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Mathematics 7 Online
OpenStudy (sasogeek):

what is a homogeneous equation and how do you know if it displays constant returns to scale?

OpenStudy (unklerhaukus):

?

OpenStudy (unklerhaukus):

homogenous means a few things

OpenStudy (sasogeek):

ok i'll ask the real question so that you can see where i'm coming from, i'd like to figure out somethings about it on my own though..... one sec.

OpenStudy (sasogeek):

Show that the production equation \(\huge Q=A[bK^a+(1-b)L^a]^\frac{1}{a} \) is homogeneous and displays constant returns to scale

OpenStudy (unklerhaukus):

Q(?,?)

OpenStudy (amistre64):

spose you scale the variables by some constant amount (t is the usual generic that ive seen); if you can factor out the scalar completely, then the equation is homogenous

OpenStudy (amistre64):

if f(tx,ty) = t*f(x,y) its homogenous

OpenStudy (amistre64):

hmm, that kind of reminds me of the definition of an odd function .... i wonder if they are related

OpenStudy (unklerhaukus):

Q(A,K,L) ?

OpenStudy (sasogeek):

i do not know, i wasn't at the college when this assignment was given but it was given to me today and it's due on monday.

OpenStudy (unklerhaukus):

looks like expansion of length due to heat

OpenStudy (sasogeek):

it'd be nice to know which letters are variables and which ones are constants in this given function :/

OpenStudy (amistre64):

I think convention has it that capitals are constants \[\large Q=A[tbK^{ta}+(1-tb)L^{ta}]^\frac{1}{ta}\] \[\large Q=A[tbK^{ta}+L^{ta}-L^{ta}tb]^\frac{1}{ta}\] \[\large Q=A[tbK^{ta}+L^{ta}-L^{ta}tb]^\frac{1}{ta}\]

OpenStudy (amistre64):

and im just going on an idea here, not really sure if itll pan out

OpenStudy (amistre64):

\[\large Q=[tb(AK)^{ta}+(AL)^{ta}-tb(AL)^{ta}]^\frac{1}{ta}\] \[\large Q^{ta}-(AL)^{ta}=tb(AK)^{ta}-tb(AL)^{ta}\]

OpenStudy (unklerhaukus):

oou,

OpenStudy (amistre64):

if A not=0 i wonder if another route would have been easier ...

OpenStudy (amistre64):

any idea if im even on teh right track with this idea?

OpenStudy (sasogeek):

how come \(\large Q^{ta}\) ?

OpenStudy (amistre64):

say ta=3 Q = N^(1/3) [Q = N^(1/3)]^3 Q^3 = N^(3/3) Q^3 = N

OpenStudy (sasogeek):

oh, i see :)

OpenStudy (amistre64):

but i wonder if it would be prudent to separate t and a in that .... hard to tell

OpenStudy (unklerhaukus):

aahh,

OpenStudy (sasogeek):

but from what you have there, what are you to factor out to confirm if it's homogeneous or not? :/

OpenStudy (amistre64):

some exponential factor of t; if I can get rid of any semblense of the "t" such that it becomes a scalar instead ... then the equation would be definined as homogenous. Assuming i have the right definition of homogeneity to begin with

OpenStudy (sasogeek):

what if from the beginning, t wasn't even supposed to be mentioned and maybe your t, is same as the b, or a ? :/ i'm not sure cos i have no idea about homogeneity and i was just presented with this exercise lol, i've got quite some reading to do :/

OpenStudy (unklerhaukus):

where are the numbers ; (

OpenStudy (amistre64):

\[\large Q=A[tbK^{ta}+L^{ta}-L^{ta}tb]^\frac{1}{ta}\] \[\large \frac QA=[tbK^{ta}+L^{ta}-L^{ta}tb]^\frac{1}{ta}\] \[\large \left(\frac QA\right)^{ta}=tbK^{ta}+L^{ta}-L^{ta}tb\] \[\large \left(\frac QA\right)^{ta}=tb(K^{ta}-L^{ta})+L^{ta}\] t is just a generic setup, it doesnt matter what it equals to. If we make it more specific, than all we do is prove that it works or does not work for a specific case.

OpenStudy (amistre64):

im trying to recall ways that logs might be useful to us .... since ive got t stuck in an exponent

OpenStudy (sasogeek):

why did t go into the exponent in the first place?

OpenStudy (amistre64):

because im assume that a and b are variables in this setup; so we have to attach a generic scalar to the variables and see if we can pull it out

OpenStudy (sasogeek):

interesting :)

OpenStudy (amistre64):

but then again, Q would be variable as well since it is defined by the inputs ....

