Question

Need help with First Yr University Calculus Business Problem.

"Economists use production functions to describe how the output of a system varies with another variable such as labor or capital. For ex) the production f(x) P(L)=200L+10L^2-L^3 gives the output of a system as function of the number of laborers L. The avg product A(L) is the avg output per laborer when L laborers are working, that is A(L)=P(L)/L. The marginal product M(L) is the approximate change in output when one additional laborer is added to L laborers, that is M(L)=dP/dL.

Question: con't on bottom

Answer 1

(a) For the production function given here, compute and graph P, A and
M.

b) Suppose the peak of the average product curve occurs at L = L0,
so that A' (L0) = 0. Show that for a general production function, M(L0) = A(L0).

Answer 2

I suppose part a is just graph what you have, divide everything by l, graph again, take thederivative, graph again.

Answer 3

as in first graph P(L) = 200L + 10L^2 -L^3

Answer 4

?

Answer 5

yes. Theres no other way to define P without any values, so we have to assume calculating and graphing P is just to graph the original function.

Answer 6

ok and then A is equal to P(L)/L...that's why you were saying that the graph of A would just be the original graph divided by l, right?

Answer 7

and the last graph is M(L) = derivative of the previous graph...

Answer 8

Correct.

Answer 9

Do you think you could help me with just graphing the original graph? I'm sure i can get the second and third after that. Where should I start?

Answer 10

Well, we have a odd degree function function where the leading coefficient is negative. This means the graph starts from quadrant 2 and goes down and off the graph through quadrant 4. When you test for any sort of symmetry there is none. If you factor it, you get x-intercepts of x = -10 and x = 20. You can do quadratic formula to TRY and find critical points. The critical points (where the graph turns basically) are at around x = -5.5 and x = 12 I mean, thats about as much as you can get out of it since the numbers arent perfect. That would be the information to graph the first function, btw.

Answer 11

how did you know the graph starts at quadrant 2? I realize the graph has to go down, because as you said the leading coefficient is less than 0. just don't know why it starts at 2

Answer 12

Its just common graph behavior:



Think of it this way. Unless there is an asymptote or hole or something, a cubic function has a domain and range of all real numbers. So no matter what it HAS to come from quadrant 2.