7.3. How to identify global extrema? (part 3)

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Из курса от партнера National Research University Higher School of Economics

Mathematics for economists

1 оценки

National Research University Higher School of Economics

Mathematics for economists

1 оценки

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.

Из урока

Global extrema. Constrained optimization with inequality constraints.

Week 7 of the Course is devoted to identification of global extrema and constrained optimization with inequality constraints. This week students will grasp the concept of binding constraints and complementary slackness conditions.

7.1. How to identify global extrema? (part 1)11:30

7.2. How to identify global extrema? (part 2)10:14

7.3. How to identify global extrema? (part 3)12:31

7.4. How to identify global extrema? (part 4)10:03

7.5. Constrained optimization with inequality constraints.8:47

7.6. Binding constraints. Complementary slackness conditions.5:40

7.7. Constrained optimization problem with inequalities for n-dimensional space.7:14

7.8. Constrained optimization problem with inequalities. Theorem.6:48

7.9. Example of solving constrained optimization problem with inequalities (part 1).9:11

7.10. Example of solving constrained optimization problem with inequalities (part 2).10:27

7.11. Example of solving constrained optimization problem with inequalities (part 3).3:53

Познакомьтесь с преподавателями

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

[MUSIC]

So we need to answer the question whether are the found local extrema.

We have found two points of local maximum, two points of local

minimum with their global make them a minimum at the same time.

This is not true here and to clarify the issue, we need to

think about the behavior of this function z at infinity.

So if we draw the coordinate plane

and choose the direction of travel.

So for instance, I can, choosing the bisector

straight line which passes through the origin at 45 degrees angle.

So if for instance I'm moving along this bisector and

go to infinity, then z(x, x),

becomes x squared and then 2x squared, it leads to positive infinity.

That means that this function has no global maximum, because

it never takes the greatest value.

At the same time if choose another bisector.

And this time, we take z( x-x),

And the function takes the negative value.

And when a x moves to a negative infinity,

the function itself gives to negative infinity.

That means the global minimum also doesn't exist.

Let's consider another problem.

This time the function z is like this.

We need to find its global extrema if they exist.

We follow the same route.

We start with identifying the critical points firstly.

In order to solve the system let me solve for y in terms for x.

So then it becomes and

I will substitute into the second equation.

So that will be a resulting equation having only one x no y.

So I'm simply writing, substituting, in this term the y value.

So I have y being placed here.

Now we need to collect all the powers of number two.

I have 2 cubed and this is two squared,

two squared makes 2 to the 6th,

x nine power, so

well clearly I get several points.

The first point being A simply ( 0,0 ).

It was clear from the start that ( 0,0 ) for

xy is a solution, one of the solutions of the system.

After that I can cross out x in the left-hand side,

and the right-hand side, then I I get equation

x to the 8th = 1 over 2 to the 9th power.

And that gives me two solutions.

One positive, another one negative.

For positive I have 1/2 to

the power of 9/8.

And for negative the same in absolute value but

I have a minus in front.

So these are my values of x.

We need to also find y.

So if I choose that,

the y will be positive, so

I have 2 squared.

And I raise this to the third power,

then I have something

like, 27 over 8.

I need to subtract 2.

27 8ths minus 2 makes 1 over

11 eighths.

And a similar y value of four, another one also, but I teach minus sign.

Okay, so I won't do it right here.

Now so we have three points and

we proceed with their asymmetrics again.

In order to calculate

it we need to find Zxx,

which is 12x squared,

Zyy which is 24y squared and

Zxy is simply one -1 okay.

Now my choice will be a first 00 point

At A.

Is a matrix, starting with the mixed derivatives.

And here I have two 0s because these derivatives,

these 2, term 0 indefinite.

That's what I can tell.

Is a saddle point.

Saddle point.

Now, I need to be very careful with calculating the values

at the rest of the points.

So I have x, which is 1 over

2 to the 9 8ths power.

And I have y value,

which is 1, 11 8ths.

Our Zxx derivative becomes

2 squared times 3,

1/4th only.

Looks reasonable as a fraction.

Now Zyy.

Or simply that's one fourth.

Now, now we are ready to substitute into the matrix,

at let it be I point in the first quadrat being b point.

So D squared z and

B Becomes

We put -1.

-1, for mixed derivatives.

Here, we have.

3.

What we can tell, a positive leading principle minor.

Now, calculation, or the determinant, gives us,

9- 1 = 8, also positive.

So B is a strict local,

Minimum because it's positive.

So B- local min.

So local min.

What will change if we substitute c point with minuses.

A point which belongs to the third quadrant and we see that these derivatives

they have squares so nothing will change.

So c is also local min.

And the question arises whether these two points can be points of global minimum.

What about global maximum?

Does it exist?

We'll answer these questions immediately.

[SOUND]

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