

Answer 3

Sure,

The answer is $2.03 to the nearest cent.

The price "p" will be a function of x = the number of units sold.

The "integral" or area under the demand curve p = D(x) = 1000/x
between x=400 and 600 is equal to:

1000(log(600)-log(400)) = $405.5

where log(x) is the natural logarithm of x (on most calculators key = LN).

This is because the integral of 1/x dx = log(x)

So integral of D(x) = 1000*log(x)

Evaluated over the interval [400,600] it equals:

1000*log(600) - 1000*log(400)

To find the average price per unit over the interval [400,600] we simply divide $405.5 by 600-400=200.

So the average price per unit to the nearest cent is:

$405.5/200 = $2.03

Answer 4

600
∫1000dx/x = 1000(ln(600) - ln(400))
.400
wow nice...
average price = (1000/(600-400))Ln(600/400)
average price = (1000/200)Ln(1.5) = 5*0.4054651
average price = $2.03

arithmetic average = $(2.50 + 1.67)/2 = $2.08

geometric average = sqrt(2.50*1.67) = $2.04

Answer 5

..600
∫1000dx/x = 1000(ln(600) - ln(400))
.400

average price = (1000/(600-400))Ln(600/400)
average price = (1000/200)Ln(1.5) = 5*0.4054651
average price = $2.03

arithmetic average = $(2.50 + 1.67)/2 = $2.08

geometric average = sqrt(2.50*1.67) = $2.04