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Lagrange Multipliers

Slide Duration:

Section 1: Vectors
Points & Vectors

28m 23s

Intro
0:00
Points and Vectors
1:02
A Point in a Plane
1:03
A Point in Space
3:14
Notation for a Space of a Given Space
6:34
Introduction to Vectors
9:51
14:51
Example 1
16:52
18:24
Example 2
21:01
Two More Properties of Vector Addition
24:16
Multiplication of a Vector by a Constant
25:27
Scalar Product & Norm

30m 25s

Intro
0:00
Scalar Product and Norm
1:05
Introduction to Scalar Product
1:06
Example 1
3:21
Properties of Scalar Product
6:14
Definition: Orthogonal
11:41
Example 2: Orthogonal
14:19
Definition: Norm of a Vector
15:30
Example 3
19:37
Distance Between Two Vectors
22:05
Example 4
27:19
More on Vectors & Norms

38m 18s

Intro
0:00
More on Vectors and Norms
0:38
Open Disc
0:39
Close Disc
3:14
Open Ball, Closed Ball, and the Sphere
5:22
Property and Definition of Unit Vector
7:16
Example 1
14:04
Three Special Unit Vectors
17:24
General Pythagorean Theorem
19:44
Projection
23:00
Example 2
28:35
Example 3
35:54
Inequalities & Parametric Lines

33m 19s

Intro
0:00
Inequalities and Parametric Lines
0:30
Starting Example
0:31
Theorem 1
5:10
Theorem 2
7:22
Definition 1: Parametric Equation of a Straight Line
10:16
Definition 2
17:38
Example 1
21:19
Example 2
25:20
Planes

29m 59s

Intro
0:00
Planes
0:18
Definition 1
0:19
Example 1
7:04
Example 2
12:45
General Definitions and Properties: 2 Vectors are Said to Be Paralleled If
14:50
Example 3
16:44
Example 4
20:17
More on Planes

34m 18s

Intro
0:00
More on Planes
0:25
Example 1
0:26
Distance From Some Point in Space to a Given Plane: Derivation
10:12
Final Formula for Distance
21:20
Example 2
23:09
Example 3: Part 1
26:56
Example 3: Part 2
31:46
Section 2: Differentiation of Vectors
Maps, Curves & Parameterizations

29m 48s

Intro
0:00
Maps, Curves and Parameterizations
1:10
Recall
1:11
Looking at y = x2 or f(x) = x2
2:23
Departure Space & Arrival Space
7:01
Looking at a 'Function' from ℝ to ℝ2
10:36
Example 1
14:50
Definition 1: Parameterized Curve
17:33
Example 2
21:56
Example 3
25:16
Differentiation of Vectors

39m 40s

Intro
0:00
Differentiation of Vectors
0:18
Example 1
0:19
Definition 1: Velocity of a Curve
1:45
Line Tangent to a Curve
6:10
Example 2
7:40
Definition 2: Speed of a Curve
12:18
Example 3
13:53
Definition 3: Acceleration Vector
16:37
Two Definitions for the Scalar Part of Acceleration
17:22
Rules for Differentiating Vectors: 1
19:52
Rules for Differentiating Vectors: 2
21:28
Rules for Differentiating Vectors: 3
22:03
Rules for Differentiating Vectors: 4
24:14
Example 4
26:57
Section 3: Functions of Several Variables
Functions of Several Variable

29m 31s

Intro
0:00
Length of a Curve in Space
0:25
Definition 1: Length of a Curve in Space
0:26
Extended Form
2:06
Example 1
3:40
Example 2
6:28
Functions of Several Variable
8:55
Functions of Several Variable
8:56
General Examples
11:11
Graph by Plotting
13:00
Example 1
16:31
Definition 1
18:33
Example 2
22:15
Equipotential Surfaces
25:27
Isothermal Surfaces
27:30
Partial Derivatives

23m 31s

Intro
0:00
Partial Derivatives
0:19
Example 1
0:20
Example 2
5:30
Example 3
7:48
Example 4
9:19
Definition 1
12:19
Example 5
14:24
Example 6
16:14
20:26
Higher and Mixed Partial Derivatives

30m 48s

Intro
0:00
Higher and Mixed Partial Derivatives
0:45
Definition 1: Open Set
0:46
Notation: Partial Derivatives
5:39
Example 1
12:00
Theorem 1
14:25
Now Consider a Function of Three Variables
16:50
Example 2
20:09
Caution
23:16
Example 3
25:42
Section 4: Chain Rule and The Gradient
The Chain Rule

28m 3s

Intro
0:00
The Chain Rule
0:45
Conceptual Example
0:46
Example 1
5:10
The Chain Rule
10:11
Example 2: Part 1
19:06
Example 2: Part 2 - Solving Directly
25:26
Tangent Plane

42m 25s

Intro
0:00
Tangent Plane
1:02
Tangent Plane Part 1
1:03
Tangent Plane Part 2
10:00
Tangent Plane Part 3
18:18
Tangent Plane Part 4
21:18
Definition 1: Tangent Plane to a Surface
27:46
Example 1: Find the Equation of the Plane Tangent to the Surface
31:18
Example 2: Find the Tangent Line to the Curve
36:54
Further Examples with Gradients & Tangents

47m 11s

Intro
0:00
Example 1: Parametric Equation for the Line Tangent to the Curve of Two Intersecting Surfaces
0:41
Part 1: Question
0:42
Part 2: When Two Surfaces in ℝ3 Intersect
4:31
Part 3: Diagrams
7:36
Part 4: Solution
12:10
Part 5: Diagram of Final Answer
23:52
Example 2: Gradients & Composite Functions
26:42
Part 1: Question
26:43
Part 2: Solution
29:21
Example 3: Cos of the Angle Between the Surfaces
39:20
Part 1: Question
39:21
Part 2: Definition of Angle Between Two Surfaces
41:04
Part 3: Solution
42:39
Directional Derivative

41m 22s

Intro
0:00
Directional Derivative
0:10
Rate of Change & Direction Overview
0:11
Rate of Change : Function of Two Variables
4:32
Directional Derivative
10:13
Example 1
18:26
Examining Gradient of f(p) ∙ A When A is a Unit Vector
25:30
Directional Derivative of f(p)
31:03
33:23
Example 2
34:53
A Unified View of Derivatives for Mappings

39m 41s

Intro
0:00
A Unified View of Derivatives for Mappings
1:29
Derivatives for Mappings
1:30
Example 1
5:46
Example 2
8:25
Example 3
12:08
Example 4
14:35
Derivative for Mappings of Composite Function
17:47
Example 5
22:15
Example 6
28:42
Section 5: Maxima and Minima
Maxima & Minima

