Mathematics in Economics – lecture 6

Extremes of a function of two real variables

Necessary condition for the extreme

A point satisfying equalities above is called a stationary (critical) point. However, this
condition is not sufficient.

In a critical point can be maximum, minimum or an inflection point. To decide which situation
occurs, we use the second derivatives and a matrix called hessian H(C)

Determinant calculation:

we multiply the numbers on the main diagonal and subtract the product of the numbers on the
secondary diagonal.


Then we use Sylvester´s theorem.

We denote: D1 =     and D2 = H(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.


Problem 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].

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.

Problem 2

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].


Now we compute all second derivatives and hessian:



We substitute point C into hessian:  Hf(1/2;  1/2) =

           .

Because D2<0, the point C is an inflection point.


Problem 3

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].


Now we compute all second derivatives and hessian:



We substitute point C into hessian:  Hf(0; 0) =



            .

Because D2 > 0, we have extreme at the point C; because D1 < 0, we have a maximum.


Problem 4

Find extremes of the function                        .

Solution:   We start with the first derivatives:


Both derivatives must be 0, which yields the critical point C [7,  –3].


Now we compute all second derivatives and hessian:


We substitute point C into hessian:  Hf(7; –3) =                 .

Because D2 > 0, we have extreme at the point C; because D1 > 0, we have a minimum.


Problem 5

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].


Now we compute all second derivatives and hessian:



We substitute point C into hessian:  Hf(12.5; 2) =                 .

Because D2 > 0, we have extreme at the point C; because D1 < 0, we have a maximum.


HOMEWORK

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