00:01
All right, so here we have a function that shows the price of a certain brand of chocolate p and dollars, and this person's preference for buying q boxes of chocolate and leaving a non -negative amount right in here in her bank account.
00:23
So it's a utility function.
00:25
It's like the most i could get while maintaining this amount of that i can.
00:31
And so we're going to find the first order condition for utility maximizing quantity.
00:38
So maximizing, we're going to think differentiation.
00:41
So we've got this u function.
00:44
So we're going to take d, u, d, d, q.
00:49
We have this is going to be, remember, this is the same as q to a half.
00:59
This is going to be one -half q to the minus one -half by the power rule.
01:05
This is going to, remember, this is 50 minus p to the q to the one -half, minus p -q to the one -half.
01:13
So this is going to become one -half times 50 minus p -q to the negative one -half, all times negative by the chain rule.
01:27
So we go.
01:28
That's our utility maximizing quantity of a box of chocolate, is our condition.
01:40
And we're going to solve this condition, is a or one, i guess, two.
01:45
We want to solve this thing in order to express the utility maximizing quantity q star as a function of p.
01:51
So maximizing, what that means, this is the derivative.
01:56
And remember, the derivative allows us to see where the highs and low parts are, right, where the maxes and midpoints are.
02:04
So here's our function, right? here's the u function, right? u would have tangent points, in other words, zero right here.
02:12
And right here.
02:14
So that means you have some graphs.
02:16
This is a positive.
02:17
So it'd be like, come like this.
02:20
Cross here.
02:22
And then it might come around here.
02:24
This is a negative slope.
02:25
But then it starts coming back up here.
02:28
And it comes up like that.
02:33
Something like that.
02:34
The point being is this is going to give us that.
02:37
But what is zero.
02:38
So what we do is you set this whole thing equal to zero.
02:42
Let me do it another color to differentiate what we got here.
02:46
This is two.
02:47
We're going to set this whole thing from 0, 1 1 1ā2 q to negative 1 1ā2.
02:54
50 plus p q to negative 1ā2.
02:58
10 .0 p equals 0.
03:01
So i'm going to move this chunk here on the right over to the right side because i want to get q by itself.
03:07
And let's see.
03:08
Let's going to work through it.
03:11
So here we get 1 over 2, q to 1ā2.
03:17
Plus and you know i you might want to clean this up because i know some professors like to have positive exponents so you'd want to move this into the denominator so just be by that and i'll i'm doing that here because it's going to help me simplify this thing oh and this equals well this is going to become a positive piece so you get p over 2 times 50 minus p q to the one -half um let's see well the the one -halfs, we can get rid of those.
03:48
Those cancel out.
03:49
It's nice.
03:52
I'm going to get q up here.
03:55
So i'm going to, let's see.
03:55
I'm going to multiply by q to the one -half here.
04:03
My q to the one -half here.
04:06
And then we get 1 equals p -q to the one -half over 50 minus p -q -2 -1 -half.
04:17
Now we're going to multiply by this guy here.
04:20
So we're going to end up with, let's see if my use of technology.
04:24
Will help me here.
04:32
So this whole thing is going to be multiplied by this on both sides.
04:37
So we're going to basically end up with 50 minus pq, which is going to one half.
04:43
This is gone now.
04:45
We have a half and a half.
04:47
The best way to get rid of this is to square the, square everything.
04:56
So then we get 50 minus pq equals p squared times q.
05:03
Get the q's to one side.
05:06
And let's see, that's going to give us 50.
05:07
Equals p squared q plus pq divide up by q p squared plus p q q is 50 divide by p squared plus p so we get 50 50 over p squared plus p equals q and there you go this is q star look at that good algebra that's two uh what condition will guarantee your quantity q -star really is truly, in fact, a maximum.
05:43
Well, that's going to be taking the second derivative.
05:56
I'm going to copy this.
06:03
Face this down here.
06:06
This up here.
06:17
So this is going to be three.
06:21
We're going to take the second derivative of this.
06:23
So d squared u, dq squared.
06:31
And this is where i left.
06:32
This in this way, you get one quarter, q to the negative three quarters, plus, no, minus, minus one quarter.
06:47
We have a big fraction here.
06:50
That will be like that...