Raffi Hovasapian

More Lagrange Multiplier Examples

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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• ## Related Books

 1 answerLast reply by: Professor HovasapianSat Jan 10, 2015 7:11 PMPost by owais khan on January 8, 2015in example 2, did you use the critical points values, because its a closed boundary, as in other example you did not use the critical method ??? 1 answerLast reply by: Professor HovasapianThu Oct 10, 2013 1:44 AMPost by yaqub ali on October 9, 2013professor why did you divide 2x/2y in example 1 when z=0 1 answerLast reply by: Professor HovasapianMon Jun 10, 2013 3:22 PMPost by Josh Winfield on June 8, 2013Example 2. Can be solved without Lagrange. Just using techniques form previous lecture using extreme value theorem.Find Critical points of f on E=region, gradf(x0,y0)=(0,0)Examine the boundary by:Parameterising the boundary c(t)=(cost,sint)Composite Function f(c(t))Finding critical points of f(c(t)) by taking the derivative and letting it =0 s/t f'(c(t))=0 (find all -t- which satisfy)FInd all values for (f(c(-t-)) and f(x0,y0) and highlight max and min values 1 answerLast reply by: Professor HovasapianSat Sep 1, 2012 4:01 PMPost by Mohammed Alhumaidi on September 1, 2012You had switched the Grad(g) when multiplying by Lambda !! and x and y !!

