Average Delivered Price


Apr 2010
I am a little confused on how to calculate part a, please could some one assist me on this.

The question is as follows:
A manufacturer who sells to customer located within R km of his mill charges a per unit price m plus a shipping charge of s per kilometre. Thus a customer located x km from the mill would pay

p:= m + s.x 0≤ x ≥ R
per unit.

Our manufacturer now wants to know the average delivered price per unit. This clearly depends on the distance of each customer from the mill and how much each customer buys.

This information is summarised by a function f(x) that gives the distribution of sales by distance. In other words, f(x) gives the number of units sold to customers located x km from the mill.

(a) Using the simple idea that the average delivered price (A) is total revenue divided by total number of units sold for customers located within R km of the mill, derive an expression for A in terms of m, s, R and f(x).
(b)Calculate a numerical value for A for the case m=100, s=1, R=30 and f(x)=900 – x2
Mar 2010
The total revenue is given by integrating:

revenue = \(\displaystyle \int_0^R(m+sx)f(x)\ dx\)

and the number of units sold is also given by integrating:

units sold = \(\displaystyle \int_0^Rf(x)\ dx\)

So the average price is given by:

\(\displaystyle A=\frac{\int_0^R(m+sx)f(x)\ dx}{\int_0^Rf(x)\ dx}\)

I'm sure you can handle part (b). Post again in this thread if you have any trouble.

- Hollywood
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