36m 41s

Intro
0:00
Maxima and Minima
0:35
Definition 1: Critical Point
0:36
Example 1: Find the Critical Values
2:48
Definition 2: Local Max & Local Min
10:03
Theorem 1
14:10
Example 2: Local Max, Min, and Extreme
18:28
Definition 3: Boundary Point
27:00
Definition 4: Closed Set
29:50
Definition 5: Bounded Set
31:32
Theorem 2
33:34
Further Examples with Extrema

32m 48s

Intro
0:00
Further Example with Extrema
1:02
Example 1: Max and Min Values of f on the Square
1:03
Example 2: Find the Extreme for f(x,y) = x² + 2y² - x
10:44
Example 3: Max and Min Value of f(x,y) = (x²+ y²)⁻¹ in the Region (x -2)²+ y² ≤ 1
17:20
Lagrange Multipliers

32m 32s

Intro
0:00
Lagrange Multipliers
1:13
Theorem 1
1:14
Method
6:35
Example 1: Find the Largest and Smallest Values that f Achieves Subject to g
9:14
Example 2: Find the Max & Min Values of f(x,y)= 3x + 4y on the Circle x² + y² = 1
22:18
More Lagrange Multiplier Examples

27m 42s

Intro
0:00
Example 1: Find the Point on the Surface z² -xy = 1 Closet to the Origin
0:54
Part 1
0:55
Part 2
7:37
Part 3
10:44
Example 2: Find the Max & Min of f(x,y) = x² + 2y - x on the Closed Disc of Radius 1 Centered at the Origin
16:05
Part 1
16:06
Part 2
19:33
Part 3
23:17
Lagrange Multipliers, Continued

31m 47s

Intro
0:00
Lagrange Multipliers
0:42
First Example of Lesson 20
0:44
Let's Look at This Geometrically
3:12
Example 1: Lagrange Multiplier Problem with 2 Constraints
8:42
Part 1: Question
8:43
Part 2: What We Have to Solve
15:13
Part 3: Case 1
20:49
Part 4: Case 2
22:59
Part 5: Final Solution
25:45
Section 6: Line Integrals and Potential Functions
Line Integrals

36m 8s

Intro
0:00
Line Integrals
0:18
Introduction to Line Integrals
0:19
Definition 1: Vector Field
3:57
Example 1
5:46
Example 2: Gradient Operator & Vector Field
8:06
Example 3
12:19
Vector Field, Curve in Space & Line Integrals
14:07
Definition 2: F(C(t)) ∙ C'(t) is a Function of t
17:45
Example 4
18:10
Definition 3: Line Integrals
20:21
Example 5
25:00
Example 6
30:33
More on Line Integrals

28m 4s

Intro
0:00
More on Line Integrals
0:10
Line Integrals Notation
0:11
Curve Given in Non-parameterized Way: In General
4:34
Curve Given in Non-parameterized Way: For the Circle of Radius r
6:07
Curve Given in Non-parameterized Way: For a Straight Line Segment Between P & Q
6:32
The Integral is Independent of the Parameterization Chosen
7:17
Example 1: Find the Integral on the Ellipse Centered at the Origin
9:18
Example 2: Find the Integral of the Vector Field
16:26
Discussion of Result and Vector Field for Example 2
23:52
Graphical Example
26:03
Line Integrals, Part 3

29m 30s

Intro
0:00
Line Integrals
0:12
Piecewise Continuous Path
0:13
Closed Path
1:47
Example 1: Find the Integral
3:50
The Reverse Path
14:14
Theorem 1
16:18
Parameterization for the Reverse Path
17:24
Example 2
18:50
Line Integrals of Functions on ℝn
21:36
Example 3
24:20
Potential Functions

40m 19s

Intro
0:00
Potential Functions
0:08
Definition 1: Potential Functions
0:09
Definition 2: An Open Set S is Called Connected if…
5:52
Theorem 1
8:19
Existence of a Potential Function
11:04
Theorem 2
18:06
Example 1
22:18
Contrapositive and Positive Form of the Theorem
28:02
The Converse is Not Generally True
30:59
Our Theorem
32:55
Compare the n-th Term Test for Divergence of an Infinite Series
36:00
So for Our Theorem
38:16
Potential Functions, Continued

31m 45s

Intro
0:00
Potential Functions
0:52
Theorem 1
0:53
Example 1
4:00
Theorem in 3-Space
14:07
Example 2
17:53
Example 3
24:07
Potential Functions, Conclusion & Summary

28m 22s

Intro
0:00
Potential Functions
0:16
Theorem 1
0:17
In Other Words
3:25
Corollary
5:22
Example 1
7:45
Theorem 2
11:34
Summary on Potential Functions 1
15:32
Summary on Potential Functions 2
17:26
Summary on Potential Functions 3
18:43
Case 1
19:24
Case 2
20:48
Case 3
21:35
Example 2
23:59
Section 7: Double Integrals
Double Integrals

29m 46s

Intro
0:00
Double Integrals
0:52
Introduction to Double Integrals
0:53
Function with Two Variables
3:39
Example 1: Find the Integral of xy³ over the Region x ϵ[1,2] & y ϵ[4,6]
9:42
Example 2: f(x,y) = x²y & R be the Region Such That x ϵ[2,3] & x² ≤ y ≤ x³
15:07
Example 3: f(x,y) = 4xy over the Region Bounded by y= 0, y= x, and y= -x+3
19:20
Polar Coordinates

36m 17s

Intro
0:00
Polar Coordinates
0:50
Polar Coordinates
0:51
Example 1: Let (x,y) = (6,√6), Convert to Polar Coordinates
3:24
Example 2: Express the Circle (x-2)² + y² = 4 in Polar Form.
5:46
Graphing Function in Polar Form.
10:02
Converting a Region in the xy-plane to Polar Coordinates
14:14
Example 3: Find the Integral over the Region Bounded by the Semicircle
20:06
Example 4: Find the Integral over the Region
27:57
Example 5: Find the Integral of f(x,y) = x² over the Region Contained by r= 1 - cosθ
32:55
Green's Theorem

38m 1s

Intro
0:00
Green's Theorem
0:38
Introduction to Green's Theorem and Notations
0:39
Green's Theorem
3:17
Example 1: Find the Integral of the Vector Field around the Ellipse
8:30
Verifying Green's Theorem with Example 1
15:35
A More General Version of Green's Theorem
20:03
Example 2
22:59
Example 3
26:30
Example 4
32:05
Divergence & Curl of a Vector Field