### More Lagrange Multiplier Examples

Find the point of the surface closest to the origin of x2 + y2 + (z − 4)2 = 9.
• Recall that the distance from a point in space to the origin is d(x,y,z) = √{x2 + y2 + z2} .
• We can use Lagrange multipliers to find the least distance, f:√{x2 + y2 + z2} = 0 from a point in g:x2 + y2 + (z − 4)2 − 9 = 0 to the origin, that is subject to g.
• Note that √{x2 + y2 + z2} = 0 is equivalent to x2 + y2 + z2 = 0. Now
 ∇f = l∇g
 g(x,y,z) = 0
is
 (2x,2y,2z) = l(2x,2y,2z − 8)
 x2 + y2 + (z − 4)2 − 9 = 0
• (2x,2y,2z) = l(2x,2y,2z − 8) is equivalent to
 2x = 2xl
 2y = 2yl
 2z = (2z − 8)l
, if x0 or y0, then l = 1. But this causes a contradiction on the third equation as 0 − 8.
• Hence x = 0 and y = 0, x2 + y2 + (z − 4)2 − 9 = 0 yields (z − 4)2 = 9 so that z = − 7 or z = 1.
• Note that z0 or else 2z = (2z − 8)l yields 0 = l which we don't allow.
• Hence l = [2z/(2z − 8)] and is defined at z = − 7 or z = 1. Our value points are then f(0,0, − 7) = 7 and f(0,0,1) = 1.
The point on the surface x2 + y2 + (z − 4)2 = 9 closest to the origin is (0,0,1).
Find the distance of the point of the surface closest to the origin of − x + y + z = 2.
• Recall that the distance from a point in space to the origin is d(x,y,z) = √{x2 + y2 + z2} .
• We can use Lagrange multipliers to find the least distance, f:√{x2 + y2 + z2} = 0 from a point in g: − x + y + z − 1 = 0 to the origin, that is subject to g.
• Note that √{x2 + y2 + z2} = 0 is equivalent to x2 + y2 + z2 = 0. Now
 ∇f = l∇g
 g(x,y,z) = 0
is
 (2x,2y,2z) = l( − 1,1,1)
 − x + y + z − 1 = 0
• (2x,2y,2z) = l( − 1,1,1) is equivalent to
 2x = − l
 2y = l
 2z = l
, so x = − [l/2], y = [l/2] and z = [l/2].
• Substituting into − x + y + z − 1 = 0 yields − ( − [l/2] ) + [l/2] + [l/2] − 1 = 0 soliving for l results in l = [2/3].
• Then x = − [l/2] = − [1/3], y = [l/2] = [1/2] and z = [l/2] = [1/3] and f( − [1/3],[1/2],[1/3] ) = [(√{17} )/6].
Hence the point on the surface − x + y + z = 2 closest to the origin is ( − [1/3],[1/2],[1/3] ) whose distance is [(√{17} )/6].
Use Lagrange multipliers to set up a system of equations \protect
 ∇f = l∇g
 g(x,y,z) = 0\protect
that finds the point of the surface closest to the origin of xy2 − z = 5.
• Recall that the distance from a point in space to the origin is d(x,y,z) = √{x2 + y2 + z2} .
• We find the least distance, f:√{x2 + y2 + z2} = 0 from a point in g:xy2 − z − 5 = 0 to the origin, that is subject to g.
• Note that √{x2 + y2 + z2} = 0 is equivalent to x2 + y2 + z2 = 0. Now
 ∇f = l∇g
 g(x,y,z) = 0
is
 (2x,2y,2z) = l(y2,2xy, − 1)
 xy2 − z − 5 = 0
subsection*(2x,2y,2z) = l(y2,2xy, − 1) is equivalent to
 2x = y2l
 2y = 2xyl
 2z = − l
, so that our system of equations is \protect
 2x = y2l
 y = xyl
 2z = − l
 xy2 − z − 5 = 0\protect
Use Lagrange multipliers to set up a system of equations \protect
 ∇f = l∇g
 g(x,y,z) = 0\protect
that finds the distance of the point of the surface closest to the origin of 1 = [((x − 2)2)/4] + [((y − 3)2)/9] + [((z − 2)2)/4].
• Recall that the distance from a point in space to the origin is d(x,y,z) = √{x2 + y2 + z2} .
• We find the least distance, f:√{x2 + y2 + z2} = 0 from a point in g:[((x − 2)2)/4] + [((y − 3)2)/9] + [((z − 2)2)/4] − 1 = 0 to the origin, that is subject to g.
• Note that √{x2 + y2 + z2} = 0 is equivalent to x2 + y2 + z2 = 0. Now
 ∇f = l∇g
 g(x,y,z) = 0
is
 (2x,2y,2z) = l( [(x − 2)/2],[(2(y − 3))/9],[(z − 2)/2] )
 [((x − 2)2)/4] + [((y − 3)2)/9] + [((z − 2)2)/4] − 1 = 0
(2x,2y,2z) = l( [(x − 2)/2],[(2(y − 3))/9],[(z − 2)/2] ) is equivalent to \protect
 2x = [(x − 2)/2]l
