Lagrange multipliers

chrisheaven · Feb 26, 2008

One of the most common problems in calculus is that of finding maxima or minima (in general, "extrema") of a function, but it is often difficult to find a closed form for the function being extremized. Such difficulties often arise when one wishes to maximize or minimize a function subject to fixed outside conditions or constraints. The method of Lagrange multipliers is a powerful tool for solving this class of problems without the need explicitly to solve the conditions and use them to eliminate extra variables.

Thought I should post a quick overview before the posting FRENZY begins in this thread 😱

milagros317 · Feb 26, 2008

This may be the best thread you've ever posted. 😀

An example of how to use Lagrange multipliers is below.

To maximize xyz where x, y, and z are non-negative and satisfy:

x^2 + 2y^2 + 3z^2 = 144,

note that Lagrange's equation becomes < yz, xz, xy > = K<2x, 4y, 6z> where K is the Lagrange multiplier.

It is clear that the resulting four equations in four unknowns has the unique solution:

K=√2, x=4√3, y=2√6, z=4

and the maximum product is then xyz = 96√2

😀

chrisheaven · Feb 29, 2008

best thread eh???-- you jest, my friend, surely LOL

kake · Feb 29, 2008

I think I may give this one a miss chris!, get Stephen Hawkins on the case.

"taxi 4 kake"

milagros317 · Feb 29, 2008

chrisheaven said:
best thread eh???-- you jest, my friend, surely LOL

It looks like your best thread to me. 😀

Bugman · Feb 29, 2008

It is clear that the resulting four equations in four unknowns has the unique solution:

Of course it is,everybody knows that. :shock:

chrisheaven · Mar 1, 2008

bugman said:
It is clear that the resulting four equations in four unknowns has the unique solution:

Of course it is,everybody knows that.

I don't understand, bugman- can you clarify, please

milagros317 · Mar 1, 2008

bugman said:
It is clear that the resulting four equations in four unknowns has the unique solution:

Of course it is,everybody knows that.

Well, I can make it clear in a few minutes at the chalkboard. 😀

Bohemianne · Mar 2, 2008

milagros317 said:
This may be the best thread you've ever posted. 😀

An example of how to use Lagrange multipliers is below.

To maximize xyz where x, y, and z are non-negative and satisfy:

x^2 + 2y^2 + 3z^2 = 144,

note that Lagrange's equation becomes < yz, xz, xy > = K<2x, 4y, 6z> where K is the Lagrange multiplier.

It is clear that the resulting four equations in four unknowns has the unique solution:

K=√2, x=4√3, y=2√6, z=4

and the maximum product is then xyz = 96√2

😀

^ PWNed! 😱 😀

milagros317 · Mar 2, 2008

It's easier than it seems at first. 😀

chrisheaven · Mar 3, 2008

much easier actually

milagros317 · Mar 3, 2008

I'm glad that you like it. 😀

chrisheaven · Mar 4, 2008

I wondered if any of the readers of this thread might be interested in this little explanation: :jester:

If I want to find the
stationary values of f(x,y) = 2x^2 + 3y^2 subject to the constraint
2x + y = 1.

The general problem is to find stationary points of f(x,y) subject to
constraint g(x,y) = 0 [Note that the constraint must be written in
this form]. So for our problem g(x.y) = 2x + y - 1.

At stationary points of f(x,y) we have

df = part(df/dx)*dx + part(df/dy)*dy = 0

This implies that the vector [part(df/dx), part(df/dy)] is
perpendicular to the vector [dx, dy]

Since g(x,y) = 0 we can write

dg = part(dg/dx)*dx + part(dg/dy)*dy = 0

Thus the vector [part(dg/dx), part(dg/dy)] is also perpendicular to
the vector [dx, dy]. This implies that the vector [part(df/dx),
part(df/dy)] is parallel to the vector [part(dg/dx), part(dg/dy)] and
that we can find a number 'k' such that

[part(df/dx), part(df/dy)] - k[part(dg/dx), part(dg/dy)]
= [0, 0]

This can be summarized by writing

phi(x,y) = f(x,y) - kg(x,y)

Then f(x,y) will have a stationary point subject to constraint
g(x,y) = 0 when part(d(phi)/dx) = 0, part(d(phi)/dy) = 0 and
g(x,y) = 0

This gives three equations to find x, y and k.

k is called the Lagrange multiplier and phi is called the auxiliary
function.

Applying these ideas to our problem, we have

f(x,y) = 2x^2 + 3y^2 and g(x,y) = 2x + y - 1

The auxiliary function is

phi(x,y) = f(x,y) - kg(x,y)
= 2x^2 + 3y^2 - k(2x + y -1)

Then:

part(d(phi)/dx) = 4x - 2k = 0 (1)

part(d(phi)/dy) = 6y - k = 0 (2)

g(x,y) = 2x + y -1 = 0 (3)

Solving (1), (2) and (3) for k, x and y we get

k = 6/7 x = 3/7 y = 1/7

So at the stationary point x = 3/7 y = 1/7 we have

f(x,y) = 2*(3/7)^2 + 3*(1/7)^2
= 18/49 + 3/49
= 21/49
= 3/7

To determine the nature of the stationary point you look at
neighbouring points to see if f(x,y) has increased or decreased or not
changed (saddle point). :jester:

milagros317 · Mar 4, 2008

In the three dimensional case, as I did above, you have four unknowns (x, y, z, and K) and four equations. 😀

chrisheaven · Mar 6, 2008

fun isn't it!

milagros317 · Mar 6, 2008

I've always thought it was fun, actually. 😀

chrisheaven · Mar 7, 2008

me too

milagros317 · Mar 7, 2008

That may be why we're considered strange. 😀

Bohemianne · Mar 7, 2008

LOL, love the thread as inspiring reminder: the "3-dimensional planes" earned me serious livelihood in the real world; and it keeps me back in track after being rusted with formula derivation...As I told Mils before, maxima/minima has been a favorite...😱

milagros317 · Mar 8, 2008

And this is Chris' maximally intelligent thread. 😀

Bohemianne · Mar 8, 2008

milagros317 said:
And this is Chris' maximally intelligent thread. 😀

^ Of all the places, the silliest people are mathematical maniacs! No intention of plugging, it took my interest because in minor sense, my eroticomics character "Miss Charlotte" is actually a calculus teacher (without too much emphasis on it...I just love the idea)! I just find it spooky that ticklishness and calculus are directly proportional. 😱 :3poke: