# Thread: locating a point closest to the origin using Lagrange multipliers?

1. ## locating a point closest to the origin using Lagrange multipliers?

Here is the question...
Locate the point on the line which is the intersection of the planes $\displaystyle y+2z=12$ and $\displaystyle x+z=6$ which is closest to the origin. Can someone solve this and please tell me how to do it using Lagrange multipliers? Thanks in advance.

2. Originally Posted by Infernorage
Here is the question...
Locate the point on the line which is the intersection of the planes $\displaystyle y+2z=12$ and $\displaystyle x+z=6$ which is closest to the origin. Can someone solve this and please tell me how to do it using Lagrange multipliers? Thanks in advance.
Let f(x, y, z) = x^2 + y^2 + z^2, g(x, y, z) = y + 2z, h(x, y, z) = x + z. Then you want to minimize f under the constraints g = 12 and h = 6. To do so, solve the system of five equations generated by grad(f) = lambda*grad(g) + mu*grad(h), g = 12, h = 6.

3. Originally Posted by rn443
Let f(x, y, z) = x^2 + y^2 + z^2, g(x, y, z) = y + 2z, h(x, y, z) = x + z. Then you want to minimize f under the constraints g = 12 and h = 6. To do so, solve the system of five equations generated by grad(f) = lambda*grad(g) + mu*grad(h), g = 12, h = 6.
Hi, yea I did that part. For the 5 equation I got the following...
1)$\displaystyle 2x=\mu$
2)$\displaystyle 2y=\lambda$
3)$\displaystyle 2z=2\lambda+\mu$
4)$\displaystyle y+2z=12$
5)$\displaystyle x+z=6$
I can't figure out to solve for the variables though. Can you help me out with this? Thanks.

4. Originally Posted by Infernorage
Hi, yea I did that part. For the 5 equation I got the following...
1)$\displaystyle 2x=\mu$
2)$\displaystyle 2y=\lambda$
3)$\displaystyle 2z=2\lambda+\mu$
4)$\displaystyle y+2z=12$
5)$\displaystyle x+z=6$
I can't figure out to solve for the variables though. Can you help me out with this? Thanks.
Since $\displaystyle \mu= 2x$ and $\displaystyle \lambda= 2y$, 2z= 2(2x)+ 2(2y) so z= 2x+ 2y. Putting that into y+ 2z= 12 gives y+ 2(2x+2y)= 4x+ 5y= 12.
Putting z= 2x+ 2y into x+ z= 6 gives x+ (2x+2y)= 3x+ 2y= 6.

Now, you have two equations to solve for x and y.

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# find the point on the plane which is closest to the origin using lagrange

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