 2y = [(2(y − 3))/9]l
 2z = [(z − 2)/2]l\protect
, so that our system of equations is \protect
 4x = (x − 2)l
 9y = (y − 3)l
 4z = (z − 2)l
 [((x − 2)2)/4] + [((y − 3)2)/9] + [((z − 2)2)/4] − 1 = 0\protect
Use Lagrange multipliers to set up a system of equations \protect
 ∇f = l∇g
 g(x,y,z) = 0\protect
that finds the point of the surface closest to the origin of z = [(y2)/9] − [(x2)/4] + 3.
• Recall that the distance from a point in space to the origin is d(x,y,z) = √{x2 + y2 + z2} .
• We find the least distance, f:√{x2 + y2 + z2} = 0 from a point in g:[(y2)/9] − [(x2)/4] − z − 3 = 0 to the origin, that is subject to g.
• Note that √{x2 + y2 + z2} = 0 is equivalent to x2 + y2 + z2 = 0. Now
 ∇f = l∇g
 g(x,y,z) = 0
is
 (2x,2y,2z) = l( − [x/2],[2y/9], − 1 )
 [(y2)/9] − [(x2)/4] − z − 3 = 0
(2x,2y,2z) = l( − [x/2],[2y/9], − 1 ) is equivalent to \protect
 2x = − [x/2]l
 2y = [2y/9]l
 2z = − l\protect
, so that our system of equations is \protect
 4x = − xl
 9y = yl
 2z = − l
 [(y2)/9] − [(x2)/4] − z − 3 = 0\protect
Find the maximum and minimum of the f(x,y) = − 3x2 + y2 on the closed set D = { (x,y)|x2 + y2 ≤ 4} .
• First we note that D is a closed and bounded set and f is continuous and defined on D. Hence a maximum and minimum occur at D.
• We find our critical points first, that is [df/dx] = − 6x = 0 which gives x = 0 and [df/dy] = 2y = 0 which gives y = 0. We have a possible extrema at f(0,0) = 0.
• Next we use Lagrange multipliers to find a maximum or minimum at the boundary g:x2 + y2 − 4 = 0.
• Now
 ∇f = l∇g
 g(x,y) = 0
is
 ( − 6x,2y) = l( 2x,2y )
 x2 + y2 − 4 = 0
and ( − 6x,2y) = l( 2x,2y ) is equivalent to
 − 6x = 2xl
 2y = 2yl
so our complete system of equations is
 − 3x = xl
 y = yl
 x2 + y2 − 4 = 0
• Note that if x = 0 then y2 = 4 and l is defined. So that y = ±2 and our possible extrema are f(0,2) = 4 and f(0, − 2) = 4.
• Similarly if y = 0 then x2 = 4 and l is defined. So that x = ±2 and our possible extrema are f(2,0) = − 12 and f( − 2,0) = − 12.
• If x0 then − 3 = l but y − 3y unless y = 0. Similarly if y0 then 1 = l but − 3x − 3x unless x = 0. Note that (0,0) does not satisfy g.
Hence our possible extrema are f(0,0) = 0, f(0,2) = f(0, − 2) = 4 and f(2,0) = f( − 2,0) = − 12. Our maximum value is 4 while our minimum value is − 12.
Find the maximum and minimum of f(x,y) = − x2 − y2 + 1 on the closed set D = { (x,y)|[(x2)/9] + [(y2)/4] ≤ 1} .
• First we note that D is a closed and bounded set and f is continuous and defined on D. Hence a maximum and minimum occur at D.
• We find our critical points first, that is [df/dx] = − 2x = 0 which gives x = 0 and [df/dy] = − 2y = 0 which gives y = 0.
• We have possible extrema along f( 0,0 ) = 1.
• We use Lagrange multipliers to find a maximum or minimum at the boundary g:[(x2)/9] + [(y2)/4] − 1 = 0.
• Now
 ∇f = l∇g
 g(x,y) = 0
is
 ( − 2x, − 2y) = l( [2x/9],[y/4] )
 [(x2)/9] + [(y2)/4] − 1 = 0
and ( − 2x, − 2y) = l( [2x/9],[y/4] ) is equivalent to
 − 2x = [2x/9]l
 − 2y = [y/4]l
so our complete system of equations is
 − 9x = xl
 − 8y = yl
 [(x2)/9] + [(y2)/4] − 1 = 0
• Note that if x0 and y0, then l = − 9 and l = − 8 which is a contradiction. So we have two cases, x = 0 or y = 0.
• If x = 0, then [(x2)/9] + [(y2)/4] − 1 = 0 yields y2 = 4 so y = ±2 and l is defined. Similarly if y = 0, then [(x2)/9] + [(y2)/4] − 1 = 0 yields x2 = 9 so x = ±3 and l is defined.
• Our possible extrema are then f(0,2) = (0, − 2) = − 3, f(3,0) = ( − 3,0) = − 8 and f(0,0) = 1.
Hence 1 is our maximum and − 8 is our minimum.
Find the maximum and minimum of f(x,y) = x2 + y2 on the closed set D = { (x,y)|(x − 1)2 + (y − 1)2 ≤ 1} .
