8/11/2019 3.2c Lagrange Multipliers

1/18

THE METHOD OFLAGRANGE

MULTIPLIERS

Chapter 3 Section 6

8/11/2019 3.2c Lagrange Multipliers

2/18

METHOD OFLAGRANGEMULTIPLIERS

8/11/2019 3.2c Lagrange Multipliers

3/18

THEOREM.

Suppose f and g are functions of x and y with

continuous first partial derivatives.

If f has a relative extremum value at the point

subject to the constraint and

then there is a constant such that

0 0x y

, 0g x y , 0g x y

0 0 0 0, , 0f x y g x y

8/11/2019 3.2c Lagrange Multipliers

4/18

HOW TO FIND THE RELATIVE EXTREMA OF A

FUNCTION f OF x AND y SUBJECT TO THE

CONSTRAINT g(x,y)=0:

1. Form the auxiliary function F of three

variables x, y, and where

, , , ,F x y f x y g x y

2. Set the first derivatives of F to zero and obtain

the system of equations:

, ,x yf x y g x y

, ,

y yf x y g x y

, 0g x y

8/11/2019 3.2c Lagrange Multipliers

5/18

3. Find the critical points of F by solving the

system of equations in STEP2.

4. The first 2 coordinates of a critical point F will give

the x and y values of the desired relative extrema.

NOTE: This method may be extended to a function

of more than 2 variables.

8/11/2019 3.2c Lagrange Multipliers

6/18

Use Lagrange multipliers to find the greatest and smallest

values that the function on the ellipse

EXAMPLE.

,f x y xy2 2

1.8 2

x y

SOLUTION. ,f x y xy 2 2

, 18 2

x yg x y

,xf x y ,xg x y y 4x

4

xy

8/11/2019 3.2c Lagrange Multipliers

7/18

,yf x y ,yg x y x y x y

,f x y xy 2 2

, 18 2

x yg x y

,g x y

2 2

1 08 2

x y

4y x

x y

2 2

1

8 2

x y

8/11/2019 3.2c Lagrange Multipliers

8/18

4y x

x y

2 2

18 2

x y

4yx

(1)

(2)

(3)

(4)

in :(4) (2)4y

x yx

2 2

4x y (5)

(5) in :(3)2 2

4 18 2y y 2 1y 1y

(5)Using : 2x

8/11/2019 3.2c Lagrange Multipliers

9/18

8/11/2019 3.2c Lagrange Multipliers

10/18

Use Lagrange multipliers to find the point closest to the

origin on the plane

EXAMPLE.

2 5 0.x y z

SOLUTION. 2 2 2, ,f x y z x y z

, , 2 5g x y z x y z

,xf x y ,xg x y 2x 2

2 2x

8/11/2019 3.2c Lagrange Multipliers

11/18

,yf x y ,yg x y 2y 1 2y

x

2y

2z

2 2 2, ,f x y z x y z

, , 2 5g x y z x y z

,zf x y ,yg x y 2z 1 2z

2 5 0x y z

8/11/2019 3.2c Lagrange Multipliers

12/18

x

2y

2z

2 5 0x y z

(1)

(2)

(3)

(4)

in :(1), (2), (3) (4) 2 5 02 2

5

3

5

3x

5

6y

5

6z

8/11/2019 3.2c Lagrange Multipliers

13/18

Thus, the point closest to the origin and is contained

on the plane is the point

The minimum distance from the origin is

2 2 25 5 5

3 6 6

2 5 0x y z 5 5 5

, ,

3 6 6

6

6

8/11/2019 3.2c Lagrange Multipliers

14/18

Find the dimensions of the closed right circular cylindrical

can of smallest surface area whose volume is

EXAMPLE.

316 .cm

SOLUTION.

2

, 2 2f r h rh r

2, 16g r h r h

,rf r h

,rg r h

2 4h r

2 rh

2h r rh

8/11/2019 3.2c Lagrange Multipliers

15/18

2, 2 2f r h rh r 2, 16g r h r h

,hf r h ,hg r h 2 r 2r 2 r

,g r h 2 16 0r h 2 16r h

2 r

216r h

2h r rh

8/11/2019 3.2c Lagrange Multipliers

16/18

8/11/2019 3.2c Lagrange Multipliers

17/18

Thus, and

The minimum surface area is

4h 2r

are the dimensions of the closed right circular cylindrical

can of smallest surface area whose volume is3

16 .cm

2

2 2 4 2 2

2, 4f

24

8/11/2019 3.2c Lagrange Multipliers

18/18