37m 16s

Intro
0:00
Divergence & Curl of a Vector Field
0:18
Definitions: Divergence(F) & Curl(F)
0:19
Example 1: Evaluate Divergence(F) and Curl(F)
3:43
Properties of Divergence
9:24
Properties of Curl
12:24
Two Versions of Green's Theorem: Circulation - Curl
17:46
Two Versions of Green's Theorem: Flux Divergence
19:09
Circulation-Curl Part 1
20:08
Circulation-Curl Part 2
28:29
Example 2
32:06
Divergence & Curl, Continued

33m 7s

Intro
0:00
Divergence & Curl, Continued
0:24
Divergence Part 1
0:25
Divergence Part 2: Right Normal Vector and Left Normal Vector
5:28
Divergence Part 3
9:09
Divergence Part 4
13:51
Divergence Part 5
19:19
Example 1
23:40
Final Comments on Divergence & Curl

16m 49s

Intro
0:00
Final Comments on Divergence and Curl
0:37
Several Symbolic Representations for Green's Theorem
0:38
Circulation-Curl
9:44
Flux Divergence
11:02
Closing Comments on Divergence and Curl
15:04
Section 8: Triple Integrals
Triple Integrals

27m 24s

Intro
0:00
Triple Integrals
0:21
Example 1
2:01
Example 2
9:42
Example 3
15:25
Example 4
20:54
Cylindrical & Spherical Coordinates

35m 33s

Intro
0:00
Cylindrical and Spherical Coordinates
0:42
Cylindrical Coordinates
0:43
When Integrating Over a Region in 3-space, Upon Transformation the Triple Integral Becomes..
4:29
Example 1
6:27
The Cartesian Integral
15:00
Introduction to Spherical Coordinates
19:44
Reason It's Called Spherical Coordinates
22:49
Spherical Transformation
26:12
Example 2
29:23
Section 9: Surface Integrals and Stokes' Theorem
Parameterizing Surfaces & Cross Product

41m 29s

Intro
0:00
Parameterizing Surfaces
0:40
Describing a Line or a Curve Parametrically
0:41
Describing a Line or a Curve Parametrically: Example
1:52
Describing a Surface Parametrically
2:58
Describing a Surface Parametrically: Example
5:30
Recall: Parameterizations are not Unique
7:18
Example 1: Sphere of Radius R
8:22
Example 2: Another P for the Sphere of Radius R
10:52
This is True in General
13:35
Example 3: Paraboloid
15:05
Example 4: A Surface of Revolution around z-axis
18:10
Cross Product
23:15
Defining Cross Product
23:16
Example 5: Part 1
28:04
Example 5: Part 2 - Right Hand Rule
32:31
Example 6
37:20
Tangent Plane & Normal Vector to a Surface

37m 6s

Intro
0:00
Tangent Plane and Normal Vector to a Surface
0:35
Tangent Plane and Normal Vector to a Surface Part 1
0:36
Tangent Plane and Normal Vector to a Surface Part 2
5:22
Tangent Plane and Normal Vector to a Surface Part 3
13:42
Example 1: Question & Solution
17:59
Example 1: Illustrative Explanation of the Solution
28:37
Example 2: Question & Solution
30:55
Example 2: Illustrative Explanation of the Solution
35:10
Surface Area

32m 48s

Intro
0:00
Surface Area
0:27
Introduction to Surface Area
0:28
Given a Surface in 3-space and a Parameterization P
3:31
Defining Surface Area
7:46
Curve Length
10:52
Example 1: Find the Are of a Sphere of Radius R
15:03
Example 2: Find the Area of the Paraboloid z= x² + y² for 0 ≤ z ≤ 5
19:10
Example 2: Writing the Answer in Polar Coordinates
28:07
Surface Integrals

46m 52s

Intro
0:00
Surface Integrals
0:25
Introduction to Surface Integrals
0:26
General Integral for Surface Are of Any Parameterization
3:03
Integral of a Function Over a Surface
4:47
Example 1
9:53
Integral of a Vector Field Over a Surface
17:20
Example 2
22:15
Side Note: Be Very Careful
28:58
Example 3
30:42
Summary
43:57
Divergence & Curl in 3-Space

23m 40s

Intro
0:00
Divergence and Curl in 3-Space
0:26
Introduction to Divergence and Curl in 3-Space
0:27
Define: Divergence of F
2:50
Define: Curl of F
4:12
The Del Operator
6:25
Symbolically: Div(F)
9:03
Symbolically: Curl(F)
10:50
Example 1
14:07
Example 2
18:01
Divergence Theorem in 3-Space

34m 12s

Intro
0:00
Divergence Theorem in 3-Space
0:36
Green's Flux-Divergence
0:37
Divergence Theorem in 3-Space
3:34
Note: Closed Surface
6:43
Figure: Paraboloid
8:44
Example 1
12:13
Example 2
18:50
Recap for Surfaces: Introduction
27:50
Recap for Surfaces: Surface Area
29:16
Recap for Surfaces: Surface Integral of a Function
29:50
Recap for Surfaces: Surface Integral of a Vector Field
30:39
Recap for Surfaces: Divergence Theorem
32:32
Stokes' Theorem, Part 1

22m 1s

Intro
0:00
Stokes' Theorem
0:25
Recall Circulation-Curl Version of Green's Theorem
0:26
Constructing a Surface in 3-Space
2:26
Stokes' Theorem
5:34
Note on Curve and Vector Field in 3-Space
9:50
Example 1: Find the Circulation of F around the Curve
12:40
Part 1: Question
12:48
Part 2: Drawing the Figure
13:56
Part 3: Solution
16:08
Stokes' Theorem, Part 2

20m 32s

Intro
0:00
Example 1: Calculate the Boundary of the Surface and the Circulation of F around this Boundary
0:30
Part 1: Question
0:31
Part 2: Drawing the Figure
2:02
Part 3: Solution
5:24
Example 2: Calculate the Boundary of the Surface and the Circulation of F around this Boundary
13:11
Part 1: Question
13:12
Part 2: Solution
13:56

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 1 answerLast reply by: Professor HovasapianTue Apr 7, 2015 10:39 PMPost by Alvi Akbar on April 7, 2015Is there any way to change the video playback speed ? 4 answersLast reply by: Professor HovasapianSun Apr 14, 2013 1:29 PMPost by Jawad Hassan on April 12, 2013Could you explain why in ex2 there isnt 4 points, you said in the video because x and y have both posetiv signs, but they also have posetiv signs in example 1. Got me confused.. 0 answersPost by Shahaz Shajahan on September 4, 2012Sorry I realised how stupid that question was :S, It works out exactly the same, obviously. 0 answersPost by Shahaz Shajahan on September 4, 2012Ok, instead of find a direct value for lambda, could we not find the values of lambda in terms of x and y and equal them together?

