View Question
Q: Calculus (Marginal Product) ( Answered ,   1 Comment )
 Question
 Subject: Calculus (Marginal Product) Category: Science > Math Asked by: eddiehosa-ga List Price: \$20.00 Posted: 17 Oct 2004 11:07 PDT Expires: 16 Nov 2004 10:07 PST Question ID: 416076
 ```I am having some problems understanding this math question related to calculus and marginal product. Assume that the marginal product of an additional skilled worker is twice the marginal product of an additional unskilled worker, and that unskilled workers are paid 2/3 the amount that skilled workers receive. With a fixed overall budget, a company that wishes to maximize its quantity of output from its workers should do which of the following (choose one from below)? A - Hire equal numbers of skilled and unskilled workers. B ? Hire more skilled and unskilled workers. C ? Hire more skilled workers and discharge unskilled workers. D ? Discharge skilled workers and hire more unskilled workers. E ? Discharge all skilled workers and half of the unskilled workers. - What formula do you actually begin with? (i.e. the formula before differentiation) - Which one of the above is the correct answer - Please show working out using calculus / derivatives / etc. - Could you also give short explainations how you did it, or perhaps, links to similar questions on the website? (Note: I have Grd 12 level calculus therefore teaching derivatives, etc insn't reqired) Some extra details: If P(n) is the product, P?(n) would be the marginal product. Thanks a lot!```
 ```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```
 eddiehosa-ga rated this answer: and gave an additional tip of: \$2.00 ```Very detailed and very good explanation - even if you didn't understand something he/she'll explain it very well to you. 5 star service!```
 ```I'd say that's definately \$20 worth of math :) Fine work Mathtalk!```