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

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En provenance du cours de National Research University Higher School of Economics

Mathematics for economists

6 notes

National Research University Higher School of Economics

Mathematics for economists

6 notes

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.

À partir de la leçon

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

Rencontrer les enseignants

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

[MUSIC]

Now we'll proceed with the topic how to identify global extrema.

The thing is that there are no clear-cut methods for

this, but we'll consider two problems.

We'll provide ideas which we need to follow dealing with

this optimization problem.

So usually we start with solving a problem

of optimization which will give us the set of critical points.

And we need to identify them using known techniques.

But in order to answer the question whether the found local extrema

are global ones, we need to do some extra investigation.

So we start with a problem in two-dimensional case.

There is a function z in the form of.

We exclude the 0 point because the logarithm of 0 doesn't exist at

the same time.

This is a removable discontinuity.

So (0,0) is a removable,

Discontinuity point.

Because the limit of z when x and

y turns 0 is 0.

It's quite easy to show.

Now, the problem is to find global extrema, or this function z,

and as always we start with the first order conditions.

It's quite convenient to change cartesian variables x,

y into polar variables, the radius r and the angle phi.

So from (x,y) We move to (r,phi) using two formulas,

Where r takes the non-negative values and

phi the angle ranges from 0, to 2 pi.

After the change of variables, the system can be written again.

And it becomes,

Since r is strictly positive here, we can cross it out.

And the system becomes a bit simpler.

If I take out sin phi in the first equation,

And cos phi in the second.

Now it's clear that if,

r = 1 then logarithm r squared = 0.

And for the values,

0, pi over 2, pi, 3 over 2, pi,

the system is satisfied and easily seen.

Because for instance,

if you choose 0 for phi, sin turns 0.

At the same time sin is here, also turns 0.

If you choose pi over 2, then cos is 0 and

cos here is also 0, so we got four points.

And all these four points belong to the unit circle.

Let's draw it.

So these are the points when plotting, Them.

Can we tell whether these critical points

are points of extremum of the function?

And we can do it without even checking the Hessian matrix

because it's clear that they are not points of extrema.

Where it's clear, let's consider the very first one, this one, point 1.

If we move slightly up in that direction.

So y goes up and x is still 1.

So from point (1,0), we move to the point

(1,epsilon), where epsilon is greater than 0.

Let's find the value of the function at this new point.

X is still 1, Y is epsilon.

We get a positive value.

But if we move down.

So this time from we move to (1,-epsilon).

We'll get a negative result.

So this is clearly a saddle point.

This analysis can be repeated again and

again at any of these four points.

Now, let's proceed and finish solution of this system.

Now, when we exclude all these four points,

we can cross out sin phi and cos phi.

So after crossing out sin and cos, we need to add up two equations.

When we add

them up,

we get,

This being 0, then, This is 1.

And we get

That gives us the value of r,

1 over the square root of e.

But now we need to take this value and plug in two of the equations again.

What do we get then? We get then a system.

It's not a system, it's a collection.

Either cos squared phi = one-half or

sin squared phi = one-half.

And after solving,

we'll get another four critical points.

This time the 5 values are pi over 4,

three-fourths pi,

five-fourths pi, and

the final, seven-fourths pi.

[MUSIC]

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