9c. The Production Function with Two Inputs
One of the main objectives of a seller is to transform inputs into outputs to sell to consumers for profit. Economists created an abstract model that shows the relationship between inputs and outputs
q = f(K,L)
-Where q is the total output and K, L… are the inputs required to make the output
-Real world example: Fast Food restaurant needs inputs of meat, lettuce, oven, grill, French fry maker, cash register, soda dispenser, employees, etc. to make outputs of hamburgers, fries, desserts, drinks, etc.
-In the model, we restrict the number of possible types of output into one. Using our example above, we assume the firm only produces hamburgers
In the simplified case of one inputs, all else constant, we use a production function q = f(K,L) where:
q is output
L represents the unit of labor
K usually represents the unit of capital
Economists are interested in the variation of these inputs, or the marginal physical product of input L.
Marginal Physical Product of X input (MPX): the additional output that can be produced by employing one more unit of that input while holding all other inputs constant.
MPL = Marginal Physical Product of Labor = dq/dL = fL
MPK = Marginal Physical Product of Capital = dq/dK = fK
*The use of partial derivatives has practical implications:
Consider a lemonade stand. If a little girl wants to know how much more lemonade she can make with X more amount of lemons, she can find the MPlemons assuming she only has one lemonade maker and she is the only worker. (Lets say she can make 42 ounces of lemonade with 10 lemons. An 11th lemon allows her to make 45 ounces of lemonade. So the MPlemons of the 11th lemon is 3 ounces). This allows for observation of how a certain input (and only that input) can affect the output level.
Diminishing Marginal Productivity
The example of the little girl and her lemonade stand assumes about fixed capital and labor quantities in the lemonade example (the sole employer is the little girl and the sole unit of capital is her lemonade maker), intuitively we predict that she won’t be able to make as much lemonade with more lemons in a certain period (a decline in productivity)
Mathematically, we must show that the slope of Rate of Technical Substitution (RTS) is always less than 0 or d(RTS)/dL > 0 Geometrically, this means that it is convex.
We assume fLand fKare positive (fL and fK > 0) and that fLLand fKKare negative.
d(MPL /MPK)/dL = d(fL/fK)/dL = d(RTS)/dL
=( fK(fLL+ fLK*dK/dL) – fL(fKL + fKK*dK/dL))/(fK)2
Using the fact that (fLK = fKL) by Young’s Theorem and that dK/dL = -fL/fK
= d(RTS)/dL = (f2K*fLL – 2fKfLfLK +f2fKK)/(fK)2
Evaluating this simplified term, we know that the denominator is always positive. So if the whole expression in the numerator is negative, we have our proof. Also, since we assumed that fLLand fKK is negative, we can be sure that the numerator is negative if fLK is positive.
On why fLK is positive and other fundamental assumptions:
Using our intuition, it is reasonable to say that it is positive because we assume that if workers had more capital, they would have higher marginal productivities. However, this is not always the case because fLK can be negative for some ranges of input (learning how to use a new machine in the beginning of a change in technology). So when we assume that RTS is diminishing, we are saying that marginal productivities diminish rapidly enough to override any possible negative cross-productivity effects.
*This is a fundamental assumption of production functions
Historical side note: Famous economist Thomas Malthus proposed this fundamental assumption when he predicted that a rapid increase in population leads to a decline in productivity (However, he did not take into account the increase in capital inputs that have equalized the diminishing marginal productivity in the last century).
Average Physical Productivity
Often called labor productivity, it is simply the ouput per unit of labor. Its importance comes from its relative ease of calculation and because marginal productivities can be deduced from it. It is seen to be an efficient means of measuring productivity.
Average Physical Product of Labor = APL = q/L = f (K, L)/L
Returns to Scale
Pertains to the response of output growth due to an increase usage of inputs
Definition: a given production function q = f(L) has its inputs multiplied by the same positive constant R (where R > 1). We classify returns to scale as one of the three:
1. Constant: if (RK,RL) = Rf(K,L) = Rq
- Decreasing: if f(RK,RL) < Rf(K,L) = Rq
- Increasing if f(RK,RL) > Rf(K,L) = Rq
Historical side note: Adam Smith studied the notion of increasing scales of production with pins. He noticed that doubling a scale of production promoted greater division of labor (increased specialization). So, there is reason to believe that there will be an increase in efficiency. However, there was also reason to assume a loss in efficiency as a result of doubling the amount of inputs since management would have a hard time overseeing a larger scale firm. A naturally arising question is which of the two tendencies had the greater effect.
