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

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来自 National Research University Higher School of Economics 的课程

Mathematics for economists

3 个评分

National Research University Higher School of Economics

Mathematics for economists

3 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

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

与讲师见面

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

We'll use the theorem stated above for solving this particular problem.

We need to maximize the function X times Y,

subject to many constraints.

So let us count how many constraints do we have here.

The number of constraints is the same as the number of the inequality signs one,

two, three, four, five, altogether five.

That will involve using five Lagrange multipliers.

So let's think maybe we can save some Lagrange multipliers,

so spare some at least one, we'll see.

But first of all, let us draw the constraints set, in coordinate plane.

As we see, X and

Y are non-negative variables,

here we see X and here we have Y.

So, somewhere here we have eight and I'm starting writing down

the inequality which is given by this linear function X plus two Y.

This straight line will cross the x-axis at point eight,

and it will cross y-axis at the point four,

and it looks kind of realistic.

Then, here we have two,

here we have three,

and also we have one on the y-axis.

Y equals three, that's the horizontal line,

so we get a trapezium here and what's left?

Drawing the parabola which is given by the equation Y equals X squared over 16 plus one.

So this curve starts at one and it will cross

the slant line somewhere here in this region,

so in order to find the point we need to solve equation.

I'll put it solving equation.

That's Y value, and Y value equals

eight minus X over two and we're looking for the positive roots.

A simple quadratic equation,

which will give us X star equals four.

So here we have K,

this time it's not that's realistic but it will do.

Okay, now I'll shade the constraints set.

It's clear that the curve will cross the slant line here

because we have found the square root and also when we

substitute for X four we get Y equals two,

so this is our constraint set.

Now I'll start labeling the vertices.

This one can be also found easily if we substitute here Y equals three and we get two.

So, I'll be dotting them,

the points, the vertices, in clockwise motion.

Here we have vertex A that's B,

that's C, that's D, okay.

Now, if we recall the theorem recently stated,

we need to check NDCQ, how it is done.

It is done by calculating the rank of a specific Jacobian matrix.

In this matrix, we fill in the gradients

of the constraints and we choose only binding ones.

So, if we look at the boundary of the constraints set,

this boundary consists of some parts

and some vertices and we have labeled the vertices already, a, b, c,

d. So we stop checking whether NDCQ holds between the vertices,

for instance here on the vertical axis X equals

zero and the rank

or this Jacobian matrix is full, is of course one.

Because the gradient X is one, zero.

Now, horizontal boundary Y equals three, the same.

Okay, now, slant line,

this time the gradient is a constant vector one,

two, rank is one.

The curve which is a parabola,

never turns zero.

So rank is also one,

and we have finished with these pieces of

the boundary and now we proceed with the vertices.

So this time the Jacobian matrix will be a two by two matrix,

because we have two variables and we have two constraints.

For instance, at this point A.

At A, this point is a point of intersection of the vertical axis and the parabola.

So J will look like,

so that will be the first line,

one, and here x is zero,

so this is that.

For any X values,

the rank is two, full rank.

At B,

even easier because B is a vertex based on the intersection of two straight lines.

At B, rank clearly is two.

The same true of C, at C as well,

and finally D. At D we

have J which is at the first row will be the gradient of this function,

and here we have a constant vector one, two.

Let's think whether it's possible for this matrix to have a smaller rank.

The smaller means rank will be one.

We know from linear algebra that the easiest way to

check what are the values of X for which this rank is one,

is to find determinants of this matrix.

Determinant is well, clearly it's not zero,

because x is positive.

So we have checked NDCQ conditions completely at all points of the boundary.

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