• First we note that D is a closed and bounded set and f is continuou and defined on D. Hence a maximum and minimum occur at D.
• We find our critical points first, that is [df/dx] = 2x = 0 which gives x = 0 and [df/dy] = 2y = 0 which gives y = 0.
• Note that the point ( 0,0 ) is not in our region D.
• We can use Lagrange multipliers to find the maximum or minimum at the boundary g:(x − 1)2 + (y − 1)2 − 1 = 0.
• Then
 ∇f = l∇g
 g(x,y) = 0
is
 (2x,2y) = l( 2(x − 1),2(y − 1) )
 (x − 1)2 + (y − 1)2 − 1 = 0
and (2x,2y) = l( 2(x − 1),2(y − 1) ) is equivalent to
 2x = 2l(x − 1)
 2y = 2l(y − 1)
so our complete system of equations is
 x = l(x − 1)
 y = l(y − 1)
 (x − 1)2 + (y − 1)2 − 1 = 0
• Now, x0 or y0 as x = [l/(l − 1)] and y = [l/(l − 1)] and so x = y. Utilizing (x − 1)2 + (y − 1)2 − 1 = 0 we obtain 2(x − 1)2 = 1 so x = 1 ±[1/(√2 )].
• We have our possible extrema at f( 1 + [1/(√2 )],1 + [1/(√2 )] ) = [3/2] + [2/(√2 )] and f( 1 − [1/(√2 )],1 − [1/(√2 )] ) = [3/2] − [2/(√2 )].
Hence [3/2] + [2/(√2 )] is our maximum and [3/2] − [2/(√2 )] is our minimum.
Find the maximum and minimum of f(x,y) = 5x − y on the closed set D = { (x,y)| − 1 ≤ x ≤ 1, − 1 ≤ y ≤ 1} .
• First we note that D is a closed and bounded set and f is continuous and defined on D. Hence a maximum and minimum occur at D.
• We find our critical points first, that is [df/dx] = 5 = 0 which gives 5 = 0 and [df/dy] = − 1 = 0 which gives − 1 = 0 both which are not possible. Our extrema are located at the boundary.
• We note that f is an increasing plane and so our maximum and minimum are achieved at the endpoinst of D.
• Our possible extrema are then f( − 1, − 1) = − 4, f(1, − 1) = 6, f(1,1) = 4, f( − 1,1) = − 6.
Hence 6 is our maximum and − 6 is our minimum.
Find the point at the surface of x2 + y2 + z2 = 1 closest and farthest to the point (3, − 1,2).
• Recall that the distance from two points (x,y,z) and (3, − 1,2) in space is d(x,y,z) = √{(x − 3)2 + (y + 1)2 + (z − 2)2} .
• We can use Lagrange multipliers to find the least or greatest distance, f:√{(x − 3)2 + (y + 1)2 + (z − 2)2} = 0 from a point in g:x2 + y2 + z2 − 1 = 0 to the origin, that is subject to g.
• Note that √{(x − 3)2 + (y + 1)2 + (z − 2)2} = 0 is equivalent to (x − 3)2 + (y + 1)2 + (z − 2)2 = 0. Now
 ∇f = l∇g
 g(x,y,z) = 0
is
 (2(x − 3),2(y + 1),2(z − 2)) = l(2x,2y,2z)
 x2 + y2 + z2 − 1 = 0
• (2(x − 3),2(y + 1),2(z − 2)) = l(2x,2y,2z) is equivalent to
 x − 3 = xl
 y + 1 = yl
 z − 2 = zl
, note that x0, y0, and z0 as x − 3 = xl yields x = [3/(1 − l)], y + 1 = yl yields y = [( − 1)/(1 − l)] and z − 2 = zl yields z = [2/(1 − l)].
• Utilizing x2 + y2 + z2 − 1 = 0 we can substitute and obtain ( [3/(1 − l)] )2 + ( [( − 1)/(1 − l)] )2 + ( [2/(1 − l)] )2 = 1 to solve for l and get l = 1 ±√{14} .
• Then for l = 1 + √{14} we have x = [3/(1 − l)] = [3/(1 − (1 + √{14} ))] = [3/(√{14} )], y = [( − 1)/(1 − l)] = [( − 1)/(1 − (1 + √{14} ))] = [( − 1)/(√{14} )] and z = [2/(1 − l)] = [2/(1 − (1 + √{14} ))] = [2/(√{14} )].
• Similarly for l = 1 − √{14} we have x = [3/(1 − l)] = [3/(1 − (1 − √{14} ))] = [3/( − √{14} )], y = [( − 1)/(1 − l)] = [( − 1)/(1 − (1 − √{14} ))] = [1/(√{14} )] and z = [2/(1 − l)] = [2/(1 − (1 − √{14} ))] = [2/( − √{14} )].
• Then our value points are f( [3/(√{14} )], − [1/(√{14} )],[2/(√{14} )] ) ≈ 1.82 and f( − [3/(√{14} )],[1/(√{14} )], − [2/(√{14} )] ) ≈ 4.74
Thus our closest point on x2 + y2 + z2 = 1 to (3, − 1,2) is ( [3/(√{14} )], − [1/(√{14} )],[2/(√{14} )] ) while the furthest is ( − [3/(√{14} )],[1/(√{14} )], − [2/(√{14} )] ).

