00:01
A consumer with a utility function, we are provided with the utility function u of x, y it is equal to x raised to power 1 by 3 into y raised to power 1 by 2 and income is given as 16 dollars.
00:16
After that price of x goods are given as it is nothing but equal to 2 dollars.
00:26
After that price of y goods is equal to 4 dollars.
00:33
Then this was the given part in the question.
00:38
In the solution of a part we have to find the utility maximization problem subject to its budget constraint.
00:50
So for this let us suppose that px be the price of x goods, it is the price of x goods and let us suppose that py be the price of y goods.
01:04
After that we will get here i that is income.
01:09
It will be equal to price of x goods multiplied by the number of x goods plus price of y goods multiplied by the number of y goods that means it is equal to this implies that income is 16 so 16 is equal to px that is 2 multiplied by x plus py that is 4 multiplied by y.
01:31
So this is the first constraint we have made here.
01:36
After that above problem can be written in utility maximization problem as this is the problem for maximize and we will write maximum maximize of u, y it is equal to x to the power 1 by 3 y raise to power 1 by 2 and it is subject to constraints i over x, y it will be equal to twice x plus 4y and this is equal to 16 we can write this implies that twice x plus 4y equal to 16.
02:08
Now in order to solve the above problem we have to use the lagrangian method.
02:14
We will use now the here lagrangian method to solve this problem.
02:21
Then we will move to the b part of a question.
02:24
In order to find we will use here the lagrangian method.
02:28
According to lagrangian method in the above problem l of x, y, lambda this will be equal to u of x, y plus lambda into i of x, y.
02:42
Then here we have number of variables.
02:46
We have the two number of variables x and y and therefore number of constraints will also be equal to 2 and number of constraints is 1.
02:58
Here we get only one constraint in the form of x and y.
03:02
So we have two variables and one constraint.
03:04
After that now find the partial derivatives in i plus j time where i is nothing but the valuable in main function which is 2, j is the number of constraints that is equal to 1.
03:19
So this i is equal to 2 since number of variables are 2 and number of constraints are 1 therefore j is equal to 1.
03:26
Now we have to find the three partial derivatives 2 in variable of main function and 1 for the constraint.
03:31
1 is for the constraint and 2 are for the main function.
03:37
So we will find this total three partial derivatives that is l y of x, y and lambda it is equal to 0.
03:46
L y of x, y, lambda it is equal to 0 and third is nothing but the l dot of lambda of x, y, lambda it is equal to 0.
04:03
Here may be x, lx, ly and l dot over lambda.
04:08
We have to find this three.
04:09
Now we write the lagrangian function l of x, y, z that will be equal to x raised to power 1 by 3, y raised to power 1 by 2 plus lambda multiplied by 16 minus twice x minus 4y.
04:23
Now first we find lx of x, y, z it is equal to 0.
04:28
We have to find this.
04:29
That means we have to take the equation 1's partial derivative with respect to x.
04:36
So we get lx of x, y, z will be equal to 1 by 3 x raised to power minus 2 by 3 and y raised to 1 by 2 is constant plus lambda is constant and here is minus 2 derivative.
04:51
So this is lx and we will take it is equal to 0 since this is the equation...