Practice Questions
1. For a production function q = AKbL1-b
- What is the MPL? The MPK?
- Let b = ½ and A = 10, now what is the MPLand MPK?
- If K was fixed at 100, in terms of L, what is the new production function?
2. If a firm increases the amount of capital and labor by amount R, what would be the necessary condition for decreasing returns to scale?
3. In the case of a linear production function q = f(K,L) = aK + bL, what type of scale of production will you expect?
4. For q = f(K,L) = 5KL – 4K2 – L2,
a. what is the MPL and the MPK?
b. Now suppose K = 10, when does the APL reach its maximum? How many units of output are
produced at this point?
c. From (b), is this at the level of output where marginal productivity is maximum?
d. Now let K = 20. How would your answers to (b) change?
e. Does the production function exhibit constant, increasing, or decreasing returns to scale?
5. The production of hammers can be described by the production function q = 2K2L2 – K3L3
- what is the APL and the APK?
- what is the APL for K = 5?
- For K = 5, when does APL = MPL?
Answers:
1a. MPL = dq/dL = A(1-b)KbL1-b-1 = A(1-b)KbL-b
MPK = dq/dK = A(b)Kb-1L1-b
1b. New q is 10K1/2L1/2. From the equations from 1a., MPL = 10(.5)K1/2L-(1/2), MPK = 10(.5)K-(1/2)L1/2
1c. if K=100, q = A100bL1-b. If assumptions from 1b) are used, q = 100L1/2
2. if f(RK,RL), then in order for decreasing returns to scale, f(RK,RL) must be less than Rf(K,L). This means the output grew by less than the factor R.
3. It can be shown that increasing the scale of production of q = f(K,L) = aK + bL, by R is :
Rq = f(RK,RL) = RaK + RbL = R(aK + bL) = Rf(K,L). This shows that the linear production function exhibits constant returns to scale.
4a. MPL = dq/dL = 5K – 2L
MPK = dq/dK = 5L – 8K
4b. If K = 10, q = 50L – L2 – 400
So, APL = q/L = 50 –L – 400/L
APLmaz = max of d(APL)/dL = -1+400/L2 = 0
Critical point is where L2 = 400 or L = {-20,20} real number is our solution so L =20
At this level, q = 50(20) – (20)2 – 100/(20) = 1000 – 400 – 5 = 595 units of output
4c. Maximal output is where MPL = 0
So, for K = 10, q = 50L – 400 – L2
MPL = 50 – 2L = 0
L for qmax = 25 which is q = 50(25) – 400 – (25)2 = 1250 – 400 – 625 = 225
225 is not equal to 595 so no, the level at which APL is max is not equal to the level at which MPL is max.
4d. for K = 20, q = 100L – 1600 – L2, so q/L = 100 – L – 1600/L
max of d(APL)/dL = -1 +1600/L2 = 0 so L = {-40,40} the real solution is L = 40
at this level, q = 100(40) – 1600 – (40)2 = 4,000– 1600 – 1600 = 800 units of output
4e. Notice how doubling the units of input ( K from 10 to 20 and L from 20 to 40) only returned a 34.4% increase in productivity (205/595). Hence, the production function exhibits decreasing returns to scale (f(RK,RL) < Rf(K,L))
5a. APL = q/L = 2K2L2 – K3L3/(L) = 2K2L – K3L2
APK = q/K= 2K2L2 – K3L3 = 2KL2 – K2L3
5b. for K = 5, APL = 50L – 125L2
5c. MPL = 2(2K2L – 3K3L2) = 4K2L – 3K3L2; for K = 5, 100L – 375L2
Set APL = MPL
50L – 125L2 = 100L – 375L2
375L2 – 125L2 = 100L – 50L
250L2 = 50L
250L2 - 50L = 0
50L(50L – 1) = 0
L = {0, (1/50)}