### Lagrange Multipliers

Find the greatest and least values of f(x,y) = y2 − x2 subjected to g(x,y):0 = x − y2.
• To find the greatest and least values of a function f subjected to g, we solve the system of equations
 ∇f = l∇g
 g(x,y) = 0
where l is a constant.
• Now, ∇f = ( − 2x,2y) and ∇g = (1, − 2y), let g(x,y) = x − y2. Then our system of equations is
 ( − 2x,2y) = l(1, − 2y)
 x − y2 = 0
• ( − 2x,2y) = l(1, − 2y) is equivalent to
 − 2x = l
 2y = − 2yl
this implies that x = − [l/2] and if y0 then l = − 1.
• For y = 0 we have x − y2 = x − (0)2 = 0 so x = 0 but then l = 0, which we don't allow.
• Otherwise x = − [l/2] = − [(( − 1))/2] = [1/2]. Utilizing x − y2 = 0, we have [1/2] − y2 = 0 so y = ±[1/(√2 )].
• Our value points are then f( [1/2],[1/(√2 )] ) = [1/4] and f( [1/2], − [1/(√2 )] ) = [1/4].
Since f(0,0) = 0 (note that (0,0) satisfies g) we conclude that [1/4] is our greatest value. There is no least value.
Find the greatest and least values of f(x,y) = 2x3 − 4y2 + 3 subjected to g(x,y):0 = x2 − y.
• To find the greatest and least values of a function f subjected to g, we solve the system of equations
 ∇f = l∇g
 g(x,y) = 0
where l is a constant.
• Now, ∇f = (6x2, − 8y) and ∇g = (2x, − 1), let g(x,y) = x2 − y. Then our system of equations is
 (6x2, − 8y) = l(2x, − 1)
 x2 − y = 0
• (6x2, − 8y) = l(2x, − 1) is equivalent to
 6x2 = 2xl
 − 8y = − l
this implies that y = [l/8] and if x0 then x = [l/3].
• For x = 0 we have x2 − y = (0)2 − y = 0 so y = 0 but then l = 0, which we don't allow.
• Otherwise, utilizing x2 − y = 0, we have ( [l/3] )2 − ( [l/8] ) = 0 or l( [l/9] − [1/8] ) = 0 and so l = 0 (which we omit) or l = [9/8].
• For l = [9/8] we have x = [l/3] = [1/3]( [9/8] ) and y = [l/8] = [1/8]( [9/8] ) which give x = [3/8] and y = [9/64].
• Our value point is f( [3/8],[9/64] ) = [3,099/1,024] ≈ 3.02.
Since f(0,0) = 3 (note that (0,0) satisfies g) we conclude that [3,099/1,024] is our greatest. There is no least value.
Find the greatest and least values of f(x,y) = xy subjected to g(x,y):1 = x2 + y2.
• To find the greatest and least values of a function f subjected to g, we solve the system of equations
 ∇f = l∇g
 g(x,y) = 0
where l is a constant.
• Now, ∇f = (y,x) and ∇g = ( 2x,2y ), let g(x,y) = x2 + y2 − 1. Then our system of equations is
 (y,x) = l( 2x,2y )
 x2 + y2 − 1 = 0
• (y,x) = l( 2x,2y ) is equivalent to
 y = 2xl
 x = 2yl
we can substitute y from the top equation to x = 2yl.
• We obtain x = 2(2xl)l which gives l = ±[1/4]. We can again substitute y = 2xl into x2 + y2 − 1 = 0.
• We obtain x2 + (2xl)2 − 1 = 0 or x2(1 + 4l2) = 1. When l = ±[1/4], we have x = ±[2/(√5 )].
• To solve for y we utilize y = 2xl, but note that we have four cases: x = [2/(√5 )] with l = [1/4], x = [2/(√5 )] with l = − [1/4], x = − [2/(√5 )] with l = [1/4], and x = − [2/(√5 )] with l = − [1/4].
• Our first case gives y = 2xl = 2( [2/(√5 )] )( [1/4] ) = [1/(√5 )]. Similarly our last case gives y = 2xl = 2( − [2/(√5 )] )( − [1/4] ) = [1/(√5 )].
• Our second case gives y = 2xl = 2( [2/(√5 )] )( − [1/4] ) = − [1/(√5 )]. Similarly our third case gives y = 2xl = 2( − [2/(√5 )] )( [1/4] ) = − [1/(√5 )].
• We obtain four value points f( [2/(√5 )],[1/(√5 )] ) = [2/5], f( − [2/(√5 )],[1/(√5 )] ) = − [2/5], f( [2/(√5 )], − [1/(√5 )] ) = − [2/5], and f( − [2/(√5 )], − [1/(√5 )] ) = [2/5].
Our least value is − [2/5] while our greatest value is [2/5].
Find the greatest value of f(x,y) = 4x − 3y subjected to g(x,y):1 = [(x2)/9] − [(y2)/4] for x > 0 and y > 0.
• To find the greatest value of f subjected to g, we solve the system of equations
 ∇f = l∇g
 g(x,y) = 0
where l is a constant.
• Now, ∇f = (4, − 3) and ∇g = ( [2x/9], − [y/2] ), let g(x,y) = [(x2)/9] − [(y2)/4] − 1. Then our system of equations is
 (4, − 3) = l( [2x/9], − [y/2] )
 [(x2)/9] − [(y2)/4] − 1 = 0
• (4, − 3) = l( [2x/9], − [y/2] ) is equivalent to
 4 = [2x/9]l
 − 3 = − [y/2]l
and so x = [18/l] and y = [6/l].
• Utilizing [(x2)/9] − [(y2)/4] − 1 we substitute our values for x and y to obtain [36/(l2)] − [9/(l2)] − 1 = 0. We now solve for l.
• Now, [36/(l2)] − [9/(l2)] − 1 = 0 is equivalent to 27 = l2 and so l = ±3√3.
• Solving for x yields x = [18/l] = [18/( ±3√3 )] since one of our conditions is x > 0 our solution is x = [6/(√3 )].
• Similarly y = [6/l] = [6/( ±3√3 )] and our condition y > 0 makes it so that our solution is y = [2/(√3 )].
We obtain the greatest value at f( [6/(√2 )],[2/(√3 )] ) = [(24√3 − 6√2 )/(√6 )].
Find the least value of f(x,y) = [(x2)/9] + [(y2)/16] subjected to g(x,y):x3 + y2 = 10 for x > 0 and y > 0.
• To find the least value of f subjected to g, we solve the system of equations
 ∇f = l∇g
 g(x,y) = 0
where l is a constant.
• Now, ∇f = ( [2x/9],[2y/8] ) and ∇g = ( 3x2,2y ), let g(x,y) = x3 + y2 − 10. Then our system of equations is
 ( [2x/9],[2y/8] ) = l( 3x2,2y )
 x3 + y2 − 10 = 0
• ( [2x/9],[2y/8] ) = l( 3x2,2y ) is equivalent to
 [2x/9] = 3x2l
 [2y/8] = 2yl
and so x = [2/27l] and l = [1/8].
• We now have x = [16/27]. Utilizing x3 + y2 − 10 = 0 we substitute our values for x and obtain ( [16/27] )3 + y2 − 10 = 0. We can now solve for y.
• Hence y = ±[1/27]√{[192,734/27]} , we only take the positve value as our condition states y > 0.
