Maths13

Országok listájaHungaryBudapesti Corvinus EgyetemGazdálkodástudományi KarNemzetközi gazdálkodás (angol nyelven)Mathematics1JegyzetekMaths13

Multivariable Optimization
13.1 Balls

Denition 13.1 The closed ball around the origin with radius 1 in the Euclidean n-space Rn is dened by the set
B = {x Rn : x 1}

It is obvious that then the closed ball around the point a Rn with radius r > 0 is the set
a + rB = {x Rn : x - a r}.

13.2

Local extreme

Take an f : Rn R function. It is said that f has a local minimum at the point a of its domain, if there is a real number > 0, such that
f (x) f (a)

for all points x of the domain satisfying that x a + B , that is, x - a . The local maximum of a function is analogously dened and the concept of global minimum and maximum can be given similarly. Hereafter we suppose that the partial derivatives of the function f : Rn R exist and they are continuous.

First order condition
Theorem 13.1 If the function f has a local minimum at point a Rn then f1 (a) = . . . = fn (a) = 0.
Let us consider the unit vectors ek Rn and denote
g(t) = f (a + tek )

By the denition of the local minimum the function g has a local minimum at the point t = 0. On the other hand g is dierentiable and
g (t) = f (a + tek ), ek .

From this it follows that for each index k = 1, . . . , n :
0 = g (0) = f (a), ek = fk (a)

According to the above theorem the points of local extremes of the function can be looked for among the solutions of the system of equations fk (x = 0 (k = 1, . . . , n). 1

13.2.

LOCAL EXTREME

But the vanishing of the partial derivatives is only a necessary condition. For instance, for the function
f (x, y) = x3 y 2

the point (x, y) = (0, 0) is a solution of the system of equations f1 (x, y) = f2 (x, y) = 0 At this point f (0, 0) = 0, but f has no local extreme at the origin, because it takes both positive and negative values in the disk around the origin of any radius. Therefore we need second order (sucient) conditions.

Second order conditions
Hereafter it is assumed that the second order partial derivatives of f exist and they are continuous.

Theorem 13.2 If the function f has a local minimum at point a then the Hessematrix of f at point a is positive semidenite.
Let v be an arbitrary point in Rn and take the function
g(t) = f (a + tv).

The second order derivative of g exists, by the conditions of the theorem and if the function f has a local minimum at a then g has a local minimum at 0, therefore g (0) 0. Hence
0 g (0) = f (a)v, v

But v R was an arbitrary point, thus the Hesse-matrix is positípositive semidefinite. This theorem does not provide a sucient (only necessary) condition to the existence of local extreme. For instance, the partial derivatives of the function
n

f (x, y) = x5 + y 4

are equal to zero at the point (0, 0) and the Hesse-matrix is the zero matrix (hence it is both positive and negative semidenite), but f has no local extreme at this point. The subsequent theorem already provides a sucient condition.

Theorem 13.3 Let us assume that the partial derivatives of the function f equal to zero at point a and the Hesse-matrix is positive denite at this point. Then f has a local minimum at a.
Similar statements can be given for the existence of local maximum.

Determining local extremes
To nd the local extremes of a function in n variables require to carry out the following steps: 1. Find the partial derivatives of the functions, 2. Set up and solve a system of equations, by setting the partial derivatives equal to zero, 3. Find the Hesse-matrix of the function at each solution point, 2

13.2.

LOCAL EXTREME

4. If the Hesse-matrix is positive denite at a point, then the function has a local minimum at that point, 5. If the Hesse-matrix is negative denite at a point, then the function has a local maximum at that point, 6. If the Hesse-matrix is indenite at a point, then the function has no local extreme at that point, 7. If the Hesse-matrix is semidenite, but not denite at a point, then can not be decide by the condition of the theorem, if the function has a local extreme at that point, the decision require special investigation.

Example 13.1 Take rst the function
f (x, y) = x4 + y 2 .