OpenStudy (amistre64):

maybe Q, A, K, and L are the variables?

OpenStudy (amistre64):

which is what unkle alluded to at the start :)

OpenStudy (unklerhaukus):

lower case are scalars

OpenStudy (sasogeek):

well we never know until we try it out to find out how things work out :/ i'm new to this anyway so anything to simplify the situation :)

OpenStudy (amistre64):

what class is this for?

OpenStudy (sasogeek):

computational mathematics

OpenStudy (amistre64):

... never heard of it :/ what have you been learning in prior chapters and do they relate to this?

OpenStudy (sasogeek):

I haven't had that class at all, I spent the whole week with the admissions and faculty office. I just received this exercise though so I'm yet to read about homogeneous functions but thought i'd ask here to start with :/

OpenStudy (amistre64):

i hope my framework is at least on the right track :) Itd prolly take me about a week trying to read thru the material for the class to be sure tho. good luck with it

OpenStudy (sasogeek):

thanks :) i'll try to do what you say and see what comes off it. attach t to the variables and try to factor it out. if it works, it's homogeneous, if not, it's not :) right?

OpenStudy (amistre64):

correct

OpenStudy (sasogeek):

this is some sort of calculus, right?

OpenStudy (amistre64):

if you can get rid of all the ts you put in; spose you end up with t^2 after factoring it all, that is acceptable as well. Not to sure how much of this has to do with calculus.

OpenStudy (sasogeek):

how about if you end up with t^a?

OpenStudy (unklerhaukus):

t ?

OpenStudy (amistre64):

if "a" was one of the variables to begin with ... im not sure.

OpenStudy (amistre64):

for example: f(x,y) = x + y^2 f(tx,ty) = tx + (ty)^2 = tx + t^2y^2 = t(x + ty^2) since we cant get rid of all the ts in the original setup, this equation would not be considered homogenous

OpenStudy (sasogeek):

ohhhhh :) nice! i think i'm getting the hang of this, so all that matters is if you know what the variables are.... :)

OpenStudy (amistre64):

that does help, yes

OpenStudy (sasogeek):

ok so usually, the function would have 2 variables right?

OpenStudy (unklerhaukus):

at least

OpenStudy (sasogeek):

ahhh, i was going to go ahead and say that since _b and _a are the only small letters, they're possibly the variables cos there's only 2 small letters :/ if we should consider AKL, that's 3 and rather odd, i think :/

OpenStudy (sasogeek):

is this correct? \(\large logAB^c=clogAB \) ?

OpenStudy (amistre64):

\[\large Q(A,K,L)=tA[tbK^{a}+(1-b)tL^{a}]^\frac{1}{a}\] \[\large Q(A,K,L)=tA[t(bK^{a}+(1-b)L^{a})]^\frac{1}{a}\] \[\large Q(A,K,L)=tt^aA[bK^{a}+(1-b)L^{a}]^\frac{1}{a}\] \[\large Q(A,K,L)=t^{(a+1)}~[A[bK^{a}+(1-b)L^{a}]^\frac{1}{a}]\]

OpenStudy (amistre64):

if A is a base, then yes

OpenStudy (sasogeek):

no, A is not a base :/ and oh, looks like you solved it and it appears homogeneous :D

OpenStudy (amistre64):

that google book helped me to get the variables right :)

OpenStudy (amistre64):

but i might have pulled out the wrong t exponent

OpenStudy (sasogeek):

I'd have to purchase it though :(

OpenStudy (amistre64):

t^(1/a) pulls out, not t^a t*t^(1/a) = t^(1+1/a) =t^((a+1)/a) typoes it :)

OpenStudy (sasogeek):

yh i just noticed :) thanks for pointing it out xD seems like an interesting topic though

OpenStudy (amistre64):

"displays constant returns to scale" the google book seems to be saying that: when the exponent value of t is less than 1, it displays a decreasing scale when the exponent value of t is equal 1, it displays a constant scale when the exponent value of t is greater than 1, it displays an increasing scale

OpenStudy (sasogeek):

so it doesn't display a constant scale.. :/

OpenStudy (amistre64):

recheck my math to make sure theres not a mistake :)

OpenStudy (sasogeek):

yes i'm trying to solve it myself on paper right now :) thanks again though xD

OpenStudy (amistre64):

good luck, thats about all i can do for it ;)

OpenStudy (sasogeek):

with the rest of the work I believe their questions i can solve on my own, basic algebra and statistics. thanks again though, can't thank you enough :)))) <3

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