Hi, eddiehosa-ga:
In economics "marginal product" has the meaning of the rate of
additional output per unit of additional input.
Here we have a problem involving two inputs (the number of skilled
workers and the number of unskilled workers), with a constraint on
costs (a fixed budget determined by the relative cost of the skilled
vs. unskilled workers).
According to the given information, "the marginal product of an
additional skilled worker is twice the marginal product of an
additional unskilled worker." Furthermore the cost of a skilled
worker is 3/2 times the cost of an unskilled worker (because
"unskilled workers are paid 2/3 the amount that skilled workers
receive").
We are not told the actual budget or how many skilled or unskilled
workers are currently employed by the company, so there is no way to
pin down an actual number of workers to hire or fire.
But we can argue that the correct answer is C: Hire more skilled
workers and discharge unskilled workers.
To maintain the current budget this would require hiring two new
skilled workers for every three unskilled workers who are discharged.
Of course this can only continue until there are no unskilled workers
(or perhaps, fewer than 3) left.
But for every exchange of 3 unskilled workers for 2 skilled one, the
gross product (output) increases by the marginal output of one
unskilled worker. Let's see how to prove this using algebra.
Let P denote the gross product, and P' denote the marginal product
with respect to adding an unskilled worker.
Then 2P' is the marginal product with respect to adding a skilled worker.
Let s be the number of additional skilled workers and t the number of
additional unskilled workers (negative values would of course mean
decreases in one labor pool or the other).
A fixed budget constrains us to have:
s + (2/3)t = 0
or to put it differently:
t = -(3/2)s
So if there were a unit increase in skilled workers, offset by the
corresponding change t = -(3/2)s in unskilled workers to keep the
budget constant, then production output would change by:
s * (2P') + t * P' = (2s - (3/2)s) * P'
= (1/2)s * P'
In particular for the case we described, s = 2 and t = -3, the
increase in output is P' the marginal product per one more unskilled
worker.
You may wonder why there is no calculus in this solution. The reason
is that we are not given an explicit formula for the gross product P
that relates to the number of skilled and unskilled workers, nor do we
have any information about the "operating point" (existing levels of
skilled and unskilled workers). Because of this all we can discuss is
whether the net result of increasing skilled workers and decreasing
unskilled workers (in proportion to their relative wage costs) is
going to increase or decrease product output.
Since the limited calculation we are able to do with the given
information does show an increase in production when (for example) 3
unskilled workers are replaced by 2 skilled ones, the conclusion C
(hire more skilled workers and discharge unskilled workers) is
justified. Careful reading of the four other answer choices shows
that none of the alternatives is tenable.
A and B offer to increase both the numbers of skilled and unskilled
workers, which conflicts with the requirement to maintain a fixed
budget.
D would decrease the number of skilled workers and increase the number
of unskilled workers, which if it were done in a budget neutral manner
would have precisely the opposite effect of our solution, i.e. would
decrease production output.
E would decrease both the skilled and unskilled workers, so obviously
production would decrease (contrary to our goal).
I'd be pleased to offer further Clarification if my solution to this
problem is not clear enough.
regards, mathtalk-ga |
Request for Answer Clarification by
eddiehosa-ga
on
17 Oct 2004 15:18 PDT
Hey mathtalk-ga. Thanks a lot for your help. I understand how you got
the answer at the end by using logic, but I am still having
difficulties with trying to solve it algebraiclly - and I'm not
understanding 100% your algebraic method... I'm still trying to relate
it with calculus as it's a calculus chapter assignment.
Two questions:
1. What does P' stand for? (Can you write it in leibniz notation?)
2. This is what I'm thinking:
Let P be the productivity
Let x be the production of skilled worker, y be the production of
unskilled worker, therefore y = 0.5x - where they are both constants.
Again, as before, s is the no. of skilled workers, t is the no. of
unskilled workers.
