Math Help - Lagrange Multipliers Confusion

1. Lagrange Multipliers Confusion

I have this question that I am lost on:

Use lagrange multipliers to find the points on the surface z^2 - xy = 49
that are closest to the origin.

This is my thinking so far:

d = sqrt ((x-0)^2 + (y-0)^2 + (z-0)^2

or
d^2 = (x-0)^2 + (y-0)^2 + (z-0)^2

so
del f = 2(x-0)i + 2(y-0)j + 2(z-0)k

The constraint is
g(x,y,z) = z^2 - xy = 49

so lambda del g(x,y,z) = lambda(k^2 - jk)

so
2(x-0)i + 2(y-0)j + 2(z-0)k = lambda(k^2 - jk), which leads to

2(z - 0) = lambda^2 and 2(y-0)*2(x-0) = lambda^2

therefore,
lambda = sqrt(2(z-0)) = sqrt(2(y-0)*2(x-0))

so sqrt(2*z) = sqrt(4*x*y)
so 2*z = 4*x*y

Im not even sure if this is correct, but if it is where do I go from here?

I have made many attempts, and with one of those I came up with 3 points:

x=7, y =7
x=1 , y=49
x=49, x=1

but again, I dont think this is correct.

2. Originally Posted by woody198403
I have this question that I am lost on:

Use lagrange multipliers to find the points on the surface z^2 - xy = 49
that are closest to the origin.

This is my thinking so far:

d = sqrt ((x-0)^2 + (y-0)^2 + (z-0)^2

or
d^2 = (x-0)^2 + (y-0)^2 + (z-0)^2

so
del f = 2(x-0)i + 2(y-0)j + 2(z-0)k

The constraint is
g(x,y,z) = z^2 - xy = 49

so lambda del g(x,y,z) = lambda(k^2 - jk)

so
2(x-0)i + 2(y-0)j + 2(z-0)k = lambda(k^2 - jk), which leads to

2(z - 0) = lambda^2 and 2(y-0)*2(x-0) = lambda^2

therefore,
lambda = sqrt(2(z-0)) = sqrt(2(y-0)*2(x-0))

so sqrt(2*z) = sqrt(4*x*y)
so 2*z = 4*x*y

Im not even sure if this is correct, but if it is where do I go from here?

I have made many attempts, and with one of those I came up with 3 points:

x=7, y =7
x=1 , y=49
x=49, x=1

but again, I dont think this is correct.
Here is how I do them. You should get the same equations that need solving.

$L = x^2 + y^2 + z^2 + \lambda (z^2 - xy - 49)$.

$\frac{\partial L}{\partial x} = 2x - \lambda y = 0$ .... (1)

$\frac{\partial L}{\partial y} = 2y - \lambda x = 0$ .... (2)

$\frac{\partial L}{\partial z} = 2z - 2 \lambda z = 0 \Rightarrow z(1 - \lambda) = 0 \Rightarrow z = 0 ~ \text{or} ~ \lambda = 1$ .... (3)

$\frac{\partial L}{\partial \lambda} = z^2 - xy - 49 = 0$ .... (4)

Solve these equations simultaneously.

Substitute $\lambda = 1$ into (1) and (2): x = y = 0. From (4): $z = \pm 7$.

So you have (0, 0, 7) and (0, 0, -7). They require testing.

Substitute z = 0 into (4): $xy = -49 \Rightarrow y = - \frac{49}{x}$. Substitute into (1) and (2) and solve for x. Then substitute back to get y. More solutions to test .....

3. Lagrange Multipliers

Hey
Im currently trying to use lagrange multipliers to find the points on the surface $z^2-x*y=49$ that are closest to the origin.

im saying $d^2=x^2+y^2+z^2$ for the distance for the origin

but i just cant grasp the whole lagrange multipliers concept...

all help much appreciated

4. We want to extremize $f(x,y,z)=x^{2}+y^{2}+z^{2}$

subject to the constraint $g(x,y,z)=z^{2}-xy=49$

Therefore,

${\nabla}f(x,y,z)=x^{2}+y^{2}+z^{2}={\lambda}{\nabl a}g(x,y,z)$

at a constrained relative extremum; that is,

$2xi+2yj+2zk={\lambda}(-yi-xj+2zk)$

$2x=-y{\lambda}, \;\ 2y=-x{\lambda}, \;\ 2z={\lambda}2z$

Now continue?.

What a lagrange multiplier tells us is that extrema occur where the gradients are parallel. ${\lambda}$ is the lagrange multiplier that represents a multiple such that
${\nabla}f(x,y,z)={\lambda}g(x,y,z)$.

5. Hey thanks
that all makes sense but do I now sub:

$2x=-y{\lambda}, \;\ 2y=-x{\lambda}, \;\ 2z={\lambda}2z$

into the constraint?

and if I do how do i then solve for the points with the minimum distance??

6. Solve them for lambda, then equate and solve for x, y or z, then sub into the constraint.

7. hey just with the final bit how can you Substitute into (1) and (2) and solve for x when there is a third unknown of $\lambda$??

it seems you have 2 equations and still 3 unknowns?

or am i just not thinking clearly?

Thanks again

8. Originally Posted by edeffect
hey just with the final bit how can you Substitute into (1) and (2) and solve for x when there is a third unknown of $\lambda$??

it seems you have 2 equations and still 3 unknowns?

or am i just not thinking clearly?

Thanks again
After the substitution of $y = - \frac{49}{x}$ into (1) and (2), the unknowns are x and $\lambda$. Two equations, two unknowns .......