We obtain the least value at f( [16/27],[1/27]√{[192,734/27]} ) = [102,511/157,464].
Find the greatest and least values of f(x,y,z) = 3x − 2y + z subjected to g(x,y,z):1 = x2 − 2y2 − z2.
• To find the greatest and least values of a function f subjected to g, we solve the system of equations
 ∇f = l∇g
 g(x,y,z) = 0
where l is a constant.
• Now, ∇f = (3, − 2,1) and ∇g = (2x, − 4y, − 2z), let g(x,y,z) = x2 − 2y2 − z2 − 1. Then our system of equations is
 (3, − 2,1) = l(2x, − 4y, − 2z)
 x2 − 2y2 − z2 − 1 = 0
• (3, − 2,1) = l(2x, − 4y, − 2z) is equivalent to
 3 = 2xl
 − 2 = − 4yl
 1 = − 2zl
this implies that x = [3/2l], y = [1/2l] and z = − [1/2l].
• Utilizing x2 − 2y2 − z2 − 1 = 0 and subsituting x, y and z yields [3/(2l2)] = 1 so l = ±√{[3/2]}
• For l = √{[3/2]} we have x = [3/2l] = √{[3/2]} , y = [1/2l] = [1/3]√{[3/2]} and z = − [1/2l] = − [1/3]√{[3/2]} and the value f( √{[3/2]} ,[1/3]√{[3/2]} , − [1/3]√{[3/2]} ).
• For l = − √{[3/2]} we have x = [3/2l] = − √{[3/2]} , y = [1/2l] = − [1/3]√{[3/2]} and z = − [1/2l] = [1/3]√{[3/2]} and the value f( − √{[3/2]} , − [1/3]√{[3/2]} ,[1/3]√{[3/2]} ).
Hence 2√{[3/2]} is our greatest value and − 2√{[3/2]} is our least value.
Find the greatest and least values of f(x,y,z) = 4x2 − z2 subjected to g(x,y,z):1 = 5x + y − z.
• To find the greatest and least values of a function f subjected to g, we solve the system of equations
 ∇f = l∇g
 g(x,y,z) = 0
where l is a constant.
• Now, ∇f = (8x,0, − 2z) and ∇g = (5,1, − 1), let g(x,y,z) = 5x + y − z − 1. Then our system of equations is
 (8x,0, − 2z) = l(5,1, − 1)
 5x + y − z − 1 = 0
(8x,0, − 2z) = l(5,1, − 1) is equivalent to
 8x = 5l
 0 = l
 − 2z = − l
since l = 0 we have no greatest or least value.
Find the greatest and least values of f(x,y,z) = xyz subjected to g(x,y,z):4 = x2 + 3y2 + z2 for x > 0, y > 0 and z > 0.
• To find the greatest and least values of a function f subjected to g, we solve the system of equations
 ∇f = l∇g
 g(x,y,z) = 0
where l is a constant.
• Now, ∇f = (yz,xz,xy) and ∇g = (2x,6y,2z), let g(x,y,z) = x2 + 3y2 + z2 − 4. Then our system of equations is
 (yz,xz,xy) = l(2x,6y,2z)
 x2 + 3y2 + z2 − 4 = 0
• (yz,xz,xy) = l(2x,6y,2z) is equivalent to
 yz = 2xl
 xz = 6yl
 xy = 2zl
solving the top equation for x results in x = [yz/2l]. We substitute this result in the other two equations.
• So ( [yz/2l] )z = 6yl gives z2 = 12l2 while ( [yz/2l] )y = 2zl gives y2 = 4l2. To find x2 we can solve for z in the bottom equation and substitute on the middle equation.
• So z = [xy/2l] and x( [xy/2l] ) = 6yl gives x2 = 12l2. Utilizing x2 − 3y2 + z2 − 4 = 0 and our previous results yields ( 12l2 )2 − 3( 4l2 )2 + ( 12l2 )2 = 4 hence l = [1/(4√{60})].
• Then x2 = 12l2 = [12/(√{60} )] = [6/(√{15} )] so x = ±√{[6/(√{15} )]} , y2 = 4l2 = [4/(√{60} )] = [2/(√{15} )] so y = ±√{[2/(√{15} )]} and z2 = 12l2 = [12/(√{60} )] = [6/(√{15} )] so z = ±√{[6/(√{15} )]} .
• We therefore have eight value points four with the greatest value, f( √{[6/(√{15} )]} ,√{[2/(√{15} )]} ,√{[6/(√{15} )]} ) = f( √{[6/(√{15} )]} , − √{[2/(√{15} )]} , − √{[6/(√{15} )]} ) = f( − √{[6/(√{15} )]} , − √{[2/(√{15} )]} ,√{[6/(√{15} )]} ) = f( − √{[6/(√{15} )]} ,√{[2/(√{15} )]} , − √{[6/(√{15} )]} ) = [(6√2 )/(4√{153})]
And four with the least value, f( − √{[6/(√{15} )]} ,√{[2/(√{15} )]} ,√{[6/(√{15} )]} ) = f( √{[6/(√{15} )]} , − √{[2/(√{15} )]} ,√{[6/(√{15} )]} ) = f( √{[6/(√{15} )]} ,√{[2/(√{15} )]} , − √{[6/(√{15} )]} ) = f( − √{[6/(√{15} )]} , − √{[2/(√{15} )]} , − √{[6/(√{15} )]} ) = − [(6√2 )/(4√{153})].
Find the greatest value of f(x,y,z) = x + 3y + 8z subjected to g(x,y,z):1 = x2 + y2 + z2 for x > 0, y > 0 and z > 0.
• To find the greatest value of f subjected to g, we solve the system of equations
 ∇f = l∇g
 g(x,y,z) = 0
where l is a constant.
• Now, ∇f = (1,3,8) and ∇g = (2x,2y,2z), let g(x,y,z) = x2 + y2 + z2 − 1. Then our system of equations is
 (1,3,8) = l(2x,2y,2z)
 x2 + y2 + z2 − 1 = 0
• (1,3,8) = l(2x,2y,2z) is equivalent to
 1 = xl
 3 = 2yl
 8 = 2zl
which implies x = [1/l], y = [3/2l] and z = [4/l].
• Utilizing x2 + y2 + z2 − 1 = 0 we substitute to get ( [1/l] )2 + ( [3/2l] )2 + ( [4/l] )2 = 1 or 77 = 4l2.
• Hence l = ±[(√{77} )/2], but since x, y and z have to be greater than zero, we only use the positive answer.
Thus x = [1/l] = [2/(√{77})], y = [3/2l] = [3/(√{77})] and z = [4/l] = [8/(√{77} )] and our greatest value is f( [2/(√{ 77})],[3/(√{77})],[8/(√{77})] ) = [75/(√{77})].
Find the least value of f(x,y,z) = (x − 1)2 + y2 + (z − 1)2 subjected to g(x,y,z):1 = x + y + z.
• To find the least value of f subjected to g, we solve the system of equations
 ∇f = l∇g
 g(x,y,z) = 0
where l is a constant.
• Now, ∇f = (2x − 2,2y,2z − 2) and ∇g = (1,1,1), let g(x,y,z) = x + y + z − 1. Then our system of equations is
 (2x − 2,2y,2z − 2) = l(1,1,1)
 x + y + z − 1 = 0
• (2x − 2,2y,2z − 2) = l(1,1,1) is equivalent to
 2x − 2 = l
 2y = l
 2z − 2 = l
which implies x = [(l + 2)/2], y = [l/2] and z = [(l + 2)/2].
• Utilizing x + y + z − 1 = 0 we substitute to get [(l + 2)/2] + [l/2] + [(l + 2)/2] = 1 or 3l + 4 = 2.
• Hence l = − [2/3] and so x = [(l + 2)/2] = [2/3], y = [l/2] = − [1/3] and z = [(l + 2)/2] = [2/3].
Thus our least value is f( [2/3], − [1/3],[2/3] ) = [1/3].