The only point, where the partial derivatives equal to zero is the origin. At this point the Hesse-matrix
H= 0 0 0 2

is obviously positive semidenite. It can be easily seen, that the function has a (global) minimum at the origin. The calculations are almost the same if the function is
g(x, y) = -x4 + y 2 .

Again the origin is the only critical point and the Hesse-matrix is the same as above, but now the origin can not be a place of local extreme, because the function takes both positive and negative values in arbitrary small neighborhood of the origin.

The origin and the like is said to be the saddle point of the function g. Similar examples can be given, when the Hesse-matrix of the function is negative semidenite. It can be summarized, that if the Hesse-matrix of the function is semidenite, (but is not denite) then everthing may occur: the given point can be a place of minimum, maximum or saddle point as well. Let us assume, that the Hesse-matrix is not zero (that is it has nonzero eigenvalue).Then the following cases may occur:
· If the Hesse-matrix is positive semidenite, then the given point can be a

place of local minimum or saddle point only.

· If the Hesse-matrix is negative semidenite, then the given point can be a

place of local maximum or saddle point only.

Let us take a somewhat more complicated example with a function in three variables. Let
f (x, y, z) = xy 2 z 3 (7 - x - 2y - 3z)

Determining the partial derivative the system of equations
f1 (x, y, z) f2 (x, y, z) f3 (x, y, z) = y 2 z 3 (7 - 2x - 2y - 3z) = 0 = 2xyz 3 (7 - x - 3y - 3z) = 0 = xy 2 z 2 (7 - x - 2y - 4z) = 0

is obtained, and one of the solution is the point (1, 1, 1) (there are other solutions, but those are not checked here ). 3

13.2.

LOCAL EXTREME

Find out if there is a local extreme at the point (1, 1, 1)! The Hesse-matrix of the function at this point:
-2 H = -2 -3 -2 -6 -6 -3 -6 -12

and its characteristic polynomial: det (H - I) = 3 + 202 + 59 + 42 The roots of this polynomial are real numbers (the eigenvalues of a symmetric matrix are real numbers), but the polynomial is of degree three. But to nd the roots is not necessary, because each coecient of the polynomial is positive, therefore the roots must be negative. Thus the point (1, 1, 1) is a place of local maximum.

Example 13.2 (A typical exam problem) Find the extreme points of the function
f (x, y, z) = (x2 - 4y)e-(x+y+z ) .
2

Then the system of equations obtained for the rst order partial derivatives:
f1 (x, y, z) f2 (x, y, z) f3 (x, y, z) = (2x - x2 + 4y)e-(x+y+z
2 2
2

)

=0 =0

= (-4 - x + 4y)e = -2z(x - 4y)e

-(x+y+z )

2

-(x+y+z 2 )

=0

whose only solution is (x, y, z) = (2, 0, 0) Let us apply the second order condition. It is a simple calculation to verify that the Hesse-matrix at the point (2, 0, 0) is
-2 H= 4 0 4 0 12 0 · e-2 0 0

The characteristic equation of the Hesse-matrix is
k() = -(2 - 10 - 40)e-2 = 0

One of the roots is zero, the other two have opposite sign, therefore (2, 0, 0) is not an extreme point.

4



Hasonló témájú dokumentumok

Hozzászólások

10. 13 2009 9. a mű adóigazgatás aggregált kínálat alkotmányjog állatszervezettan arc bevezetés dia dimat dinamika eloadas esszék fejlődés fizika 1 fizkémgyak forgó mozgás gazd.töri gazdasági jog intézményi gyakorlat ítéletlogika jogi kefo kiadott anyag korai csecsemőkor könyv 2 kőzet log makro marketing tétel médiakutatás munkaerőpiac munkafüzet növények olasz rugó sql szám szerves kémia szerződés szte-btk dr. simon józsef tanári jegyzet tanirodai technika teszt tűzjelző vizsgakérdés