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 .......