*These practice questions are only helpful when you work on them offline on a piece of paper and then use the solution steps function to check your answer.

### Lagrange Multipliers

Lecture Slides are screen-captured images of important points in the lecture. Students can download and print out these lecture slide images to do practice problems as well as take notes while watching the lecture.

• Intro 0:00
• Lagrange Multipliers 1:13
• Theorem 1
• Method
• Example 1: Find the Largest and Smallest Values that f Achieves Subject to g
• Example 2: Find the Max & Min Values of f(x,y)= 3x + 4y on the Circle x² + y² = 1

### Transcription: Lagrange Multipliers

Hello and welcome back to educator.com and Multivariable Calculus.0000

Today's topic is going to be Lagrange multipliers. This is a very, very important topic.0004

Because it is very important, we are actually going to be spending several lessons on it. In this particular lesson, what I am going to do is write out the theorem and the method.0009

Then we will start with a couple of basic examples, just to get the underlying technique under our belts. Then the following lesson, I am actually going to be spending some time on just more examples, slightly more complicated, a little bit more involved.0019

Again, this a very powerful technique for finding maxima, minima, on a restrained... the maximum and minimum of a function constrained in some sort of way.0033

So, In and of itself, it is not that difficult because, really all you are doing is you are solving a bunch of simultaneous equations. Whether it be 2 equations, 3, 4, or 5 depending on how many variables you are working with.0044

The problem with this is there just tends to be a lot of things on the page, so there is a lot to keep track of, so again, just go slowly, make sure everything is written out, do not take any short cuts, aside from that let us just jump right on in and get a sense of what is going on.0055

I am just going to start out by writing the theorem and then from the theorem I am going to use that just to write out the quick method, then we will go ahead and start the examples.0074

The theorem is going to be a bit on the long side, but it is important, so let g be a continuously differentiable function -- all that means is when you differentiate it, you get a continuous function -- continuously differentiable function, on an open set u.0085

Now, let a be the set of points in u. Let me actually give the points a name, the set if points x in u, so we are talking about a vector, a point in n space, such that g of x equals 0.0131

This is a very, very important hypothesis, but the gradient of g at x does not equal 0. So it is just really, really important that on this particular set of points that we are dealing with that this gradient of this function g not be 0.0163

Essentially we are just saying that it is smooth. That is all. Now, let f be a continuously differentiable function on u, and let p be a point in a, such that p is an extremum, in other words it is a maximum or minimum, an extremum for f.0185

That is, let p... let us write it this way... p is an extremum for f subject to -- let us not use the word subject, let us use the word constrained, because that is exactly what we are doing for f -- constrained by g.0245

Then... and again do not worry about this, this is just a formality to write it, it is just something that we might occasionally refer to, it is the method that is important, but I want you to see the formality.0287

Then, there exists a number -- traditionally we use the Greek letter lambda for that -- such that... here is the important part... the gradient of f evaluated at p = λ × the gradient of g evaluated at p.0297

So, as far as what this means geometrically, I will be talking about that a little bit later in subsequent lessons.0326

Probably not the next lesson, but the one after when we do... after we have done all of our examples, we are actually going to go back and think about them geometrically and I am going to talk about what this really means.0332

But again, right now I just want to get the technique under our belts. Just getting used to how we use Lagrange multipliers to find the maxima and minima on functions constrained by a certain other function, g.0343

Now, since this is the formal theorem, so these are the important parts right here, the gradient -- and again we are only concerning ourselves with situations where the gradient g at a given x is not equal to 0 -- again, smooth g.0358

Later on, if you go on into higher mathematics, you will talk about cases where this is not the case, but for right now we want to keep it nice and straight forward.0373

So this is really the important thing right here. Gradient of f evaluated at p = λ × gradient of g evaluated at p where p happens to be the max or min for the function f.0381

Now, let us describe the actual method and let us do some examples.0395

Let, and again, we are mostly going to be working in 2 and 3 variables just to keep things normal for us, so, let f(x,y,z) and g(x,y,z) be defined and continuously differentiable.0403

To find the max and min -- or min -- values of f subject to the constraint g, find the values of x, y, z, and λ that simultaneously satisfy the equation, this thing right here.0438

Gradient of f = λ × the gradient of g, and g(x,y,z).0497

So in other words, what we are going to do is we are going to actually take... we are going to form this equation and we are going to get a series of equations from there and then we are also going to deal with the equation g(x,y,z) or g = 0.0515