Productivity = no. of skilled workers * production of skilled worker +
no. of unskilled workers * production of unskilled worker
P = sx + ty
P = sx + t0.5x
P(s) = sx + (-3/2s)0.5x
P(s) = sx - 3/4sx
P'(s) = x - 3/4x (remembering that X is a constant)
P'(s) = 1/4x
This is as far as I went... does it make sense or not? Hope you can
enlighten me on this note.
|
Clarification of Answer by
mathtalk-ga
on
17 Oct 2004 17:26 PDT
Hi, eddiehosa-ga:
I was trying to use your notation with P', as far as the problem
allows it to be pinned down. You said P(n) is the product and P'(n)
is the marginal product, without being specific about what n means.
In this problem the output P depends on two inputs, the number of
skilled workers and the number of unskilled workers. We are not told
the exact marginal product with respect to either input, but we are
told that the marginal product of an additional is skilled worker is
twice the marginal product of an additional unskilled worker. We are
also told:
a) the budget (for wages) is fixed
b) an unskilled worker is paid 2/3rds of what a skilled worker is paid
When you have a function P depending on two variables, the two
"marginal" rates are known to mathematicians as partial derivatives.
I didn't try to drag in the terminology since I'm not sure it means
something to you, but the issue of two inputs must be managed one way
or another.
The way I tried to present the problem is by reducing the number of
unknowns from two inputs to one, by using the fixed budget constraint.
That is, if s is the number of additional skilled workers and t is the
number of additional unskilled workers, we have to have:
s + (2/3)t = 0
so that the budget remains unchanged. I used this to solve for t in
terms of s, with the obvious consequence that when s is positive, t is
negative (i.e. there's actually a decrease of three unskilled workers
for every increase of two skilled workers):
t = -(3/2)s
So I explained this first in words and then in algebra.
Now in calculus there is something called the chain rule which tells
us how to compute the derivative when functions are composed. We are
not told the actual number of skilled workers or the number of
unskilled workers currently employed (there might, for all the problem
tells us, be zero unskilled workers now, which turns out to be the
optimum). But we can use the chain rule to relate the information we
do have to the question of whether increasing skilled workers is a
good thing (increases production) or a bad thing (decreases
production).
Bearing in mind the result above that an increase (hiring) of s
skilled workers must be matched by a decrease (discharge) of (3/2)s
unskilled workers (because t = -(3/2)s), we use the chain rule to
combine the marginal product for skilled workers with the marginal
rate for unskilled workers.
The notation P' that I used is the marginal product for one unskilled
worker (look back at my Answer and you'll see that I define it this
way). I chose this because it makes it easy to state the marginal
product for one skilled worker also, namely 2P'.
The chain rule says that when we change the skilled workers by s and
the unskilled workers by t, the overall change to product (output) is:
s * (marginal product of one skilled worker)
+ t * (marginal product of one unskilled worker)
Of course this is saying intuitively that the two changes (effects of
"extra" skilled workers and of "extra" unskilled workers) are
additive. When either s or t is negative, the arithmetic produces a
corresponding negative adjustment to output.
Plugging in respectively 2P' for the marginal product of one skilled
worker and P' for the marginal product of one unskilled worker, the
expression above will become simplified:
s * (2P') + t * P'
and we can further simplify it by using the relationship t = -(3/2)s to get:
s * (2P') - (3/2)s * P' = (2 - 3/2)s * P'
= 0.5 s * P'
Again the problem does not state an actual number of workers, skilled
or unskilled, nor does it state an actual formula for P in terms of
the unknowns.
Everything the problem tells us boils down to the expression above for
how the product (level of output) changes when we hire some number s
of additional skilled workers and offset those hires by discharging
(3/2)s unskilled workers.
[Naturally it's considered impolite to discharge half a worker, so I
preferred to phrase it as hiring two skilled workers and discharging
three unskilled workers. But it's only a math problem!]
