For more information, please see full course syllabus of Multivariable Calculus
For more information, please see full course syllabus of Multivariable Calculus
More Lagrange Multiplier Examples
 Recall that the distance from a point in space to the origin is d(x,y,z) = √{x^{2} + y^{2} + z^{2}} .
 We can use Lagrange multipliers to find the least distance, f:√{x^{2} + y^{2} + z^{2}} = 0 from a point in g:x^{2} + y^{2} + (z − 4)^{2} − 9 = 0 to the origin, that is subject to g.
 Note that √{x^{2} + y^{2} + z^{2}} = 0 is equivalent to x^{2} + y^{2} + z^{2} = 0. Now
∇f = l∇g g(x,y,z) = 0 (2x,2y,2z) = l(2x,2y,2z − 8) x^{2} + y^{2} + (z − 4)^{2} − 9 = 0  (2x,2y,2z) = l(2x,2y,2z − 8) is equivalent to
2x = 2xl 2y = 2yl 2z = (2z − 8)l  Hence x = 0 and y = 0, x^{2} + y^{2} + (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.
 Recall that the distance from a point in space to the origin is d(x,y,z) = √{x^{2} + y^{2} + z^{2}} .
 We can use Lagrange multipliers to find the least distance, f:√{x^{2} + y^{2} + z^{2}} = 0 from a point in g: − x + y + z − 1 = 0 to the origin, that is subject to g.
 Note that √{x^{2} + y^{2} + z^{2}} = 0 is equivalent to x^{2} + y^{2} + z^{2} = 0. Now
∇f = l∇g g(x,y,z) = 0 (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  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].
 

 Recall that the distance from a point in space to the origin is d(x,y,z) = √{x^{2} + y^{2} + z^{2}} .
 We find the least distance, f:√{x^{2} + y^{2} + z^{2}} = 0 from a point in g:xy^{2} − z − 5 = 0 to the origin, that is subject to g.
 Note that √{x^{2} + y^{2} + z^{2}} = 0 is equivalent to x^{2} + y^{2} + z^{2} = 0. Now
∇f = l∇g g(x,y,z) = 0 (2x,2y,2z) = l(y^{2},2xy, − 1) xy^{2} − z − 5 = 0
 
 

 
 
 

 

 Recall that the distance from a point in space to the origin is d(x,y,z) = √{x^{2} + y^{2} + z^{2}} .
 We find the least distance, f:√{x^{2} + y^{2} + z^{2}} = 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 √{x^{2} + y^{2} + z^{2}} = 0 is equivalent to x^{2} + y^{2} + z^{2} = 0. Now
∇f = l∇g g(x,y,z) = 0 (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
 
 

 
 
 

 

 Recall that the distance from a point in space to the origin is d(x,y,z) = √{x^{2} + y^{2} + z^{2}} .
 We find the least distance, f:√{x^{2} + y^{2} + z^{2}} = 0 from a point in g:[(y^{2})/9] − [(x^{2})/4] − z − 3 = 0 to the origin, that is subject to g.
 Note that √{x^{2} + y^{2} + z^{2}} = 0 is equivalent to x^{2} + y^{2} + z^{2} = 0. Now
∇f = l∇g g(x,y,z) = 0 (2x,2y,2z) = l( − [x/2],[2y/9], − 1 ) [(y^{2})/9] − [(x^{2})/4] − z − 3 = 0
 
 

 
 
 

 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:x^{2} + y^{2} − 4 = 0.
 Now
∇f = l∇g g(x,y) = 0 ( − 6x,2y) = l( 2x,2y ) x^{2} + y^{2} − 4 = 0 − 6x = 2xl 2y = 2yl − 3x = xl y = yl x^{2} + y^{2} − 4 = 0  Note that if x = 0 then y^{2} = 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 x^{2} = 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.
 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:[(x^{2})/9] + [(y^{2})/4] − 1 = 0.
 Now
∇f = l∇g g(x,y) = 0 ( − 2x, − 2y) = l( [2x/9],[y/4] ) [(x^{2})/9] + [(y^{2})/4] − 1 = 0 − 2x = [2x/9]l − 2y = [y/4]l − 9x = xl − 8y = yl [(x^{2})/9] + [(y^{2})/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 [(x^{2})/9] + [(y^{2})/4] − 1 = 0 yields y^{2} = 4 so y = ±2 and l is defined. Similarly if y = 0, then [(x^{2})/9] + [(y^{2})/4] − 1 = 0 yields x^{2} = 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.
 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 (2x,2y) = l( 2(x − 1),2(y − 1) ) (x − 1)^{2} + (y − 1)^{2} − 1 = 0 2x = 2l(x − 1) 2y = 2l(y − 1) 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 )].
 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.
 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:x^{2} + y^{2} + z^{2} − 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 (2(x − 3),2(y + 1),2(z − 2)) = l(2x,2y,2z) x^{2} + y^{2} + z^{2} − 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  Utilizing x^{2} + y^{2} + z^{2} − 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
*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
More Lagrange Multiplier Examples
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
 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
Multivariable Calculus
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 straightforward. 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 z^{2}  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 = x^{2} + y^{2} + z^{2}, 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: x^{2} + y^{2} + z^{2}.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 z^{2}  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... z^{2}  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 z^{2}  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 z^{2}  0 × 0  1 = 0, so I get z^{2} = 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 0^{2}  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 x^{2} = y^{2}.0725
Well, let us see. x^{2} = y^{2}, well I found that y = 1/x, therefore I am going to put this y into here and I am going to get x^{2} = 1/x^{2}, which equals 1/x^{2}. That implies that x^{4} = 1.0741
Well, if x^{4} = 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 z^{2}  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) = x^{2} + 2y^{2}  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 x^{2} + y^{2} = 1, or x^{2} + y^{2}  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 2y^{2}, I am sorry, this is 2y^{2}.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) = x^{2} + y^{2}  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 x^{2} + y^{2}  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 x^{2} + 0^{2}  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 x^{2} + y^{2}  1 = 0, when I put it into here, I am going to end up with the following.1469
It is going to be 1/2^{2} + y^{2}  1 = 0.1491
I get 1/4 + y^{2}  1 = 0, I get y^{2} = 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. Byebye.1658
1 answer
Last reply by: Professor Hovasapian
Sat Jan 10, 2015 7:11 PM
Post by owais khan on January 8, 2015
in 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 answer
Last reply by: Professor Hovasapian
Thu Oct 10, 2013 1:44 AM
Post by yaqub ali on October 9, 2013
professor why did you divide 2x/2y in example 1 when z=0
1 answer
Last reply by: Professor Hovasapian
Mon Jun 10, 2013 3:22 PM
Post by Josh Winfield on June 8, 2013
Example 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 answer
Last reply by: Professor Hovasapian
Sat Sep 1, 2012 4:01 PM
Post by Mohammed Alhumaidi on September 1, 2012
You had switched the Grad(g) when multiplying by Lambda !! and x and y !!