• February 17th 2006, 08:23 PM
bobby77
The total cost of producing x dictionaries is c(x)=10x+3000
(A) what is the average cost of producing 1000 dictionaries
(b)what is the marginal average cost at production level of 1000 units
(c) use the previous questions to estimate the average cost per dictionary, 171001 dictionaries are produced.
• February 17th 2006, 09:57 PM
CaptainBlack
Quote:

Originally Posted by bobby77
The total cost of producing x dictionaries is c(x)=10x+3000
(A) what is the average cost of producing 1000 dictionaries
(b)what is the marginal average cost at production level of 1000 units
(c) use the previous questions to estimate the average cost per dictionary, 171001 dictionaries are produced.

a)The avareage cost $AC(N)$ of producing $N$ dictionaries is the cost of producing N dictionaries divided by $N$, so is:

$AC(N)=c(N)/N$

So $AC(1000)=13000/1000=13$ units of cost per dictionary.

b)The marginal cost of production when production is at $N$ units is the
additional cost of producing the $N+1$ th unit (check the definition you have
been given to see if it agrees with this). So here the marginal cost is $10$ units of cost.
But that is not what the question asks for, it asks for the marginal avaerage cost
[Math]MAC(n)[/tex], which is a term which is unfamilliar to me, but I will assume this means
the change in average cost of producing one more item:

$
MAC(n)=AC(n+1)-AC(n)=$
$\frac{10(n+1)+3000}{n+1} - \frac{10n+3000}{n}=\frac{3000}{n(n+1)}
$

So:

$
MAC(1000)=\frac{3000}{1000 \times 1001}\approx 0.0030
$

c) The marginal average cost is an approximation to the rate of change
of the average cost, so let:

$
\frac{d}{dn}AC(n) =MAC(n)= \frac{3000}{n (n+1)}
$

hence:

$
AC(N)=AC(1000)+\int_{1000}^N \frac{3000}{n (n+1)} dn
$
$
=13+3000 \ln \left( \frac{1000}{1001} \frac{N+1}{N} \right)
$

So substituting $171001$ for $N$ gives:

$
AC(171001)\approx 13-2.981 = 10.019
$

RonL
• February 17th 2006, 10:47 PM
CaptainBlack
Quote:

Originally Posted by CaptainBlack
b)The marginal cost of production when production is at $N$ units is the
additional cost of producing the $N+1$ th unit (check the definition you have
been given to see if it agrees with this). So here the marginal cost is $10$ units of cost.
But that is not what the question asks for, it asks for the marginal avaerage cost
[Math]MAC(n)[/tex], which is a term which is unfamilliar to me, but I will assume this means
the change in average cost of producing one more item:

$
MAC(n)=AC(n+1)-AC(n)=$
$\frac{10(n+1)+3000}{n+1} - \frac{10n+3000}{n}=\frac{3000}{n(n+1)}
$

Given the use to be made of $MAC$ in the next section of
this problem, suggests that we take a central difference definition of
$MAC$:

$
MAC(n)=\frac {AC(n+1)-AC(n-1)}{2}
$
.

Then reworking part (c) gives:

$
AC(171001)\approx 13-2.982 = 10.018
$
.

RonL
• February 18th 2006, 10:08 AM
ticbol
Quote:

Originally Posted by bobby77
The total cost of producing x dictionaries is c(x)=10x+3000
(A) what is the average cost of producing 1000 dictionaries
(b)what is the marginal average cost at production level of 1000 units
(c) use the previous questions to estimate the average cost per dictionary, 171001 dictionaries are produced.

Here is one way.

a) Average Cost = (Total Cost) divided by (Total dictionaries made).

Total Cost = C(x) = 10x +3000 -----------(i)

Ave. Cost = C(x) / x
Ave. Cost = 10 +3000/x ---------------(ii)
So, for 1000 dictionaries,
Ave. Cost = 10 +3000/1000
Ave. Cost = 10 +3 = 13 monetary units per dictionary. -------answer.

-------------------------------------------------------------------------
b) Marginal Average Cost is the change in average cost due to producing one more dictionary.
In other words, marginal average cost is the first derivative of the Average Cost function.
Ave. Cost = 10 +3000/x
Marginal Ave. Cost = derivative of Ave. Cost with respect to x.
Marginal Ave. Cost = -3000/(x^2) -----------------(iii)

So, after 1000 dictionaries,
Marginal Ave. Cost = -3000/(1000^2) = -3/1000 mon.unit for the 1001st dictionary. ------answer.

That means for the 1001st dictionary, its average cost is 13 +(-0.003) = 12.997 monetary units.
It means also that the average cost for all of the 1001 dictionaries is 12.997 monetary units.
Meaning, the 1001st book lowered the average cost of every book made so far.

-------------------------------------------------------------
c)Find the ave. cost to produce 171,001 dictionaries.

c.1) Using the Ave. Cost function,
Ave. cost = [10(171,001) +3000]/(171,001) = (1,713,010)/(171,001) = 10.01754 mon.units.
Or,
= 10 +3000/x = 10 +3000/(171,001) = 10 +0.01754 = 10.01754 mon.units.

-------------------------
c.2) Using the Marginal Average Cost function,
Producing the 1st dictionary, its cost is = 10(1) +3000 = 3010 mon.units.
Its average cost? 3010 mon.units, of course.
(How can there be average cost when only one dictionary is produced? Umm, this is Math. Just roll your eyes. Or pretend that the average there is enclosed in quotations.)

The average cost of any dictionary is [3010 -3000/(x^2)], using the marginal thing.
So for 171,001 dictionaries, where x is from 1 to 171,001:
Ave. Cost = 3010 +INT.(1->171,001)[-3000/(x^2)]dx
Ave. Cost = 3010 -3000*INT.(1->171001)[x^(-2)]dx
Ave. Cost = 3010 -3000*[-1 *x^(-1)](1->171001)
Ave. Cost = 3010 +3000*[1/x](1->171001)
Ave. Cost = 3010 +3000*[1/171001 -1/1]
Ave. Cost = 3010 +3000*[0.000005848 -1]
Ave. Cost = 3010 +3000*[-0.999994152]
Ave. Cost = 3010 -2999.962456
Ave. Cost = 10.01754 mon.units ------------------answer.