*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.

### More Lagrange Multiplier Examples

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
• Example 1: Find the Point on the Surface z² -xy = 1 Closet to the Origin 0:54
• Part 1
• Part 2
• Part 3
• 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
• Part 2
• Part 3

### Transcription: More Lagrange Multiplier Examples

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

So, the last lesson, we introduced the method of Lagrange multipliers, we did a couple of basic examples.0004

In this lesson, we are just going to continue doing examples to develop a sense of how to handle this method of Lagrange multipliers.0009

Again, there is nothing particularly difficult about the method, it is pretty straight-forward. The difficulty is solving the simultaneous equations.0018

Because there is a lot going on, on the page, there are different cases when this equals 0, when this does not equal 0... so there are lots of things to keep track of.0025

These examples are going to be slightly more complicated, they are going to be handled the same way, they are just going to look like there is a lot more going on.0036

In some sense that is true, in some sense that is just the nature of Lagrange multiplier problems.0045

In any case, let us just jump right on in and see what we can do.0051

Okay. So, the first example. Example 1.0057

Find the point on the surface z2 - xy = 1, closest to the origin.0065

Notice in this particular problem, they only gave us 1 function, and they just said closest to the origin.0090

This is an example of a Lagrange multiplier or a max/min problem where we have to extract information, come up with another function for ourselves whether it is f or g, depending on which one.0097

Right now, we are not quite sure which one is f and which one is g. So, find the point on the surface closest to the origin.0107

Well, distance from the origin is just the distance function. So, distance... I will write it out, that is not a problem... so distance = x2 + y2 + z2, all under the radical.0116

As it turns out, we want the point that is closest to the origin, but the point has to be on this surface.0134

As it turns out, this is f, right here, and this is g. That is the constraint. We want to minimize, in other words, this function subject to this constraint.0144

Well, fortunately, since we are dealing with the distance function, if we minimize this square root function, it is the same as... we can also minimize the square of that: x2 + y2 + z2.0156

I do not want to deal with the square root, so I will just instead of minimizing this, I am just going to minimize the function.0170

This is actually going to be my f, right here, and this is going to be my g.0176

Let us go ahead and form what it is that we actually form. We want to form the gradient of f, we want to set it equal to λ × the gradient of g, λ is our Lagrange multiplier.0184

Again, so we are going to have this set of equations and our constraint, which is going to be z2 - xy - 1 = 0.0203

It is always going to be the constraint as a function set equal to 0. That is our other equation.0213

Now, we have to satisfy all of these simultaneously. Okay, let us just go ahead and take care of this.0222

So, the gradient of f, let us see... gradient of f, that is equal to... well, gradient of f is going to be 2x, 2y, and 2z.0230

That is going to equal... actually, let us just go ahead and do the gradient separately and then we will set it equal to each other.0245

