[email protected] MTH16_Lec-08_sec_7-5_LaGrange_Multipliers.pptx 1 Bruce Mayer, PE Chabot...

[email protected] • MTH16_Lec-08_sec_7-5_LaGrange_Multipliers.pptx 1

Bruce Mayer, PE Chabot College Mathematics

Bruce Mayer, PELicensed Electrical & Mechanical Engineer

[email protected]

Chabot Mathematics

§7.5 LaGrangeMultipliers



Review §

Any QUESTIONS About• §7.4 → Least Squares Linear

Regression

Any QUESTIONS About HomeWork• §7.4 → HW-07

7.4

http://chemwiki.ucdavis.edu/@api/deki/files/12880/Figure5.10.jpg



§7.5 Learning Goals

Study the method of Lagrange multipliers as a procedure for locating points on a graph where constrained optimization can occur

Use the method of Lagrange multipliers in a number of applied problems including utility and allocation of resources

Discuss the significance of the Lagrange multiplier λ

http://www.google.com/url?sa=i&rct=j&q=&esrc=s&frm=1&source=images&cd=&cad=rja&docid=lKezI75vAomaYM&tbnid=6SA-HBIqMkf2UM:&ved=0CAUQjRw&url=http://www.stolinsky.com/wordpress/index.php/2013/11/25/lagrange-has-a-point/&ei=2HQCU6CcD9GFogSl_oJA&psig=AFQjCNGbgErNy9cCQ18fpEBGuYfenmwUWg&ust=1392756141733815



Lagrange Multipliers Often the Domain of

an Optimization is CONSTRAINED for some Reason; that is,

• k a CONSTANT

The constraint Eqn could be solved for, say y:

In other words, the Constraint fcn describes a LINE in the xy-Plane Domain surface kyxg ,

kxhy ,Constrained Domain LINE



Lagrange Multipliers The Constrained

DOMAIN Line is then Projected Up or Down by the fcn

Functional projection produces a LINE on the Range Surface

It can be shown than any extremum on the range line must be a C.P. of

yxfz ,

Constrained Range LINE

kyxgyxfyxF ,,,



Lagrange Multipliers Where λ is a new

independent variable

To Find max/min for F(x,y) take

Solving the 3 eqns:

From the above equations determine the Critical Point (C.P.) Location:

Then

kyxgyxF

y

yxg

y

yxf

y

yxF

x

yxg

x

yxf

x

yxF

,0,

,,0

,

,,0

,

kgy

g

y

f

x

g

x

f

byax

bafz ,minmax



Lagrange Multiplier Method



Example Lagrange Multipliers

Use the method of Lagrange multipliers to find the maximum value of

Subject to the Constraint of

yxxyyxfz 22),(

10 yx




SOLUTION First find the partial derivatives of f & g:

And set each equal to the Lagrange multiplier, λ, times the partials of the left side of the constraint equation:

11and22 22

y

g

x

gxxy

y

zxyy

x

z

yxgyx ,10

12and12 22 xxyxyy




Solving the first two equations for λ:

By the Last Eqn: Now use the Constraint Eqn:

The ONLY Soln to the last eqn:

xyxxyy 2and2 22

xy

xxyxyx 10so1010

5x




Recall eqn for y(x):

Thus have Two Critical Points

Check max/min by functional evaluation

Thus the MAX value of 250 occurs at (5,−5)

55 yyxy

5,5and5,5 2,21,1 baba

0)5()5()5(5)5,5( 22 f

250)5()5()5(5)5,5( 22 f



Example Find 2Var Domain

A seller’s assigned area is the six-mile radius surrounding the center of a city.

