# lagrange multipliers

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• Jul 9th 2007, 05:58 AM
turion645344
lagrange multipliers
hey, ive got this problem, which isnt that hard, but i cannot see how to get to the solution. Any help will be very much appreciated!!

Find the maximum of f(x,y)=4xy, subject to the condition x2+4y2=1

its a lagrange multiplier, help!!
• Jul 9th 2007, 06:14 AM
topsquark
Quote:

Originally Posted by turion645344
hey, ive got this problem, which isnt that hard, but i cannot see how to get to the solution. Any help will be very much appreciated!!

Find the maximum of f(x,y)=4xy, subject to the condition x2+4y2=1

its a lagrange multiplier, help!!

We have the condition
$\displaystyle x^2 + 4y^2 = 1$

So
$\displaystyle x^2 + 4y^2 - 1 = 0$

Thus we may consider the maximum of the function
$\displaystyle f(x, y) = 4xy = 4xy + \lambda(x^2 + 4y^2 - 1)$
where $\displaystyle \lambda$ is an arbitrary constant.

Now:
$\displaystyle \frac{\partial f}{\partial x} = 4y + \lambda \cdot 2x = 0$

$\displaystyle \frac{\partial f}{\partial y} = 4x + \lambda \cdot 8y = 0$

$\displaystyle \frac{\partial f}{\partial \lambda} = x^2 + 4y^2 - 1 = 0$

The first equation says
$\displaystyle \lambda = - \frac{2y}{x}$

So the second equation now says:
$\displaystyle 4x + \left ( - \frac{2y}{x} \right ) \cdot 8y = 0$

$\displaystyle 4x - \frac{16y^2}{x} = 0$

$\displaystyle x^2 - 4y^2 = 0$

So we have two equations in two unknowns:
$\displaystyle x^2 - 4y^2 = 0$

$\displaystyle x^2 + 4y^2 - 1 = 0$

$\displaystyle 2x^2 - 1 = 0$

$\displaystyle x = \pm \frac{\sqrt{2}}{2}$

Thus
$\displaystyle \left ( \pm \frac{\sqrt{2}}{2} \right ) ^2 - 4y^2 = 0$

$\displaystyle 4y^2 = \frac{1}{2}$

$\displaystyle y = \pm \frac{\sqrt{2}}{4}$

So there are a total of 4 solution points:
$\displaystyle \left ( \pm \frac{\sqrt{2}}{2}, \pm \frac{\sqrt{2}}{4} \right )$
where the $\displaystyle \pm$ signs are independent.

I leave it to you to determine which are relative maxima or relative minima.

-Dan
• Jul 9th 2007, 06:55 AM
turion645344
cheers Dan that really helped. I am still a bit new to Lagrange multipliers and am not sure how to work out the relative maxima (I have worked it out as 1, but not entirely sure) and how do I check that this is the relative maxima not the minima?
• Jul 9th 2007, 07:36 AM
topsquark
Quote:

Originally Posted by turion645344
cheers Dan that really helped. I am still a bit new to Lagrange multipliers and am not sure how to work out the relative maxima (I have worked it out as 1, but not entirely sure) and how do I check that this is the relative maxima not the minima?

Well, there is a more complicated way using second derivatives, but two simple methods also work:
1) Graph it
2) Can you find any values of the function greater than 1?

As I get the values $\displaystyle \pm 1$ when I put in the values of x and y that were derived for possibilities, I would say that
$\displaystyle \left ( \frac{\sqrt{2}}{2}, \frac{\sqrt{2}}{4} \right )$ and $\displaystyle \left ( -\frac{\sqrt{2}}{2}, -\frac{\sqrt{2}}{4} \right )$ generate maximum values and that $\displaystyle \left ( -\frac{\sqrt{2}}{2}, \frac{\sqrt{2}}{4} \right )$ and $\displaystyle \left ( \frac{\sqrt{2}}{2}, -\frac{\sqrt{2}}{4} \right )$ generate the minimum values.

-Dan
• Jul 9th 2007, 08:00 AM
turion645344
oh right, yea that wasnt too bad! also ive managed to do most of the other questions, but this one im finding hard...

find the minimum of g(x,y) = x2+2y2+2xy+2x+3y subject to the condition x2-y=1.

again any help would be much appreciated!
• Jul 9th 2007, 08:36 AM
ThePerfectHacker
Quote:

Originally Posted by turion645344
oh right, yea that wasnt too bad! also ive managed to do most of the other questions, but this one im finding hard...

find the minimum of g(x,y) = x2+2y2+2xy+2x+3y subject to the condition x2-y=1.

again any help would be much appreciated!

