[PDF] [PDF] Practice Problems for Exam 2 (Solutions) 1 Use Lagrange

[PDF] Practice Problems for Exam 2 (Solutions) 1 Use Lagrange

Practice Problems for Exam 2 (Solutions) 1 Use Lagrange Multipliers to find the global maximum and minimum values of ( ) = 2 + 22 4 subject to the constraint 

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Lagrange method is used for maximizing or minimizing a general function Examples • Example 1: A rectangular box without a lid is to be made from 12 m2 of

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Section 7 4: Lagrange Multipliers and Constrained A constrained optimization problem is a problem of the form Example of use of Lagrange multipliers

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Such problems are called constrained optimization problems For example, suppose that the constraint g x, y" / k is a smooth closed curve parameterized by r t" 

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PracticeProblemsforExam2(Solutions)

x

2+2y24ysubjecttotheconstraintx2+y2=9.

g(x;y)=ktondthecriticalpoints: 2x=2x

4y4=2y

x

2+y2=9:

5.Thus,there

arefourcriticalpoints:(0;3)and(p

5;2).Evaluatingf(x;y)atthesepoints

gives:f(0;3)=6,f(0;3)=30,f(p

5;2)=5=F(p5;2).Therefore,theglobal

atthetwopoints(p

5;2).J

2.Compute

2 0 2y y

2(4xy)dxdy.

ISolution.

2 0 2y y

2(4xy)dxdy=

2 0 (2x2xy) x=2y x=y2dy 2 0 (6y22y4+y3)dy=

2y32y5

5+y44 2 0 =1664

5+4=365:

J

3.Compute

1 0 x 0 xy 0 xyzdzdydx.

ISolution.

1 0 x 0 xy 0 xyzdzdydx= 1 0 x 0xyz2 2 z=xy z=0dydx 1 0 x 0x 3y3

2dydx=

1 0x 3y48 y=x y=0dx 1 0x 7

7dx=x864

1 0=1 64:
J

Math20571

PracticeProblemsforExam2(Solutions)

(a)SketchR.

ISolution.

xy x=4y2x=y24 R J

Rf(x;y)dA

asaniterateddoubleintegral.

ISolution.

R f(x;y)dA= 2 2 4y2 y

24f(x;y)dxdy

J

5.Computethefollowingintegral:

1 0 1 p ysin(x3)dxdy: xRy x=1x=p y 1 0 1 p ysin(x3)dxdy= 1 0 x2 0 sin(x3)dydx= 1 0 ysin(x3) x2 0dx 1 0 x2sin(x3)dx=1

3cos(x3)

1 0=1

3(1cos1):

J

Math20572

PracticeProblemsforExam2(Solutions)

6.Computethefollowingintegral:

1 0 p 2y2 yp x2+y2dxdy: 01 01xRy dA=rdrdand 1 0 p 2y2 yp x2+y2dxdy=

Rpx2+y2dA

=4 0 p 2 0 rrdrd= =4 0 p2 0 r2drd =4 0r 3 3 r=p 2 r=0d= =4 02p2 3d 2p 2 12=p 2 6: J

ISolution.Firstdrawapictureofoneleaf:

xy 0 R

Math20573

PracticeProblemsforExam2(Solutions)

asin2,andtheareaisgivenby Area= R dA= =2 0 asin2 0 rdrd= =4 0r 2 2 asin2 0 d a2 2 =2 0 sin22d=a22 =2

01cos42d

a2 4 sin44 =2 0 a28: J planesandtheplanex+y+z=3. volumeisgivenby 3 0 3y 0 (3xy)dxdy= 3 0 3xx2 2xy x=3y x=0dy 3 0

93y96y+y2

2(3yy2)

dy 3 0 9

2+y223y

dy=92y+y363y22 3 0 =27

2+276272=276=92:

J x

2+y2=4,andz=0.Usecylindricalcoordinates.

ISolution.ThevolumeofQis

Q dV= 2 0 2 0 9r2 0 rdzdrd 2 0 2 0 zjz=9r2 z=0rdrd= 2 0 2 0 (9rr3)drd 2 0 9r2 2r44 2 0d= 2 0 14d =28:

JMath20574

PracticeProblemsforExam2(Solutions)

ofradiusp coordinates. sphericalcoordinatesby0p

2,02,0=4.Hence,thevolumeof

Qisgivenbytheintegral

Q dV= 4 0 2 0 p2 0

2sinddd

4 0 2 0 3 3sin p 2 0 dd=2p2 3 4 0 2 0 sindd =(2) 2p 2 3! 4

0sind=(2)

2p2 3! (cos)j=4 0 4

3(p21):

J

Math20575

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