Optimizing under constraints:

Cobb-Douglas productivity function and limited budgets 

 

The Cobb Douglas productivity function P = f (L,K) is given by formulas that look like P := c*L^a*K^(1-a)

where L is the number of worker hours in 1000's ("Labor") and K is investment in form of capital in 1000 $'s.

The specific values of  c and  a in the formula depend on what is being produced and on the particular manufacturing process used. In the "matress-production example" in the text we have 

c := 48.1and  a := .6, so that the specific Cobb Douglas P-function is P = 48.1*L^.6*K^.4.

 

Let's take a look at the surface (graph) of this function:

[Maple Plot] [Maple Plot]

Now suppose that we have a budget of  $98,000.
This money is to be allocated between Labor and capital investment.
Suppose further that we pay  $8  per labor hour.

For given values of  L  and  K  we will need   1000*L*8 = 8000L $ for Labor and 1000*K $ for capital.
Since our budget is limited to a total of  $98,000 our allocation options are only the ones for which

8000 L + 1000 K = 98,000 , or, dividing all by 1000:   8 L + K = 98 .

Notice that this constraint can be rewritten as  K = 98 - 8 L
 Since we have only $98000 in our budget, we have to use a combination of L and K for which K = 98 - 8 L.

It is our goal to determine that labor/capital allocation of a total $98000, for which productivity is maximized.

Adding the "constraint curve K = 98 - 8L" (a straight line) to the picture:

[Maple Plot] [Maple Plot]

The points on the red line indicate those combinations of L and K which fit our budget.

Leaving out the surface, and showing the red curve alone we see this:

[Maple Plot] [Maple Plot]

where the picture in the right is simply an enlargement of the left picture.

Goal: Find that allowable combination (L, K) at which this curve peaks (max. productivity).

The following two pictures are obtained  from rotating the last picture so that we look along the K-axis (left picture) and then along the L-axis. Both of them show the productivity of all of the allowable (L,K)'s. 
(Think about this! If we choose L, then K is determined, and we can calculate the productivity when we have both!)

[Maple Plot] [Maple Plot]

Observe: Finding the relative max. "along the constraint" looks an awful lot like finding a relative max of a function with one variable !!!

If we can find a formula for the "red-graph" (that is the productivity as a function of either L alone (left graph) or of K alone (right graph)), then this will be easy !!

The pictures indicate that such formulas exist (e.g. P is graphed as a function of L),
but how can we find such formulas?

Well, notice that the constraint allows us to express K in terms of L : K = 98 - 8L.
So, in the formula 

P = 48.1*L^.6*K^.4

we can write  98 - 8L instead of K! 
That produces the formula  P = 48.1 L.6  (98 - 8L).4 , productivity as a function of L alone!!!

To find the L  at which  P  is maximal we need to find the value of L for which the derivative  P'(L) = 0.

We showed in class that  L = 7.35  is that value. 
Because of our budget constraint we must have K = 98 - 8L = 39.2.

Altogether we see that under our budget constraint productivity is maximized when we allocate the available $98,000 as 7,350 labor hours and $39,200 of capital investment.

(Note that we could have used the fact that  L = (98 - K)/8  to express P as a function of K, and then found the value of  K  at which P'(K) = 0... we would have found K = 39.2. Try it!)

!!!  Finally: Observe that  we were able to use one-variable-techniques to find the constraint extremum. !!!
!!! The reason was that we were able to use the constraint condition to express K in terms of L !!!
!!! so that the formula for productivity could be converted into a formula of L alone. !!!