For more information, please see full course syllabus of Multivariable Calculus
For more information, please see full course syllabus of Multivariable Calculus
Lagrange Multipliers
 To find the greatest and least values of a function f subjected to g, we solve the system of equations
where l is a constant.∇f = l∇g g(x,y) = 0  Now, ∇f = ( − 2x,2y) and ∇g = (1, − 2y), let g(x,y) = x − y^{2}. Then our system of equations is
( − 2x,2y) = l(1, − 2y) x − y^{2} = 0  ( − 2x,2y) = l(1, − 2y) is equivalent to
this implies that x = − [l/2] and if y0 then l = − 1.− 2x = l 2y = − 2yl  For y = 0 we have x − y^{2} = 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 − y^{2} = 0, we have [1/2] − y^{2} = 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].
 To find the greatest and least values of a function f subjected to g, we solve the system of equations
where l is a constant.∇f = l∇g g(x,y) = 0  Now, ∇f = (6x^{2}, − 8y) and ∇g = (2x, − 1), let g(x,y) = x^{2} − y. Then our system of equations is
(6x^{2}, − 8y) = l(2x, − 1) x^{2} − y = 0  (6x^{2}, − 8y) = l(2x, − 1) is equivalent to
this implies that y = [l/8] and if x0 then x = [l/3].6x^{2} = 2xl − 8y = − l  For x = 0 we have x^{2} − y = (0)^{2} − y = 0 so y = 0 but then l = 0, which we don't allow.
 Otherwise, utilizing x^{2} − 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.
 To find the greatest and least values of a function f subjected to g, we solve the system of equations
where l is a constant.∇f = l∇g g(x,y) = 0  Now, ∇f = (y,x) and ∇g = ( 2x,2y ), let g(x,y) = x^{2} + y^{2} − 1. Then our system of equations is
(y,x) = l( 2x,2y ) x^{2} + y^{2} − 1 = 0  (y,x) = l( 2x,2y ) is equivalent to
we can substitute y from the top equation to x = 2yl.y = 2xl x = 2yl  We obtain x = 2(2xl)l which gives l = ±[1/4]. We can again substitute y = 2xl into x^{2} + y^{2} − 1 = 0.
 We obtain x^{2} + (2xl)^{2} − 1 = 0 or x^{2}(1 + 4l^{2}) = 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].
 To find the greatest value of f subjected to g, we solve the system of equations
where l is a constant.∇f = l∇g g(x,y) = 0  Now, ∇f = (4, − 3) and ∇g = ( [2x/9], − [y/2] ), let g(x,y) = [(x^{2})/9] − [(y^{2})/4] − 1. Then our system of equations is
(4, − 3) = l( [2x/9], − [y/2] ) [(x^{2})/9] − [(y^{2})/4] − 1 = 0  (4, − 3) = l( [2x/9], − [y/2] ) is equivalent to
and so x = [18/l] and y = [6/l].4 = [2x/9]l − 3 = − [y/2]l  Utilizing [(x^{2})/9] − [(y^{2})/4] − 1 we substitute our values for x and y to obtain [36/(l^{2})] − [9/(l^{2})] − 1 = 0. We now solve for l.
 Now, [36/(l^{2})] − [9/(l^{2})] − 1 = 0 is equivalent to 27 = l^{2} 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 )].
 To find the least value of f subjected to g, we solve the system of equations
where l is a constant.∇f = l∇g g(x,y) = 0  Now, ∇f = ( [2x/9],[2y/8] ) and ∇g = ( 3x^{2},2y ), let g(x,y) = x^{3} + y^{2} − 10. Then our system of equations is
( [2x/9],[2y/8] ) = l( 3x^{2},2y ) x^{3} + y^{2} − 10 = 0  ( [2x/9],[2y/8] ) = l( 3x^{2},2y ) is equivalent to
and so x = [2/27l] and l = [1/8].[2x/9] = 3x^{2}l [2y/8] = 2yl  We now have x = [16/27]. Utilizing x^{3} + y^{2} − 10 = 0 we substitute our values for x and obtain ( [16/27] )^{3} + y^{2} − 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.
 To find the greatest and least values of a function f subjected to g, we solve the system of equations
where l is a constant.∇f = l∇g g(x,y,z) = 0  Now, ∇f = (3, − 2,1) and ∇g = (2x, − 4y, − 2z), let g(x,y,z) = x^{2} − 2y^{2} − z^{2} − 1. Then our system of equations is
(3, − 2,1) = l(2x, − 4y, − 2z) x^{2} − 2y^{2} − z^{2} − 1 = 0  (3, − 2,1) = l(2x, − 4y, − 2z) is equivalent to
this implies that x = [3/2l], y = [1/2l] and z = − [1/2l].3 = 2xl − 2 = − 4yl 1 = − 2zl  Utilizing x^{2} − 2y^{2} − z^{2} − 1 = 0 and subsituting x, y and z yields [3/(2l^{2})] = 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]} ).