We don't know the value of P', only that it is the marginal product
for one unskilled worker (presumably a positive value!). We don't
know how many unskilled workers there are to discharge (if any). All
one can deduce is that it would increase production by 0.5 s * P' if
it were possible to hire s additional skilled workers (and discharge
(3/2)s unskilled workers).
Therefore we can conclude that out of the multiple choice, only C is
true. It would be good to hire more skilled workers and discharge
unskilled workers.
The problem gives you very nearly the minimum amount of information
needed to support this conclusion. This sort of exercise seems
intended to help you grasp the application of ideas, not the mechanics
of taking derivatives, because there are no explicit formulas to
differentiate, are there?
Now let's look at your approach. You decided to create a formula for
P that matches in essential respects what we know about it. You
defined:
s := number of skilled workers
x := marginal product of a skilled worker
t := number of unskilled workers
0.5x := marginal product of an unskilled worker
This is subtly different from my definitions, in which s and t are
adjustments to the numbers of skilled and unskilled workers, resp.
But if it helps you to see the relationships, then this isn't a
terribly important point.
You then wrote:
P = sx + t(0.5x)
and proceeded to simplify things using t = -(3/2)s. Here is the
reason I was careful to define s and t and increments to the numbers
of workers. We actually cannot say the number of unskilled workers is
-3/2 times the number of skilled workers (because who wants negative
workers?). But we can get around this by using a bit of
differentiation!
Take the derivatives on both sides with respect to s (number of
skilled workers), and we get:
dP/ds = x + (dt/ds)(0.5x)
We can say that dt/ds = -3/2, ie. that the rate of change in unskilled
workers with respect to change in skilled workers is minus
three-halves. Thus:
dP/ds = x - (3/2)(0.5x) = (1/4)x
This is equivalent to what I got, even though it looks a bit
different. The difference is that my P' was the marginal product of
an _unskilled_ worker (a choice made out of laziness so I could use
2P' for marginal product of a skilled worker, rather than have x and
0.5x as you set it up). If you recall P' = 0.5x, then the two results
are seen to agree completely. Hiring an additional skilled worker
changes the overall output by:
dP/ds = (1/4)x = 0.5 P' [in my notation]
Again I think the point of this exercise is more about grasping the
idea of an application of derivatives, rather than developing facility
in differentiation.
By working the problem in your own terms, I think you've pretty well
accomplished that purpose.
regards, mathtalk-ga
|
Clarification of Answer by
mathtalk-ga
on
17 Oct 2004 18:57 PDT
[Warning: a few Unicode characters ahead, which may not display
properly in every browser!]
A bit of "Leibniz" (or Leibnitz) notation:
Let P(s,t) be the production function of two independent variables:
s := # of skilled workers
t := # of unskilled workers
The marginal product functions are then partial derivatives:
?P/?s := marginal product of one skilled worker
?P/?t := marginal product of one unskilled worker
If we further constrain s,t so that t = t(s) is a function of s, then
the combined effect of s on P, acting directly as the first argument
and also indirectly through its effect on the second argument t, is
called the "total derivative". We have this expression for the total
derivative of P with respect to s, courtesy of the chain rule:
dP/ds = ?P/?s + (?P/?t)(dt/ds)
Keeping the distinction between these different flavors of derivatives
is important not only for economic applications, but in many
engineering contexts as well (such as fluid dynamics).
regards, mathtalk-ga
|
Request for Answer Clarification by
eddiehosa-ga
on
17 Oct 2004 20:59 PDT
Thank you so much!!! Everything makes a lot of sense. One last
question to bother you... how would I be able to 'disprove' the other
choices using algebraic methods / calculus and not by logic... i.e.
Choice A, C, D, E. (like you said discharging skilled workers and half
of skilled workers doesnt work towards the goal - is there a way to
represent that?)
|
Request for Answer Clarification by
eddiehosa-ga
on
17 Oct 2004 21:15 PDT
Sorry - I meant A, B, D, E for "disproving" using calculus.
|
Clarification of Answer by
mathtalk-ga
on
18 Oct 2004 05:18 PDT
Of the five alternatives, only C is consistent with what "calculus" tells us.
