Mathematics in economics
Lecture 5
Mgr. Jiří Mazurek, Ph.D.
Mathematics in Economics/PMAT

Lecture 5
Extremes of a function of two real variables




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1.
Local vs global extremes.
Bounded vs unbounded extremes.
Necessary condition for the extreme:
A point satisfying equalities above is called a stationary
 (critical) point. However, this condition is not sufficient

Lecture 5
Extremes of a function of two real variables


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1.
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In a critical point can be maximum, minimum or an inflection
 point. To decide which situation occurrs, we use the second
derivatives and a matrix called hessian:
Hf(x,y) =
Then we use Sylvester´s theorem.

Lecture 5
Extremes of a function of two real variables




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1.
We denote: D1 =               and D2 = Hf(C). Then:
If D2>0, then we have an extreme. Moreover,
If D1>0, we have a minimum, if D1<0, we have a maximum.
IF D2<0, we have an inflection point.
If D2 = 0, we cannot decide.

Lecture 5
Extremes of a function of two real variables
 - Problem 1




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1.
Find extremes of the function                        .
Solution:
We start with the first derivatives:
Both derivatives must be 0, which yields the critical point
C [0,0].

Lecture 5
Extremes of a function of two real variables
 - Problem 1 – cont.




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1.
Now we compute all second derivatives and hessian:
We substitute point C into hessian:  Hf(0,0) =                 .
Because D2<0, the point C is an inflection point.
Hf(x,y) =
.

Lecture 5
Extremes of a function of two real variables
 - Problem 2




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1.
Find extremes of the function                             .
Solution:
We start with the first derivatives:
Both derivatives must be 0, which yields the critical point
 C [1/2,1/2].

Lecture 5
Extremes of a function of two real variables
 - Problem 2 – cont.




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1.
Now we compute all second derivatives and hessian:
Substituting point C into hessian yields the same result.
Because D2 < 0, the point C is an inflection point.
Hf(x,y) =
.

Lecture 5
Extremes of a function of two real variables
 - Problem 3




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1.
Find extremes of the function                             .
Solution:
We start with the first derivatives:
Both derivatives must be 0, which yields the critical point
 C [0,0].

Lecture 5
Extremes of a function of two real variables
 - Problem 3 – cont.




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1.
Now we compute all second derivatives and hessian:
Because D2 = 0, We cannot decide the nature of C.
But how do we know it is certainly a minimum?
Hf(x,y) =
.
Hf(C) =

Lecture 5
Problem 4




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1.
Find the maximum of the revenue function:
Solution:
We start with the first derivatives:
Both derivatives must be 0, which yields the critical point
 C [12.5,2].

Lecture 5
Problem 4 – cont.




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1.
Now we compute all second derivatives and hessian:
Because D2 > 0, we have an extreme.
Because D1 <0, we have a maximum.
Hf(x,y) =
.
Hf(C) =

Lecture 5
Problems to solve – 1 (Assignment 7)
 Find extremes of the following functions:



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1.
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Lecture 5
Indefinite integral
 Integration is a reverse procedure to differentiation.
Notation:
Legend:
          …. Integration sign
 f(x)    …. Integrated function
   C     …. Integration constant


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1.
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Lecture 5
Indefinite integral - cont
 Indefinite integral is a linear operator:

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1.
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We compute integrals with the use of formulas above, and
with the use of the table of elementary integrals, see
the next slide.

Lecture 5
Indefinite integral – elementary integrals

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Lecture 5
Indefinite integral – elementary integrals

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Lecture 5
Indefinite integral - examples

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 =
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Lecture 5
Indefinite integral - examples

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Lecture 5
Indefinite integral – integration methods

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For more complicated integration we use sutitable
integration methods:
• Method per partes
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• Partial fractions
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• Substitutions
All these methods will be demonstrated on examples.

Lecture 5
Indefinite integral – rational functions

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By a rational function we mean the function of the form:
where P(x) and Q(x) are polynomials.
In the first step we find the roots xi of the denominator in order
to rearrange the denominator into a product.

Lecture 5
Indefinite integral – rational functions – cont.

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Then, the situation splits into three possible cases:
1.All roots of a denominator are single. Then we obtain the
following partial fractions:
2. Some root, for example x1, is of order higher than 1:

Lecture 5
Indefinite integral – rational functions – cont.

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3. A denominator or its part given as a quadratic polynomial
has no roots:
Coeffcients in numerators are unknown and must be computed
by clearing a denomiantor and solving a subsequent equation.

Lecture 5
Integration of rational functions – Problem 1

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Solve:                      .
Solution: The rational function is of case 1, with roots -2 and 1.
Therefore, we obtain the following division into partial fractions:
Now we clear the denominator:
And we get two equations:

Lecture 5
Integration of rational functions – Problem 1-cont.

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Solving the equation yields:                    .
Hence:
Now, we can integrate:
We will continue in Lecture 6.

Lecture 5
Thank you for your attention!

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