When we solve those simultaneous equations, we are going to get values for x, y, z, and lambda, where the x, y, z, those are the choices that allow for the x, y, z.0525

Those are the choices that allow for max and mins. We evaluate that function at the x, y, z, to see which number is highest, which number is lowest. That is all that is going on here.0532

So let us just go ahead and jump into some examples and I think it will start to make sense, as well, when we look at examples.0540

Again, we are going to do a large number of these examples. This is very, very important. Example 1.0549

There tends to be a lot going on with Lagrange multipliers. Lots of symbols on the page and sometimes it is not very easy to keep track of, nothing difficult, just a lot to keep track of.0560

Okay. Find the largest and smallest values that f achieves subject to g.0572

Well, f(x,y) = the function xy, and g(x,y) = x2/8 + y2/2 = 1.0600

Let us just stop and think about what this means when we say find the largest and smallest values that f achieves subject to g.0618

In the previous lessons when we talked about max, min. We either talked about a closed domain, or we talked about an open set. Maybe all of -- you know -- our particular space that we were dealing with.0626

Here what we are doing is we are saying find the maximum and minimum value of the function, but we are going to put a constraint on that. We want the values of x and y to satisfy this equation also.0639

That is what we mean by a function f constrained by g.0649

In other words, this function, you know the function x, y is perfectly defined on here except... well actually no, this is defined everywhere... but here what they are saying is this is an equation of an ellipse.0653

They want us to find the values of x and y. So x and y have to be on this ellipse and then of all the points x and y on this ellipse, which ones maximize the function, which ones minimize the function. That is what we are talking about.0671

We are constraining the x and y values. We are saying that is not just any x or any y. I am going to put a constraint on the x and y. They also have to satisfy this equation, that is all that is going on.0688

In this case it is just one constraint, you can have two, three, four constraints, as many as you need depending on the problem.0699

Okay. So find the largest and smallest values of f(g) subject to g. That is our f, that is our g, well, let us just go ahead and do it.0705

What we are going to do... we are going to find the gradient of f, and then we are going to set it equal to λ × the gradient of g, and then we are going to solve a series of equations along with this equation equal to 0.0713

Alright. So let us do the gradient of f. Let us just do this systematically. Let us see, so df/dx is equal to y, and df/dy is equal to x.0732

I am going to write this as a column vector. Again, the gradient is a vector, so instead of writing it horizontally, I am going to write it vertically. You will see why in a minute... y, x.0750

I mean it is a personal choice for me simply because I like the way it looks on a page. You can arrange it in whatever way looks good to you and what is comfortable.0760

So, y, x. So now let us do dg/dx, so let us find the gradient, alright? The partial derivative with respect to x is going to be 2x/8, which is x/4.0769

dg/dy is going to equal to 2y/2, that is equal to y.0791

This one is x/4 and y. I will write that as a column vector, so now we are going to form the gradient of f equals λ × gradient of g. So we write y, x, equals λ × well x/4y, which equals... we multiply the λ across both.0794

So we get λ × x over 4, and we get λ y. Here we go. Now we have our 2 equations.0828

Our 2 equations are... let me do this in red... this thing equal to that thing, so we have y is equal to λ × x over 4, and we have this thing equal to this thing.0838

That is why I arranged it in a column, it is just a little bit easier for me to the correspondence. The first entry - first entry, second entry - second entry.0855

I also have x = λ × y, and of course my third equation which is... remember we said the third equation in this series of equations is this, this, the gradient of f = λ × gradient of g, and of course we have the have x2/8 + y2/2 - 1 = 0.0864

We want g to equal 0, so that is why I brought the 1 over onto this side. That is all I am doing.0893

Okay. So now, let us go ahead and well, let us see what we can do. Let us solve. This is the system that we have to solve, so we need to find λ and we need to find x and we need to find y.0899

So, let us see what we have got. Let me rewrite these... y = λx/4, and I have got x = λy and I have x2/8 + y2/2 equals -1 = 0.0916

So, the x, the y and the λ they have to satisfy all three of these equations.0948

Let me go ahead and take care of this one, so let me see. 1 - λx... y, I am going to move this over... so, y - λx/4 = 0.0952

Well here, x = λ y, so I am going to go ahead and put this x into here. That gives me y - λ × λy/4 = 0.0965

I get y - λ2y/4 = 0. Let me come over here so that I can use more of the page. Let me go ahead and factor out a y.0984

I get y × 1 - λ2/4 = 0, so I have two solutions for y.0994

I have y = 0 -- excuse me... let me see here -- and 1 - λ2/4 = 0, so I get 1 = λ2/4 = 0. That means that λ + or - 2.1004

Now. I have two possibilities. I have y = 0, λ = + or - 2. Let us just deal with one case at a time. That is what you are going to be doing with these Lagrange multiplier problems.1029

You are just going to be dealing with one case at a time. This is where it starts to get a little interesting. You have to be very careful to choose your cases properly.1041

So, case 1... we have y = 0. Well, if y = 0 this implies that x = λ × y, x = λ × 0. That means that x = 0.1047

Okay. That is one possibility. Now we have the point (0,0), but now the problem is we also have to satisfy this third equation.1072

If I put this equation x, 0, y, 0 into this third equation, x2/8 + y2/2, 0 and 0. What I end up with is... well, the thing is this point (0,0), is not on this ellipse. That is a problem.1079

Even though we ended up with this possibility, because it is actually not on this ellipse, it is not part of the... we cannot use this point, we cannot test this point, so this one is out. We do not have to deal with that.1097

Now we will deal with the other case. The other case where λ = + or - 2.1109

So, case 2. λ = + or - 2. Now, let us see... from the equation... from this equation right here, x = λ × y.1116

We get x = + or - 2 × y. Okay, so now, I am going to go ahead and take this and the value of y and I am going to use this equation.1140

I am going to form, so it is going to be + or - 2y, I am basically going to put this value of x into x2 into here... squared over 8 _ y2/2 - 1 = 0.1160

Here I am going to get 4y2/8 + y2/2 - 1 = 0, and I am going to get... I am going to just multiply everything by 8 here, so I am going to end up with 4... let me do it over here... going to end up with 4y2 + 4y2 - 8 = 0.1179

So, I am going to get 8Y2 - 8 = 0, I am going to get y2 = 1, which implies that y = + or - 1.1217

Now, if y = + or - 1, and I go back to one of my original equations, that implies that x is equal to + or - 2. There we go. I have found x's and y's that actually satisfy this third equation, so these values work.1233