The gradient of g is going to equal -y, that is df/dx, -x, that is df/dy, and 2z, that is df/dz of g.0255

Now, we set them equal to each other with the λ, so what we have is 2x, 2y, 2z is equal to... let me do this systematically, I do not want to get too far ahead of myself here... λ × -y, -x and 2z, which is equal to -λx, -λy, and 2λz.0268

Okay. There we go. Now, let us go to blue. This and that is one equation, this and that is another equation, this and that is a third equation, and of course we have our fourth equation. Those are our four equations and four unknowns, x, y, z, and λ.0306

Again, four equations and four unknowns. It is a bit much, but hopefully we should be able to work it out.0320

Let us see what we have got. One thing that we want... let me actually write out the equations here so that we have them in front of us... so, 2x = -λx... I am sorry, oops, I got these reversed, my apologies... this is -y, so this is y and this is x. There we go, that is a little bit better.0328

So, 2x = -λy. There you go, you can see it is very, very easy to go astray in these problems if you are not keeping track of every little thing.0359

So, -y and then we have 2y = -λx, and we have 2z = λ × 2z. Let us go ahead and actually write the other equation... z2 - xy - 1 = 0.0374

Those are our four equations. Okay. So, let us just take a look at this for a second, and see how to proceed. One of the possibilities here is... well, one thing we know, we know that λ is not going to be 0. Let us exclude this possibility.0395

λ is definitely not going to be 0 because this would imply, if λ is 0, that means 2x = 0, 2y = 0, 2z = 0, because that would imply that x, y, and z equals 0, 0, 0, but 0, 0, 0, does not satisfy this equation.0413

If I put 0's in for here, what I get is -1 = 0, which is not true, so this particular point, if λ = 0, this particular point might work, but it does not satisfy the constraint. We can eliminate that as a possibility.0439

So that is good, λ is not equal to 0. So, our first possibility is... let me see, let us go ahead and move to the next page and let me rewrite the equation so that we have it above us.0454

So, 2x = -λy, 2y = λx, 2z = 2λz, and we have z2 - xy - 1 = 0.0469

These are our four equations that we need to solve simultaneously. The one possibility here is let us take the possibility where z is not equal to 0. Okay.0490

So, when z is not equal to 0, that means we can go λ = 2z/2z, we solve this equation right over here, and we get λ = 1.0500

When λ = 1, I can put it into this equation and this equation. Then I get 2x = -y, and I get... I have 2y = -x.0519

Well, I am going to substitute one of these in here. I am going to take this, so I am going to rewrite this one as -2y = x, and I am going to substitute this x into that equation.0540

I get 2 × -2y = -y. I get -4y = y. and the only way that this is true, is that y = 0.0556

When y = 0, that means x = 0. Now, I have y = 0, and x = 0.0574

So, let us see here. That actually is a possibility, so now if y = 0 and x = 0, now let me go ahead and take these values and put them into this equation here.0590

Now I have got z2 - 0 × 0 - 1 = 0, so I get z2 = 1, I get z equals + or - 1.0605

My possibilities... so a couple of points, are (0,0,1), and (0,0,-1), right? x and y are 0, z is + or - 1.0622

Those are two possibilities. When I evaluate them at f, I end up with 1, and I end up with 1.0636

So, those are a couple of possibilities. This was the case when z was not equal to 0.0645

Now we will go ahead and do the case for z = 0, so we are just eliminating all of the possibilities here.0652

When z = 0, what we have is, if I take this equation, the constraint equation, I get 02 - xy - 1 = 0.0661

So, I get -xy = 1, or y = -1/x.0680

This is 1 relation that I have got here. Now, let me go back to my other relations involving x and y. That is 2x = -λ y, 2y = -λx.0690

I am going to divide the top and bottom equation, in other words I am going to divide the top equation by the bottom equation.0703

I get 2x/2y = -λy/-λx, the λ's cancel, and I get y/x.0712

So 2x/2y = y/x. The 2's actually end up cancelling, so I am just going to go ahead and cross multiply. I am going to get x2 = y2.0725

