# Find the closest point to the origin on the curve of intersection to a cone

1. Mar 29, 2009

### jimbo71

1. The problem statement, all variables and given/known data
find the point closest to the origin on the curve of intersection of the plane 2y+4z=5 and the cone z^2=4x^2+4y^2

2. Relevant equations

3. The attempt at a solution
see 40 attachement. I found the used f(x,y,z)=x^2+y^2+z^2 and found its gradient. found ggrad and hgrad and set fgrad=lambda*ggrad+mu*hgrad. using the two constraint equations i attempted to solve for x,y,z,lamdbda,mu. Either I messed up the setting up of this problem or my algebra is wrong some where becuase I keep getting x^2=-25. This is preventing me from solving x,y,z,lambda,mu. Please tell me which of the two possible mistakes I made and also what steps do I take after correctly solving for x,y,z,lambda,mu. Thank you!

2. Mar 29, 2009

### xaos

without seeing your work, its hard to guess what you did wrong. there should be more than one case available. if you're getting nonsense in one case, you still have the other cases to check. once you get x,y,z, plug these into sqrt(f(x,y,z)) to get a distance. if they're ALL nonsense, then you did something wrong.

varify:

constraint1=0
constraint2=0,
these give 5 equations in 5 unknowns.

3. Mar 29, 2009

### jimbo71

why is do you say constraint 1 and 2 are =0. i thought constraint 1 was =5 because it is 2y+5z and constraint 2=0 because you could subtract the z^2 over and get zero???

4. Mar 29, 2009

### xaos

well, we want the constraints in the form H(x,y,z)=0 G(x,y,z)=0 so that we can take the three dimensional gradient [d/dx,d/dy,d/dz] for the multiplier equation. you still haven't given anything for us to work with.

5. Mar 30, 2009

### HallsofIvy

Staff Emeritus
In other words, "constraint 1" is H(x, y, z)= 2x+5z- 5= 0 and "constraint 2" is $G(x,y,z)= z^2- 4x^2- 4y^2= 0$.