Now I have 4 possibilities. My first point is (+2,+1). My second point is (-2,+1). My third possibility is (+2,-1), and my fourth point is (-2,-1).1254

Now what I am going to do is I am going to evaluate the function at those points. So that is what I am doing.1277

So, f(p1) is going to equal, 2 × 1 is 2, f(p2) is going to equal -2, f(p3) is going to be -2, and f(p4) is going to equal 2.1283

That is it. Our max's are achieve there. Our mins are achieved here. A maximum at (2,1) and (-2,-1), and the minimum of that and that. That is the method of Lagrange multipliers.1303

Let us do another example. Okay. So, let us see here. Example 2... oops, let me go back to black ink, actually... well, actually, you know what? Let us do this one in blue. How is that? Okay. Example 2.1319

Find the max and min values of f, let me do it over here, f(x,y) = 3x + 4y on the circle x2 + y2 = 1.1343

okay, so one thing you should know about these max, min problems... and we will get more of a taste for this when we do some more examples in the next lesson... the problems are not written out the same way.1383

Oftentimes you are given information that looks like a Lagrange multiplier problem, but you have to extract information. In other words you have to extract what f is, extract what g is. It is not always going to say find the maximum and minimum values of f subject to g.1391

It is not going to be as explicit like this, but again we will see more examples like that.1410

okay, so find the max and min values of this function on the circle x2 + y2. So all this means is that this function, 3x + 4y is defined for the entire plane, but we want to constrain it.1415

We want to find the max and min values on the circle. So, somewhere, some point on this circle is going to maximize this function and some point on this circle is going to minimize this function.1430

So f subject to the constraint g. Okay, let us see here. What shall we do first? Well, okay. We are going to do what we always do.1441

We are going to take the gradient of f, and we are going to set it equal to λ × the gradient of g, okay?1460

We are also going to have this set of equations, and we are also going ot have g, in this case x, y = 0. That is our last equation. We have to satisfy these two equations... or this set of equations.1467

So, the gradient of f, this one is going to be... so the partial derivative with respect to x is going to be 3, the partial derivative with respect to y is 4, so we have 3, 4 = λ × gradient of g.1483

That is going to be 2x -- let me make this a little bit better -- Is going to be 2x... and this is going to be 2y, so we have λ... or I will write 2λx and 2λy. So there you go.1500

This corresponds to that, this corresponds to that, so we have the equation 3 = 2 × λx, and we have 4 = 2 × λy.1519

Yes, that is exactly right... and of course we have the g(x,y) = 0, so we are going to get x2 + y2 - 1 = 0.1536

This is our set of three equations and three unknowns, x, y, and λ.1548

We have three equations and three unknowns, theoretically this is solvable. Now we just have to find x, y, and λ... ultimately x and y, but we have to find λ along the way.1553

So let us see what we have got here. I think the best way to approach this is since we have this equation, and we have this, I am just going to go ahead and solve each of these equations, 1 for x and 1 for y.1563

So in this particular case, x is going to equal... well I am going to divide by 2λ, so it is going to be 3/2λ, and y, when I divide by 2λ, it is going to be 4/2λ, which is equal to 2/λ.1581

So now I have x... oops, these crazy lines showing up all over again, alright... λ, this is 4... so I have x and I have y, and now I am going to take these values of x and y, and I am actually going to put them into this equation to see what I get for λ.1600

So let us go ahead and do that. Let us do that on the next page. So, I have the function x2 + y2 + 1 = 0, and we said that x = 3/2λ.1619

This is going to be 3/2λ2 - 1 = 0. So let us go ahead and work all of this out. 9/4λ2 + 4/λ2 -1 = 0. We are not going to leave anything out here.1634

I am going to go ahead and multiply through by 4λ2, I think... no, just 4... yea, 4λ2.1658

So, when I multiply by 4λ2, I am going to get 9 over here, I am going to get 16 over here, -4λ2 = 0, just to get rid of the denominator, that is all I did.1672

9 + 16 is 25... that equals 4λ2, so λ2 = 25/4, therefore λ = + or - 5/2. Okay, so we found λ, now it should be not a problem.1687

Now that we have found λ, well, x = 3/2λ, so when I put that in there, that is going to end up equaling... well, let us do it all, let us not miss anything here... 3/2 × + or - 5/2. It is going to end up equaling + or - 3/5.1707

Now, y = 2/λ, which is equal to 2/ + or - 5/2, which equals -4/5, and I hope I have done my arithmetic correctly.1733

Now, let us stop and think about this. We have + and - 3/5 for x, and we have + or - 4/5 for y.1750

You are probably thinking to yourself, just like the previous problem, that we have 4 points: (3/5,3/5), (3/5,-4/5), (-3/5,4/5), (-3/5,4/5).1756

That is actually not the case. This is where you have to sort of look at other things. There is other analyses going on here.1770

x = 3/2λ, y = 2/λ. x and y have the same sign, so they are either both positive, or both negative. So, we do not have 4 points to pick, we only have 2 points to pick, 1 in the first quadrant, 1 in the third quadrant.1779

That is what is going on here. You can go ahead and use the others, it is not a problem, you will go ahead and get the answer... but you know, just something to be aware of.1793

So, that is it. Basically our points that we are going to pick are... let us see... (3/5,4/5), that is one possibility... and (-3/5,-4/5), that is the other possibility.1802

Okay. When I take... let us just call this P1, and let us call this P2... so when I take f(p1), in other words I put it back into the function 3 × 3x + 4y, 3 × 3/5... well you know what, let me work it all out. It is probably a good idea if I work it all out.1831

I will do it on the next page, so we will do f(3/5,4/5), that is going to equal 3 × 3/5 + 4 × 4/5 = 25/5... that is going to equal 5... and f(-3/5,-4/5) = 3 × -3/5 + 4 × -4/5 = -25/5 = -5.1858

So, here, at the point (3/5,4/5), it achieves a maximum -- the maximum value is 5 -- and (-3/5,-4/5), the function f achieves a minimum of -5.1899

We have found the maximum and minimum values of 3x + 4y, subject to the constraint that x and y lie on the unit circle. x2 + y2 = 1, that is what is going on.1912

So, again, in the next lesson we are going to continue on with more examples of Lagrange multipliers because we want to be very, very, very familiar with this.1926

Then, after that, we will pull back a little bit and take a look at exactly what is going on.1934

We want to make sure that you actually understand why this is the case, and why this works.1938

Again, nothing theoretical, we just want to make it plausible for you, that this is not some technique that just drops out of the sky.1943

Thank you for joining us here at educator.com, we will see you next time. Bye-bye.1949

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