Well, let us see. x2 = y2, well I found that y = -1/x, therefore I am going to put this y into here and I am going to get x2 = -1/x2, which equals 1/x2. That implies that x4 = 1.0741

Well, if x4 = 1, that implies that x = + or - 1.0768

When x = + or - 1, I go back to any one of the equations here, and I end up with y is equal to... well, let me actually move onto the next page here... so I have got x = + or - 1.0777

x = 1 implies that y = -1, and x = -1 implies that y = 1.0798

Now I have a couple of other points. z was 0, and these are the x's and y's.0812

One possibility is (1,-1,0), and the other possibility is (-1,1,0).0818

When I evaluate these at f, I get 2.0829

There we go. We have taken care of all of the possibilities. We had the four points, we had (0,0,1), (0,0,-1), and we have (1,-1,0), (-1,1,0).0841

For the other points, we found that the values were actually 1. Here the values are 2, so, the mins occur at (0,0,1) and (0,0,-1), and the value of the function at those points is 1.0855

On that given surface, on the surface z2 - xy - 1 = 0, these 2 points on this surface are going to minimize the distance to the origin. That is what we have done here.0884

Again, as you can see, there seems to be a lot going on. Lots of cases to sort of go through. I wish that there was some sort of algorithmic approach to solving these problems other than the basic equation of setting the gradient of f = λ × gradient of g + the constraint equation equal to 0.0906

We have four equations and 4 unknowns, 5 equations and 5 unknowns. Clearly this gets more difficult. You just have lots of other cases to just sort of go through.0925

Again, successful Lagrange multipliers is more about experience than anything else. You can be reasonably systematic, but clearly you saw just with these 4 equations and 4 unknowns there is a lot to keep track of.0934

That is the only difficulty. Do not let the procedure actually interfere with your view of the mathematics.0949

Okay. Let us just do another example. That is all we can do, just keep doing examples until we start to get familiar with general notions. So, example number 2. Example 2.0958

Find the max and min of f(x,y) = x2 + 2y2 - x on the closed disc of radius 1 centered at the origin.0974

Okay. So, what we want to do is... we have the unit... now, we are considering the entire region, and it is a closed region. We are considering the interior and we are considering the boundary. That is all that is happening here.1016

So, let us see... equals 0... so this is the... our function, and we want to find the maximum and minimum values on this region.1035

We have a couple of things going on here. One of the things that we are going to do is... so this is kind of a max/min problem, and a Lagrange multiplier problem.1055

The Lagrange multiplier problem will usually apply to the boundary here, because we have an equation for that boundary which is x2 + y2 = 1, or x2 + y2 - 1 = 0.1072

As far as the interior is concerned, well we just treat it like any other max/min problem. We see if we can find a critical point and take it from there.1085

Let us go ahead and do that first. Let us deal with the interior, so, let us find the gradient of f and we will set it equal to 0 to see where the critical points are.1093

The gradient of f is going to be 2x - 1. That is the derivative with respect to the first variable x, and then we have 2y2, I am sorry, this is 2y2.1105

Our derivative with respect to y is going to be 4y. Okay. Now we want to check to see where these are equal to 0, so we have 2x - 1 = 0, and we have got 4y = 0.1123

This implies that y = 0 and this implies that x = 1/2, so at the point (1/2,0), the point (1/2,0) is a critical point.1136

Let us just go ahead and find the value of f at that point. When we go ahead and evaluate at that point, f(1/2,0), I go ahead and put it into f and I end up with -1/4.1154

That is one possible value. That is a critical point, that is on the interior. Now we can go ahead and deal with the boundary.1170

Now, let us check the boundary. So, the boundary is g(x,y) = x2 + y2 - 1, and that is going to be one of the equations, we will need to set that to 0.1180

Let us go ahead and find the gradient of g. The gradient of g is equal to, well that is going to be 2x, and that is going to be 2y.1195