We need to solve,
$\displaystyle \nabla g = k\nabla f$ where $\displaystyle f=x^2-y$.
Thus,
$\displaystyle \left< 2x+2y+2, 4y+2x+3 \right> = k\left< 2x , - 1 \right>$

Thus,
$\displaystyle \left\{ \begin{array}{c}2x+2y+2 = 2xk \\ 4y+2x+3 = -k \\ x^2 - y = 1 \end{array} \right\}$
• Jul 9th 2007, 08:39 AM
turion645344
yea i managed to get to these three formulas but this is where i got stuck!
• Jul 9th 2007, 09:06 AM
ThePerfectHacker
Quote:

Originally Posted by The Greatest
$\displaystyle \left\{ \begin{array}{c}2x+2y+2 = 2xk \\ 4y+2x+3 = -k \\ x^2 - y = 1 \end{array} \right\}$

$\displaystyle 2x+2y+2 = 2x(-4y-2x-3)$

$\displaystyle 2x+2y+2=-8xy-4x^2-6x$

$\displaystyle 8xy+4x^2+8x+2y+2=0$

$\displaystyle 4xy+2x^2+4x+y+1=0$

$\displaystyle 4x(x^2-1)+2x^2+4x+(x^2-1)+1=0$

$\displaystyle 4x^3 - 4x + 2x^2 + 4x + x^2 =0$

$\displaystyle 4x^3 + 3x^2 = 0$

$\displaystyle x^2(4x+3)=0$

You are big boy you can do it from here.
• Jul 10th 2007, 03:58 AM
turion645344
cheers ive got x=-0.75 and then y= -7/16 therefore ive worked out the minimum as -1.2109 (-1,27/128)

is this right? and how do i prove its the minimum value?:confused:
• Jul 10th 2007, 04:15 AM
topsquark
Quote:

Originally Posted by turion645344
cheers ive got x=-0.75 and then y= -7/16 therefore ive worked out the minimum as -1.2109 (-1,27/128)

is this right? and how do i prove its the minimum value?:confused:

It looks to me like you have another possibility:
(x, y) = (0, -1)

So of these possibilities, which would be the minimum?

-Dan
• Jul 10th 2007, 04:54 AM
CaptainBlack
Quote:

Originally Posted by topsquark
It looks to me like you have another possibility:
(x, y) = (0, -1)

So of these possibilities, which would be the minimum?

-Dan

$\displaystyle g(0, -1) = -1$,

so assuming that the sums are right then the other is certainly smaller than this.

To show it is a minimum we substitute y=x^2-1 into the objective, and show that the resulting function has a minimum at x=-3/5 (and it does).

RonL
• Jul 10th 2007, 11:10 AM
turion645344
To show it is a minimum we substitute y=x^2-1 into the objective, and show that the resulting function has a minimum at x=-3/5 (and it does).

ive tried to this, but keep on getting the maths wrong! :mad: help!:)
• Jul 10th 2007, 05:59 PM
topsquark
Quote:

Originally Posted by turion645344
To show it is a minimum we substitute y=x^2-1 into the objective, and show that the resulting function has a minimum at x=-3/5 (and it does).

ive tried to this, but keep on getting the maths wrong! :mad: help!:)

Plug each (x, y) point into your g(x, y) function. (The thing, after all, you are trying to minimize.) The two possibilities you have are the only ones where the function is either at a relative maximum or a relative minimum. I see that
$\displaystyle g(0, -1) = 0^2 + 2(-1)^2 + 2 \cdot 0 \cdot (-1) +2 \cdot 0 + 3(-1) = 2 - 3 = -1$

$\displaystyle g \left ( -\frac{3}{5}, -\frac{16}{25} \right ) = ~ ... ~ = -\frac{733}{625}$

Since $\displaystyle -1 > -\frac{733}{625}$ the $\displaystyle \left ( -\frac{3}{5}, -\frac{16}{25} \right )$ point is your minimum.

-Dan
• Jul 10th 2007, 08:34 PM
CaptainBlack
Quote:

Originally Posted by topsquark
Plug each (x, y) point into your g(x, y) function. (The thing, after all, you are trying to minimize.) The two possibilities you have are the only ones where the function is either at a relative maximum or a relative minimum. I see that
$\displaystyle g(0, -1) = 0^2 + 2(-1)^2 + 2 \cdot 0 \cdot (-1) +2 \cdot 0 + 3(-1) = 2 - 3 = -1$

$\displaystyle g \left ( -\frac{3}{5}, -\frac{16}{25} \right ) = ~ ... ~ = -\frac{733}{625}$

Since $\displaystyle -1 > -\frac{733}{625}$ the $\displaystyle \left ( -\frac{3}{5}, -\frac{16}{25} \right )$ point is your minimum.

-Dan

What we have shown is that the curves defined by the objective and the
constraint kiss at the points we have found. A kissing point may be the
analog of a local maxima, minima or point of inflection . We still need to
classify the point, or at least justify our conclusion that one is a minimum.

In this case we can conclude that the function does have a minima as when
x goes of to infinity the objective becomes arbitrily large, also that any
minima for this objective and constraint must be smooth, so we then know
that one of the kissing points must be a minima, and we have identified which
it is.

RonL

RonL
• Jul 11th 2007, 05:42 AM
turion645344
http://www.mathhelpforum.com/math-he...4c88ddc0-1.gif

i was just wondering where this came from, as i thought that the other values were (-0.75,-7/16) which gives a value of -1,27/128. this is smaller than -1 so is the minimum value right?

also to test the minimum using the idea that as
x goes of to infinity the objective becomes arbitrily large. can you actually use this concept to test for a minimum value?!
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