 To find the greatest and least values of a function f subjected to g, we solve the system of equations
where l is a constant.∇f = l∇g g(x,y,z) = 0  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
 
 

 To find the greatest and least values of a function f subjected to g, we solve the system of equations
where l is a constant.∇f = l∇g g(x,y,z) = 0  Now, ∇f = (yz,xz,xy) and ∇g = (2x,6y,2z), let g(x,y,z) = x^{2} + 3y^{2} + z^{2} − 4. Then our system of equations is
(yz,xz,xy) = l(2x,6y,2z) x^{2} + 3y^{2} + z^{2} − 4 = 0  (yz,xz,xy) = l(2x,6y,2z) is equivalent to
solving the top equation for x results in x = [yz/2l]. We substitute this result in the other two equations.yz = 2xl xz = 6yl xy = 2zl  So ( [yz/2l] )z = 6yl gives z^{2} = 12l^{2} while ( [yz/2l] )y = 2zl gives y^{2} = 4l^{2}. To find x^{2} 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 x^{2} = 12l^{2}. Utilizing x^{2} − 3y^{2} + z^{2} − 4 = 0 and our previous results yields ( 12l^{2} )^{2} − 3( 4l^{2} )^{2} + ( 12l^{2} )^{2} = 4 hence l = [1/(^{4}√{60})].
 Then x^{2} = 12l^{2} = [12/(√{60} )] = [6/(√{15} )] so x = ±√{[6/(√{15} )]} , y^{2} = 4l^{2} = [4/(√{60} )] = [2/(√{15} )] so y = ±√{[2/(√{15} )]} and z^{2} = 12l^{2} = [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}√{15^{3}})]
 To find the greatest value of f subjected to g, we solve the system of equations
where l is a constant.∇f = l∇g g(x,y,z) = 0  Now, ∇f = (1,3,8) and ∇g = (2x,2y,2z), let g(x,y,z) = x^{2} + y^{2} + z^{2} − 1. Then our system of equations is
(1,3,8) = l(2x,2y,2z) x^{2} + y^{2} + z^{2} − 1 = 0  (1,3,8) = l(2x,2y,2z) is equivalent to
which implies x = [1/l], y = [3/2l] and z = [4/l].1 = xl 3 = 2yl 8 = 2zl  Utilizing x^{2} + y^{2} + z^{2} − 1 = 0 we substitute to get ( [1/l] )^{2} + ( [3/2l] )^{2} + ( [4/l] )^{2} = 1 or 77 = 4l^{2}.
 Hence l = ±[(√{77} )/2], but since x, y and z have to be greater than zero, we only use the positive answer.
 To find the least value of f subjected to g, we solve the system of equations
where l is a constant.∇f = l∇g g(x,y,z) = 0  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
which implies x = [(l + 2)/2], y = [l/2] and z = [(l + 2)/2].2x − 2 = l 2y = l 2z − 2 = l  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].
*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.
Answer
Lagrange Multipliers
Lecture Slides are screencaptured 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
Multivariable Calculus
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) = x^{2}/8 + y^{2}/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 x^{2}/8 + y^{2}/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 x^{2}/8 + y^{2}/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  λ^{2}y/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  λ^{2}/4 = 0, so I have two solutions for y.0994
I have y = 0  excuse me... let me see here  and 1  λ^{2}/4 = 0, so I get 1 = λ^{2}/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, x^{2}/8 + y^{2}/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 x^{2} into here... squared over 8 _ y^{2}/2  1 = 0.1160
Here I am going to get 4y^{2}/8 + y^{2}/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 4y^{2} + 4y^{2}  8 = 0.1179
So, I am going to get 8Y^{2}  8 = 0, I am going to get y^{2} = 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 x^{2} + y^{2} = 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 x^{2} + y^{2}. 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 x^{2} + y^{2}  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 x^{2} + y^{2} + 1 = 0, and we said that x = 3/2λ.1619
This is going to be 3/2λ^{2}  1 = 0. So let us go ahead and work all of this out. 9/4λ^{2} + 4/λ^{2} 1 = 0. We are not going to leave anything out here.1634
I am going to go ahead and multiply through by 4λ^{2}, I think... no, just 4... yea, 4λ^{2}.1658
So, when I multiply by 4λ^{2}, I am going to get 9 over here, I am going to get 16 over here, 4λ^{2} = 0, just to get rid of the denominator, that is all I did.1672
9 + 16 is 25... that equals 4λ^{2}, so λ^{2} = 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. x^{2} + y^{2} = 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. Byebye.1949
1 answer
Last reply by: Professor Hovasapian
Tue Apr 7, 2015 10:39 PM
Post by Alvi Akbar on April 7, 2015
Is there any way to change the video playback speed ?
4 answers
Last reply by: Professor Hovasapian
Sun Apr 14, 2013 1:29 PM
Post by Jawad Hassan on April 12, 2013
Could 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 answers
Post by Shahaz Shajahan on September 4, 2012
Sorry I realised how stupid that question was :S, It works out exactly the same, obviously.
0 answers
Post by Shahaz Shajahan on September 4, 2012
Ok, 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?