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Production and Isoquants

 
 
Reply Fri 13 Oct, 2017 01:24 pm
As with basically everything in Macroeconomics, I get the logic but can't translate it into Math.

I'm presented with the following problem:
(27 points) For each of the following production functions, sketch a representative isoquant (2 points).
Calculate the marginal product for each input, and indicate whether each marginal product is diminishing,
constant, or increasing (3 points). Calculate the marginal rate of technical substitution for each
function (2 points). Also indicate whether the function exhibits constant, increasing, or diminishing
returns to scale (2 points).
(a) F(L, K) = LK3
(b) F(L, K) = L + 3K
(c) F(L, K) = (min{L, K})^1/3

For the first part, it seems that you have to solve for K as the isoquant = dQ/dL. Is that right? If so, why?

For the second part, I have the equation MPL=∆q/∆L |K ̅ but as q isn't given, I'm unsure how to apply this rule to the above equations.

MRTS as I understand it is the equivalent of MRS in that it is the slope at any given point along the isoquant, and would thus require differentiation. I suppose if the previous confusions were cleared up, I'd have an equation at this point to find the differential of.

If anyone has a moment, could they either take me through one of these questions or provide some good online explainers as I'm utterly lost.

Many thanks,
Sean
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OverTheReminds
 
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Reply Fri 27 Oct, 2017 10:40 am
@elguapador,
1) In order to represent an isoquant you have to decide a quantity you wanna produce (you can do it randomly if not given by the problem). Thus:

• Q= LK3 (what do you mean with "K3"?) So, you know Q because you decided it randomly, and you can write it as:
Q/K3 = L.
Now, you choice different levels of K and see what happens to L, for instance, when K is zero, L is infinity, then you keep going on like that, asking yourself what happens to L for different levels of K.

You've got to do so because the isoquant tells you when you have a certain amount of L or K and you want to produce Q, how much of the other input you gotta use. So, if you use a certain level of K, you are now able to know how much L you need to produce Q, and it you do it for different levels of K you have a series of points on the isoquant.

• Q= L+3K
--> Q-3K = L
Just pretend that Q is 99.
Look for different points on the isoquant.
1) Q=99 , K = 0 L = Q = 99 (At L=99 it touches the x axis)
2) Q=99 , K = 9 L = 99-3(9) = 99-27 = 72
3) Q=99 , K = 15 L = 99-15(3) = 54
4 )Q=99 , K = 33 L = 99-33(3) = 0
AND NOW YOU CAN DRAW IT.

• Q = min {L^1/3, K^1/3}
Drawing it is quite simple, for instance, think that the quantity Q = 27. To produce one unit you'll have to use 3 units of Labour and 3 units of Capital. In order to understand why, take another example: if you have to build a bike, the function is: 1 frame, 2 tyres, which means Q= min {1F, 2T}, for producing two you'll need 2 frames and 4 tyres ad so on. Now you just need to apply it to your function.

CALCULATING THE MP
For calculating the marginal product of each input, you need to remember that it is equal to the slope of the production function depending on the single input. When the function depends on two inputs (L and K) you need to use the partial derivative.

1) Q = LK3 (I'll assume that K3 is 3 times K).
MPL = ∂Q/∂L = 3K
MPK = ∂Q/∂K= 3L
MRTS = 3K/3L = K/L
When you go down the isoquant, the Capital decreases (so K/L decreases), and L increases (so K/L decreases even faster), so MRTS is decreasing.

2) Q= L + 3K
MPL = ∂Q/∂L = 1 (K is considered as constant)
MPK= ∂Q/∂K = 3 (L is a constant)
The function has a constant MRTS (MPL/MPK), as it is a function between two perfect substitute inputs (aL+bK = Q)

3) Q = Min {L^1/3, K^1/3}.
Along the isoquant they can't be changed along the isoquant as you could produce the same quantity, so there's no MRTS.

I'll continue later, sorry
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OverTheReminds
 
  1  
Reply Fri 27 Oct, 2017 03:20 pm
@elguapador,
I'll continue here, for some reason it doesn't allow me to edit.

The MRTS in a complementary inputs function makes no sense. Economically, it tells you how much of the input K you must increase in you give up 1 unit of L (or how much K you must give up to use one more unit of L). Think about the bike example now: If you have 3 tyres and give up one, you don't need to add more capital to produce one single bike, which can't be said mathematically: that's why it doesn't exist.

Now, let's answer to the question: indicate whether each marginal product is diminishing

1)
MPL = K
MPK = L
I remember I had a similar question on my book, and after doing the exercise I checked the correct answers. What my book says about it is:

MPL Is constant when the Labour used changes
MPK Is constant when the Capital used changes
Basically, according to my book if you are asked if they are constant, you must consider how the MPx changes to the change of the x (L or K) input used.

2)
MPL = 1
MPK = 3
Of course they're both cnonstant.

3) Not defined.

RETURS TO SCALE:
The returns to scale tell you, if you increase the amount of both the inputs by a certain amount λ, how much the quantity produced will increase. Let's call the increase in the quantity produced ϕ.
What you need to do, is multiplying bot L and K by λ.

-----------------From the theory I know that:---------------------
1) Perfect substitutes production functions have constant return to scale
2) Perfect complements production functions (or Leontief function) have constant return to scale
3) Cobb-Douglas Production functions can have Constant, deminishing or increasing returns to scale.
The Cobb Douglas function must be like that:
AL^α K^β, with A, α, β > 0

If α+β = 1 Then you'll have constant returns to scale
If α+β > 1 Then you'll have increasing returns to scale
If α+β < 1 Then you'll have decreasing returns to scake

Let's check it out.

1) Q1= L3K
Q2 = (λL)3(λK)
= λL3λK
Here we have: A=1 , α = 1, β = 1, α+β= 1+1=2 -->We expect to have increasing returns to scale
= λ²(L3K)
Q1 = L3K
Q2 = λ²Q1.
Which means that by increasing both L and K to λL and λK, we now produce λ² units of good.

2) Q1 = L+3K
Q2 = (λL) + 3(λK)
= λ (L+3K)

Q1 = L+3K
Q2 = λQ1. For instance, this means that by using double Labour and double capital you produce the double of what you produced before.

3)
Q1= Min {L^1/3+K^1/3}
Q2 = Min {(λL)^1/3 + (λK)^1/3}
Q2 = Min {λ^3/3 (L^1/3+K^1/3)}
Q2 = Min {λ (L^1/3+K^1/3)}
Q1 = λQ2

You need to raise λ to the third power in order to bring it out of the brackets
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