I've done this so far:
F(x,y) = root(xy) + lambda( x + y  1) = 0
Fx = 1/2(xy)^(1/2) . y + lambda = 0
Fy = 1/2(xy)^(1/2) . x + lambda = 0
Flambda = x + y  1 = 0
x + y  1 = 0
x + y = 1
y = 1  x
I don't know how to continue from here :/
Any help would be appreciated thank you
Lagrange Multipliers Questions

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 1
 19032017 18:44

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 2
 19032017 22:21
(Original post by Fatts13)
I've done this so far:
F(x,y) = root(xy) + lambda( x + y  1) = 0
Fx = 1/2(xy)^(1/2) . y + lambda = 0
Fy = 1/2(xy)^(1/2) . x + lambda = 0
Flambda = x + y  1 = 0
x + y  1 = 0
x + y = 1
y = 1  x
I don't know how to continue from here :/
Any help would be appreciated thank you
I'd look to the first two equations; one minus the other gives you another relatinship between x and y. 
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 3
 20032017 21:54
(Original post by ghostwalker)
The bit in red shouldn't be present.
I'd look to the first two equations; one minus the other gives you another relatinship between x and y.
F(x,y, lambda) = root(xy)  lambda(x + y 1)
Fx = 1/2(root(y/x))  lambda
Fy = 1/2(root(x/y)  lambda
Flambda = 1  x  y
Fx = Fy
1/2(root(y/x))  lambda = 1/2(root(x/y)  lambda
y/x = x/y
y^2 = x^2
y = x
when x = y and Flambda = 0
1  x  y = 0
1 = x + y
Therefore x = y = 1/2
Therefore the critical point is (1/2,1/2)
Though my problem is I'm not exactly sure about how I got Fx and Fy :s 
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 4
 20032017 22:04
(Original post by Fatts13)
This is what I got for this one:
F(x,y, lambda) = root(xy)  lambda(x + y 1)
Fx = 1/2(root(y/x))  lambda
Fy = 1/2(root(x/y)  lambda
Flambda = 1  x  y
Fx = Fy
1/2(root(y/x))  lambda = 1/2(root(x/y)  lambda
y/x = x/y
y^2 = x^2
y = x
when x = y and Flambda = 0
1  x  y = 0
1 = x + y
Therefore x = y = 1/2
Therefore the critical point is (1/2,1/2)
Though my problem is I'm not exactly sure about how I got Fx and Fy :s 
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 5
 20032017 22:06
(Original post by ghostwalker)
Are you saying you differentiated, but don't know how you differentiated?
Is this alright to do? If so then I'm sorry ignore me 
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 20032017 22:16
(Original post by Fatts13)
As in I differentiated but I'm not sure whether it is right as I did root(1/x) . y and made it into root(y/x)
Is this alright to do? If so then I'm sorry ignore me
In the case of Fx for example, you did (correctly)
Which equals
Note:
In post #3 you went from
y^2=x^2
to y=x
It should have been y=x OR y=x. But the latter has no solution when you substitute into Flambda, so you're not missing a solution this time. 
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 20032017 22:43
(Original post by ghostwalker)
Actually, from your first post, you didn't do that.
In the case of Fx for example, you did (correctly)
Which equals
Note:
In post #3 you went from
y^2=x^2
to y=x
It should have been y=x OR y=x. But the latter has no solution when you substitute into Flambda, so you're not missing a solution this time.
Ok I'll add y = x and then state no solutions, thank you.
I would give you a thumbs up but it's not letting me
Also
for this questions, whilst its given us the constraint what would be the function?
Sorry for being a hassle 
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 8
 20032017 22:45
(Original post by Fatts13)
Also
for this questions, whilst its given us the constraint what would be the function? 
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 20032017 22:50
(Original post by ghostwalker)
What's the distance of a point (x,y) from (0,0) ? 
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 20032017 22:53
(Original post by Fatts13)
Oh, so the distance would be x^2 + y^2 therefore that's the function?
BUT, when the square root is a minimum, so is x^2+y^2, so yes, go with that  it's probably easier to work with.Last edited by ghostwalker; 20032017 at 22:56. 
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 20032017 23:10
(Original post by ghostwalker)
Distance is actually the square root of that.
BUT, when the square root is a minimum, so is x^2+y^2, so yes, go with that  it's probably easier to work with.
F(x,y,lambda) = x^2 + y^2  lambda( y^2  2root3xy  x^2  2)
Fx = 2x + 2root3lambday + 2lambdax
Fy = 2y + 2root3lambdax  2lambday
Flambda = 2  y^2 + 2root3xy + x^2
But I seem to be struggling to get for example what y could be by equating Fx = Fy
I've got to y = x ((1 + lambda  root3lambda)/(root3lambda  1 + lambda))Last edited by Fatts13; 20032017 at 23:15. 
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 12
 20032017 23:27
(Original post by Fatts13)
Ok so then this would be the equation:
F(x,y,lambda) = x^2 + y^2  lambda( y^2  2root3xy  x^2  2)
Fx = 2x + 2root3lambday + 2lambdax
Fy = 2y + 2root3lambdax  2lambday
Flambda = 2  y^2 + 2root3xy + x^2
But I seem to be struggling to get for example what y could be by equating Fx = Fy
I've got to y = x ((1 + lambda  root3lambda)/(root3lambda  1 + lambda))
You want ot get rid of lambda.
