00:01
Okay, number 54.
00:04
This problem is really long.
00:07
We need to feel a little bit more patient on that.
00:10
So, okay.
00:11
A study defined administrative intensity at the number of administrative personal a per production personal x or i equals a over x.
00:26
Now to seek the optimal administrative intensity of an industry, the following cobbidogne's production function was defined.
00:40
Okay, we got this function.
00:43
Where s is the number of distinct occupational facilities represented in the labor force.
00:53
Okay, those are different.
00:54
Variable represents different stuff.
00:59
So a, unlike the production function described in this chapter, for which alpha plus beta plus gamma equals 1, the study assumed that increasing cumulative control loses result in decreasing returns to scale so that, okay, we got this relation.
01:21
Alpha plus beta plus gamma equals 1 minus lambda, where lambda is the control loss parameter.
01:29
The study also defined k, small k equals capital k over x, as the capital intensity per worker, shows that the production function can then be rewriting as this one.
01:44
Okay, so for part a, we got the function originally the c, s to the sigma, a to the alpha multiply, capital k to the beta, x to the gamma.
02:06
Now we have this relation alpha plus beta plus gamma equals 1 minus nmada.
02:16
And we also have k equals capital k.
02:20
It's right k like this over x.
02:25
Now, okay, we want to show q equals c s sigma i alpha k beta x x1 minus nabeda.
02:38
Now this part, those three parts doesn't change.
02:42
So where it changes is we substitute somehow a by i and something about x.
02:53
Also in the problem we have i equals a over x.
03:00
So that's going to be important.
03:02
I equals a over x because we substitute this part.
03:09
A by i.
03:11
So a is from here, a is xi, right? so we can write a as xi.
03:18
So q, this is what do we want to prove.
03:23
It's c s sigma.
03:25
A is x i now.
03:26
To the alpha, multiply k beta x to the gamma.
03:34
Now we can combine those two.
03:39
It's just x to the alpha plus gamma.
03:47
And by this relation, wait one second.
03:56
We also have an idea, we also need this one since let's see, k, let's be raised, this.
04:14
We also need this is capital k, but here is, yeah, here is lowercase k.
04:22
You can see this is lowercase.
04:24
This part is capital.
04:25
Okay, so we want to substitute capital k by lower case somehow.
04:31
By this relation, capital k equals lower k x.
04:35
So this part we can write k as lowercase k x beta.
04:44
So we can combine those three terms.
04:48
C .s.
04:50
Sigma, i'm going to write this together.
04:54
K -beta x -arpha plus this is beta plus gamma, and from here, this part is just equals 1 minus lambda.
05:06
So the answer is going to be c -s -sigma i -a -a -k -beta, multiply x to 1 minus lambda.
05:16
And this is what we want to prove exactly the same.
05:20
Okay, so that's part a, not part b.
05:29
Let's see.
05:31
Part b, the study showed that the profit p can be right on odds.
05:37
This function profit, okay, where p is market price per unit, r is the production cost per unit, and ww1 are those meaning, neta equals xi in the question for p, show that dp over the i equals 0 when the administrative intensity is given by this function.
06:05
So we want to find dp overdii.
06:11
Let's just find dp over the i where our p is p minus r with small p q, subscribe.
06:25
W0s, subtract w1a minus f, where our a equals xi and our q, q is somewhere we have in the first part.
06:43
So we can write ps, p minus r.
06:46
Q is c s, sigma, i, alpha, small k, veda, x, 1 minus nabla.
06:56
It's dp over the i, so we want to transfer every a to i or x to i, i guess.
07:13
Or we can do x, maybe.
07:16
Subtract w0 s minus w1.
07:20
At least we want to transfer a to xi, xi minus f.
07:26
Now we just do partial derivative of dp over the i.
07:33
We got i here.
07:36
So those are like constants.
07:37
We don't need to get alpha, i, alpha, alpha, minus one.
07:42
And we write all those constant.
07:43
Although it's like long, but there doesn't matter.
07:47
K to the beta x1 minus nometer.
07:50
This part has no i.
07:52
This part has no i.
07:53
So zero for them.
07:55
This part, we got wix.
08:00
Now we want to set this equal zero in the question and we solve for i.
08:06
So if this is 0, we got this part equals w ix.
08:13
We can divide this part both sides.
08:22
So i alpha 1 minus 1 equals this part divided by the whole part.
08:33
That's how you solve for that.
08:34
And you can take an alpha -1 root, and then you got your i p minus r c s sigma k veta x1 minus nometer and here we can come by we can divide multiply x nominal minus one top and bottom what do we have w1 x to t lambda and this part is 0 so p minus r c s sigma k to the beta.
09:19
Now that's the whole part and we take a alpha minus 1's root.
09:26
So to the alpha, 1 over alpha minus 1.
09:34
So we have almost the same in this question, but the difference is he takes a 1 over 1 minus alpha the root.
09:46
So we can't do that.
09:49
It's just you take a negative 1.
09:54
In the right side.
10:00
So yeah, you take a negative 1 here, you take a negative 1...