History indicates that x miles east and y miles north of city center, his/her sales competition by other businesses is Modeled by

Find• the location(s) for minimum competition • The minimum level of competition




SOLUTION The constraint for this function is the

circle of radius six miles centered about the middle of the city. Such a circle can be described by the points (x,y) satisfying the equation:

Taking the partials of the competition function find:

222 6 yx

21.0and2.0 xy

Cxy

x

C




In this case g(x,y) = k →

ReCall the Lagrange Equation:

Then the Lagrange Multiplier Minimum System

kyxyxg 222 6,

kyxgyxfyxF ,,,

336

221.0

122.0

22

2

yxkg

yxy

g

y

f

xxyx

g

x

f




Using eqn (1) to Solve for y• To prevent

Division by ZeroSpecify x ≠ 0

Use the above result in eqn (2)

SolvingtheAbove

xxy 22.0

1021.021.0 22 xyx

2222 200201.0 xx

210200200222 xx




Combining this result with the solution for y in terms of λ and the constraint equation to solve for λ:

3610210 22

346.0300

36




Finally, use the value of λ to determine values of x & y for minimum competition:

Testing the Four (x,y) Pairs find:

Thus the minimum of 1.69 businesses occurs 3.46 miles north and 4.90 miles either east/west of the center of the city

(x,y) (−4.90, −3.46) (−4.90,3.46) (4.90,−3.46) (4.90,3.46)

C(x,y) 18.31 1.69 18.31 1.69



Lagrange Multiplier as a Rate

Thus λ is a Marginal Rate for the max or min with respect to a change in the constraint value



Example Lagrange as Rate

In the Previous the minimum value was M=1.69 Businesses, with k = 36 sq-miles

If k increased by 1 sq-mi (in context this would be increasing the radius of the seller’s route), the approximate change in the minimum value:

The min no. of competing businesses would INcrease by about 0.346

346.0dk

dM



WhiteBoard Work

Problems From §7.5• P7.5-32 →

ConstantElasticityof Substitution(CES)ProductionFunction



All Done for Today

Born: 25 January 1736 Died: 10 April 1813 (aged 77)

Professorship • École Polytechnique

Academic advisors• Leonhard Euler• Giovanni Beccaria

Doctoral students• Joseph Fourier• Giovanni Plana• Siméon Poisson

JosephLouis

Lagrange

http://www.google.com/url?sa=i&rct=j&q=&esrc=s&frm=1&source=images&cd=&cad=rja&docid=36QefxUlMAjl9M&tbnid=I89vUC-fN_VV9M:&ved=0CAUQjRw&url=http://www.artofmanliness.com/2012/11/19/how-throw-a-spiral-football/&ei=Io7VUrvACY-hogSr_IKoDA&bvm=bv.59378465,d.cGU&psig=AFQjCNFpdAGRelnFsLiPrVTkxg_YeEefqw&ust=1389813543017809



All Done for Today

Born: 25 January 1736 Died: 10 April 1813 (aged 77)

Professorship • École Polytechnique

Academic advisors• Leonhard Euler• Giovanni Beccaria

Doctoral students• Joseph Fourier• Giovanni Plana• Siméon Poisson

JosephLouis

Lagrange

http://www.google.com/url?sa=i&rct=j&q=&esrc=s&frm=1&source=images&cd=&cad=rja&docid=36QefxUlMAjl9M&tbnid=I89vUC-fN_VV9M:&ved=0CAUQjRw&url=http://www.artofmanliness.com/2012/11/19/how-throw-a-spiral-football/&ei=Io7VUrvACY-hogSr_IKoDA&bvm=bv.59378465,d.cGU&psig=AFQjCNFpdAGRelnFsLiPrVTkxg_YeEefqw&ust=1389813543017809



Bruce Mayer, PELicensed Electrical & Mechanical Engineer

[email protected]

Chabot Mathematics

Appendix

–

srsrsr 22

a2 b2











Q := 50*(0.3*K^(-1/5) + 0.7*L^(-1/5))^-5

dQdK = diff(Q, K)

dQdL = diff(Q, L)

K := 140/(5+2*(35/6)^(5/6))

Kn := float(K)

L := K*(35/6)^(5/6)

Ln := float(L)



Qmax = subs(Q, K = Kn, L = Ln)

Qmax = subs(Q, K = K, L = L)

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