AVERAGE COST Vs AVERAGE COST OVER A RANGE OF
VALUES
~One of the most confusing aspects in the beginning
stages of business calculus is treating the “average cost of producing x units
of an item” and the “average cost of producing a number of units in a range of
values” in the same way. For example, would the average cost of producing 50 units be the same
as the average of cost of producing the first 50 units? The answer is definitely
NO! This is where the confusion
occurs. I will try to explain this difference.
~The first case deals with finding the average cost
function, denoted by C(x)/x, at a given value of x.
To do this, we simply substitute the x value into this function and read the answer. The answer will be in amt per
unit.
~For example, if the cost function is C(x) = x2- x + 200, then
the average cost function is x – 1 + 200/x. If we were interested in the average
cost of producing 50 units, we would have 50 – 1 + 200/50 = 53 per
unit.
~Taking a closer look at the average cost function above, we see that there is a vertical asymptote at x=0 and a slant asymptote,
y=x-1. Since we are dividing by the number of units, x, this function approaches
infinity as x approaches 0. As the number of units increase, this function approaches the line y=x-1 from above,
since the last term in the average cost function, 200/x, approaches
zero.
~Also, since the average cost function may be represented as [C(x) – 0] / (x – 0), we see that
this gives the slope of a line connecting (0,0) to (x, C(x)), for the value of x considered. Keep this
geometric view in your mind when we investigate the second case.
~On the other hand, the average cost function over an interval of x values would
yield a very different outcome.
~In this case, we are required to find the average value of a function over an
interval. This is a popular application of integration. The formula used would be as
follows:
~The average value (y-value) of a function, y = f(x), over [a;b] is given
by 1/(b-a)∫abf(x)dx
~Geometrically, this has a completely different meaning
from that in the first case. It relates the area under the cost function to the
area of one rectangle that has exactly the same numerical value. The height of
that rectangle is defined as the average value of the function over [a;b]. In
essence, we are averaging an infinite number of functional values from x=a to
x=b, hence, the integration process.
~In the example given in this discussion, the cost function is C(x) = x2- x + 200. Instead of asking for
the average cost of producing 50 units, let’s pose our question a little
differently, “what would be the average cost of producing the first
50 units?” In this case, we would want to find the average value of the cost function over the interval
[0;50]. Using the average value integral above, we get:
1/(50-0)∫050(x2- x + 200)dx = 1008.33
The answer of 1008.33 represents the average height of the cost function over all values of x from 0 to 50.
~Note: If we simply take the average of the
values of the cost function at x=0 and x=50, we would get 1325, not the correct
answer. It will only work, if the cost function is linear (in this case it's
quadratic).