Although you've been equating Fx and Fy, I don't think that's the way to go.
You have Fx=0, so rearrange that to lambda = something.
Similarly Fy=0.
Now eliminate lambda between the two to get a relationship in x and y.....(1)
You can end up with something close to your constraint.
You should, with the constraint and (1) be able to eliminate y^2x^2, and get xy=constant.
Not gone further.
Time for bed. 
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 13
 21032017 01:24
(Original post by ghostwalker)
Yes, it's not as straight forward. I've not worked it through fully.
You want ot get rid of lambda.
Although you've been equating Fx and Fy, I don't think that's the way to go.
You have Fx=0, so rearrange that to lambda = something.
Similarly Fy=0.
Now eliminate lambda between the two to get a relationship in x and y.....(1)
You can end up with something close to your constraint.
You should, with the constraint and (1) be able to eliminate y^2x^2, and get xy=constant.
Not gone further.
Time for bed.
F(x,y,lambda) = x^2 + y^2  lambda( y^2  2root3xy  x^2  2)
Fx = 2x + 2root3lambday + 2lambdax
Fy = 2y + 2root3lambdax  2lambday
Flambda = 2  y^2 + 2root3xy + x^2
Fx = 0
2x + 2root3lambdax + 2lambdax = 0
y =  (1 + lambda) x / root3 lambda
Fy = 0
2y + 2root3lambdax  2lambday = 0
y = root3lambda x / lambda  1
y =y Therefore
 (1 + lambda) x / root3 lambda = root3lambda x / lambda  1
4lambda^2  1 = 0
(2lamda + 1) ( 2lambda  1) = 0
lambda = 1/2 or 1/2
When lambda = 1/2 we find y = root3 x
Substitue y = root3 x into Flambda = 0 giving us
x = 1/2 or 1/2
When x = 1/2 lambda = 1/2 y =(root3)/2
When x = 1/2 lambda = 1/2 y =  (root3)/2
Therefore for lambda = 1/2 there are two critical points:
(1/2,  (root3)/2) and (1/2, (root3)/2)
When lambda = 1/2 we find y = root3x
Substitue y = root3 x into Flambda = 0
2x^2 + 1 = 0 Therefore no solutions
Therefore lambda = 1/2 is not relevant, thus leaving only two critical points:
(1/2,  (root3)/2) and (1/2, (root3)/2) 
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 14
 21032017 07:16
(Original post by Fatts13)
No worries, I think I got the answer  took some time but I got there eventually after ELEMENTAL mistakes
F(x,y,lambda) = x^2 + y^2  lambda( y^2  2root3xy  x^2  2)
Fx = 2x + 2root3lambday + 2lambdax
Fy = 2y + 2root3lambdax  2lambday
Flambda = 2  y^2 + 2root3xy + x^2
Fx = 0
2x + 2root3lambdax + 2lambdax = 0
y =  (1 + lambda) x / root3 lambda
Fy = 0
2y + 2root3lambdax  2lambday = 0
y = root3lambda x / lambda  1
y =y Therefore
 (1 + lambda) x / root3 lambda = root3lambda x / lambda  1
4lambda^2  1 = 0
(2lamda + 1) ( 2lambda  1) = 0
lambda = 1/2 or 1/2
When lambda = 1/2 we find y = root3 x
Substitue y = root3 x into Flambda = 0 giving us
x = 1/2 or 1/2
When x = 1/2 lambda = 1/2 y =(root3)/2
When x = 1/2 lambda = 1/2 y =  (root3)/2
Therefore for lambda = 1/2 there are two critical points:
(1/2,  (root3)/2) and (1/2, (root3)/2)
When lambda = 1/2 we find y = root3x
Substitue y = root3 x into Flambda = 0
2x^2 + 1 = 0 Therefore no solutions
Therefore lambda = 1/2 is not relevant, thus leaving only two critical points:
(1/2,  (root3)/2) and (1/2, (root3)/2)
I'm not going to check all the details.
I notice one thing  in red , you just cancelled the "x". You can only cancel a value out of an equation in that fashion, IF you know x=0 is not a solution. At this stage, you don't know that.
And as a final point you really ought to check the distance for these points, (i.e. what does the function evaluate to?) to check they're the same distance, and are thus the closest points. One might have been closer than the other.Last edited by ghostwalker; 21032017 at 07:20. 
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 21032017 18:39
(Original post by ghostwalker)
Well done.
I'm not going to check all the details.
I notice one thing  in red , you just cancelled the "x". You can only cancel a value out of an equation in that fashion, IF you know x=0 is not a solution. At this stage, you don't know that.
And as a final point you really ought to check the distance for these points, (i.e. what does the function evaluate to?) to check they're the same distance, and are thus the closest points. One might have been closer than the other.
Whilst this isn't a Lagrange Multiplier question, I don't know how to do this :/
So what I've done so far:
df/dx = (y^2  1) / (1 + xy) ^2 when this = 0 then y = +/ 1
df/dy = (x^2  1) / (1 + xy)^2 when this = 0 then x = +/ 1
Now I'm struggling to get the second partial derivatives, I got these but I'm not 100% sure at all:
d^2f/dx^2 = (2y(y^21)) / (yx + 1)^3
d^2f/dy^2 =( 2(x+y) ) / (yx + 1)^3
Sorry for the hassle
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Updated: March 21, 2017
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