So, 2x and 2y, so now we are going to go ahead and set the gradient of f = λ × gradient of g, and what we get is 2x - 1, and 4y = λ × 2x and 2y, which is equal 2λx and 2λy.1211

The two equations that we get are... let me do this one in red actually, so that I separate the equations out that I am going to be working on, again. This is all about solving simultaneous equations.1243

I get 2x = 2λx. I get 4y = 2λ × y. Let me make sure that I make everything clear so that we do not make the same mistake we made last time, so, this is 2λy.1254

And, of course we have our equation x2 + y2 - 1 = 0. That is the constraint.1273

Now I have got 3 equations and 3 unknowns, x, y, and λ.1280

So, let us deal with case 1. Case 1, let us take λ = 0.1286

So, case 1, λ = 0. Okay. When I set λ = 0, what is actually going to end up happening is I am just going to end up getting the previous answer.1297

That is not an issue, so let us deal with case 2.1309

case 2, where λ does not equal 0. So, one possibility for here, so, when λ does not equal to 0, now there are some sub cases that I have to consider.1315

The first sub case that I am going to consider is y = 0.1327

When I set y = 0, then I am going to get x2 + 02 - 1 = 0. I am going to get x = + or - 1.1334

Therefore, I am going to have the points (1,0) and (-1,0).1349

Well, at (1,0), when I evaluate it at f, I am going to get 0, and when I evaluate it at (-1,0), when I evaluate f, I am going to get the value 2.1355

Now I have got (1,0), (-1,0), these are the values, and I also had that other point, that (1/2,0) that I found from just working on the interior of the disc and the value was I think -1/4, or something like that, or whatever it was.1370

Now, that is the case where λ does not = 0, and the case where y does equal 0. Now we want to consider the other case where y does not = 0.1385

Again, it seems like there is a lot going on but usually you can make sense of it by just spending some time with it. So, y does not equal 0.1400

Well, when y does not equal 0, 4y equals 2λ y. That means λ = 2. 4y/2y, let me go ahead and write that out actually. λ = 4y/2y = 2.1409

When λ = 2, this implies that 2x - 1 equals... so the equation is 2λx, right?1435

So 2x - 1 = λ = 2 = 4x, so I am going to get 2x = -1, x = -1/2.1454

So, when x = -1/2, now when I put that into my x2 + y2 - 1 = 0, when I put it into here, I am going to end up with the following.1469

It is going to be -1/22 + y2 - 1 = 0.1491

I get 1/4 + y2 - 1 = 0, I get y2 = 3/4, therefore y = + or - sqrt(3)/2.1500

Okay, so, now my other points. x is -1/2, +sqrt(3)/2, and I get -1/2 - sqrt(3)/2.1516

When I evaluate these at f, I am going to end up with 9/4 and 9/4, so we have got 9/4, 9/4, -1/2, 0, I check all of those points to see which one is the maximum, which one is the minimum.1531

So the max takes place at the points (-1/2,sqrt(3)/2), and (-1/2,-sqrt(3)/2). Value is 9/4.1555

The minimum takes place at the original point that we found, which was... not -1/2... at 1/2, 0 and that value was -- let me see, what was that value if we can recall -- -1/4. There we go. This is our solution. I will go ahead and put f here, when we evaluate f.1580

So, at this point and this point, our function achieves a maximum and at this point, the function achieves a minimum. That is it.1610

Clearly there is a lot going on. A lot of things to sort of keep track of, but we are solving several equations in several variables. This is just the nature of the problem, the nature of the best.1618

Okay. So, in the next lesson, we are going to actually continue discussing Lagrange multipliers. We will do some more examples.1633

We will pull back a little bit so we will discuss some of the geometry of the solutions and we will try to make sense of what is actually going on.1640

Again, not theoretically, we just want this to seem reasonable to you, that we did not just drop this in your lap and say use this technique to find max/min for a function subject to this constraint. We still want this to make sense.1645

Okay. Thank you for joining us here at educator.com, we will see you next time. Bye-bye.1658

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