8.5. Solving minimization costs problem using Kuhn-Tucker conditions (part 2).

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来自 National Research University Higher School of Economics 的课程

Mathematics for economists

1 个评分

National Research University Higher School of Economics

Mathematics for economists

1 个评分

This course is an important part of the undergraduate stage in education for future economists. It's also useful for graduate students who would like to gain knowledge and skills in an important part of math. It gives students skills for implementation of the mathematical knowledge and expertise to the problems of economics. Its prerequisites are both the knowledge of the single variable calculus and the foundations of linear algebra including operations on matrices and the general theory of systems of simultaneous equations. Some knowledge of vector spaces would be beneficial for a student. The course covers several variable calculus, both constrained and unconstrained optimization. The course is aimed at teaching students to master comparative statics problems, optimization problems using the acquired mathematical tools. Home assignments will be provided on a weekly basis. The objective of the course is to acquire the students’ knowledge in the field of mathematics and to make them ready to analyze simulated as well as real economic situations. Students learn how to use and apply mathematics by working with concrete examples and exercises. Moreover this course is aimed at showing what constitutes a solid proof. The ability to present proofs can be trained and improved and in that respect the course is helpful. It will be shown that math is not reduced just to “cookbook recipes”. On the contrary the deep knowledge of math concepts helps to understand real life situations.

从本节课中

Kunh-Tucker conditions. Homogeneous functions.

Week 8 of the Course is devoted to Kuhn-Tucker conditions and homogenous functions. This week students will find out how to use Kuhn-Tucker conditions for solving various economic problems.

8.1. Kuhn-Tucker conditions.7:28

8.2. Solving consumer choice problem using Kuhn-Tucker conditions (part 1).9:57

8.3. Solving consumer choice problem using Kuhn-Tucker conditions (part 2).11:15

8.4. Solving minimization costs problem using Kuhn-Tucker conditions (part 1).9:05

8.5. Solving minimization costs problem using Kuhn-Tucker conditions (part 2).7:51

8.6. Homogeneous functions.6:24

8.7. Homogeneous functions. Two propositions.5:24

8.8. Income-consumption curve.10:28

8.9. Euler's Theorem.12:21

与讲师见面

Kirill Bukin
Associate Professor, Candidate of sciences (phys.-math.)
Faculty of Economic Sciences, Department of Theoretical Economics

Intuition tells us that the factor,

one, is being far too expensive,

will not be employed,

so let us check that.

That may mean a corner solution and that only x_2 will be employed at the amount of y,

it comes from this equation.

If we substitute zero value for x_1,

we get immediately, x_2 equals y,

so let us check that.

So, if x_2 is y and x_1 is zero,

then according to the complimentary slackness condition, here,

we get equation, and that will give us the Lambda value.

So, from here, we get Lambda is one.

How does it comply with the first inequality?

So, let us substitute in this inequality,

the complimentary slackness is fulfilled automatically since x_1 is zero.

Here, we have -100, we substitute one,

x_2 is y, and that should be no greater than zero,

and that means that y is less than or equal to 99.

That corresponds to the previous part of the solution when we were looking

for inner optimal bundles of factors of production.

To complete the solution,

we need to check that buying only x_1 factor is not possible here. Okay, let's do it.

So, if x_1 equals y and x_2 equals 0,

that means that from the first inequality,

which turns into equality,

we get a Lambda value which is 100.

But having substituting 100 here,

we get a contradiction into this inequality.

We get minus 1 plus 100, y plus 1.

It's not possible because y is greater than zero, so impossible.

The final step which concludes the problem will be finding total cost functions here.

According to the definition,

the total costs, which is a function of the output,

that will be, and

depending on the output value,

this function consists of two pieces.

So, if the output is low,

so for Y, well,

you're not greater than 99,

only second factor is employed.

That means we get

y because x_2 star

is y and x_1 star is zero,

so we simply substitute.

Now, for greater values,

for y greater than 99,

we need to recall what was found earlier.

We'd be looking for optimal values of x_1 star and x_2 star.

So, previously was formed this value,

but we didn't calculate x_2 star,

x_2 star should be formed from the previous equation.

Remember, when we were solving a system we've got,

so we need to substitute and we get.

So, finally, I'll write it here,

that should be substituted into the formula.

So then, TC equals,

and that's how we get 20 times 1 plus y minus.

Let us check. Clearly, we need to get a continuous function of the output, y,

so if we substitute 99 into the argument of this function,

we get 200 minus 101, which makes 99,

and that is exactly the number we get when we substitute y value as 99.

That concludes the solution of the problem.

Although, we need to say something about the sufficiency condition.

We've been using on the necessary condition,

there is always a threat that we get

a critical point which actually is not either a maximizer, minimizer.

Here, everything is fine because we still can employ [inaudible] theorem since this part,

or the [inaudible] is a closed bounded set,

and we have found just one critical point to the Lagrangian,

so that will be the greatest value of

the negative objective function

because we started with a minimization problem

and we convert it into maximization problem,

and that finally completely concludes the proof.

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