Bear in mind that "calculus" showed us an output improvement comes
(under the constraint of a fixed budget) from hiring two skilled
workers for every three unskilled workers that can be discharged.
Realizing that this is consistent with C (which merely says hire
skilled workers and discharge unskilled workers) is "logic". Actually
C is less specific than the conclusion given by our calculation, but
consistent with it.
As we discussed before, the other four choices are logically
inconsistent with what "calculus" tells us. A and B involve hiring
more of both, which violates the fixed budget; E asks us to discharge
some of both, which not only violates the fixed budget but would
obviously cut productivity. D is discharging skilled workers and
hiring more unskilled works, which is contrary to what "calculus"
tells us would improve things.
So calculus gives us the direction in which improvement is achieved.
After that one has to read the five alternatives to determine if one
answer is the best, as with a standard multiple choice test.
regards, mathtalk-ga
|
Request for Answer Clarification by
eddiehosa-ga
on
18 Oct 2004 19:11 PDT
Sorry to bother you again, but I just read your Leibniz notation part
and dont understand what you wrote by saying:
dP/ds = ?P/?s + (?P/?t)(dt/ds)
Do you mind interpreting it for me?
|
Clarification of Answer by
mathtalk-ga
on
18 Oct 2004 20:15 PDT
The "curly" delta's signify the partial derivatives. As you will see
in the earlier discussion, the partial derivative of P with respect to
the number of skilled workers is the marginal product of one
additional skilled worker.
Likewise the partial derivative of P with respect to the number of
unskilled workers is the marginal product of one additional unskilled
worker.
In the expression you are asking about, I describe the "ordinary"
derivative as the "total derivative" of P with respect to s, given
that t is also treated as a function of s, i.e. t = t(s).
With this setup we can reduce P to a function of s alone:
P(s,t(s))
and ask for its ordinary (total) derivative with respect to s.
This is the meaning of the formula:
dP/ds = ?P/?s + (?P/?t)(dt/ds)
The rate at which P changes per unit change in s is the sum of these two terms:
dP/ds = (?P/?s)(ds/ds) + (?P/?t)(dt/ds)
But of course the derivative ds/ds = 1 always, so it immediately
simplifies to the expression above.
If you haven't studied partial derivatives yet, the notation and the
concepts may be unfamiliar to you. It helps to have some additional
examples to compare it to. One source of such examples would be
"related rate" problems from a first year calculus course. Remember
the pile of sand, in the shape of a cone? Sand is added to the pile
at a certain rate, and the height and circumference of the cone remain
in some fixed proportion.
Here volume V = (1/3)pi R^2 h, where R is the radius of the cone's
base and h is the height of the cone.
Both R and h are changing as a function of time t. The rate at which
the volume is changing dV/dt is "related" to the rates at which the
radius R and the height h are changing by simple differentiation:
dV/dt = (1/3)pi * [R^2 * dh/dt + h * (2R dR/dt)]
Another way to look at the problem is to treat V as a function of two variables:
V(R,h) = (1/3)pi * R^2 * h
Then ?V/?R = (2/3)pi * R * h
and ?V/?h = (1/3)pi * R^2
We can then express the total derivative of V with respect to time as:
dV/dt = (?V/?R)(dR/dt) + (?V/?h)(dh/dt)
Knowing a relationship between dR/dt and dh/dt (because of the
constant proportion between them) is pretty similar to your problem,
where we knew:
dt/ds = -3/2
and of course that ds/ds = 1.
Naturally there are infinitely many examples and almost as many topics
to which a discussion of calculus could be extended. I wasn't trying
to force you to learn every possible way to look at the problem, but
hopefully looking at it from a couple of different perspectives gives
an impression of its real depth.
regards, mathtalk-ga
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