What are the closest points to the origin on the level surface xy2z4=1?

Jamin2112
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Homework Statement



Find the points on the level surface xy2z4=1 that are closest to the origin.

Homework Equations



Lagrange's method for finding extrema

The Attempt at a Solution



If I have a level surface F(x,y,z)=c, it's points closest to the origin will be the ones in which the gradient vector points to the origin. A generic vector pointing to/from the origin is G=<x,y,z>, so F must be a scalar multiple of G.

I come up with a system of equations

ßx=y2z4
ßy=2xyz4
ßz=4x2z3
xy2z4=1.

I can first simplify a little bit.

ßx=y2z4
ß=2xz4
ß=4x2z2


I can set the 2nd and 3rd equations equal.

2xz4=4x2z2
----> x= z2/2

I can plug that x into the first 2 equations.

(y2z4)/[z2/2]=2[z2/2]z4
----> y = +/- √(z4/2)

Plugging those into the constraint xy2z4=1

----> z=4(1/10).

Am I right? What is the most straight-forward way of solving such a problem?
 
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looking at xy^2z^4=1 can make the following observations that may help later
x =\frac{1}{y^2z^4}
so x > 1 and x,y,z \neq 0
 
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then i think you're doing the correct thing with the lagrange multipliers

the gradient point to the origin comes from minimising g(x,y,z) = x^2 + y^2 + z ^2 subject to the constraint f(x,y,z) = xy^2z^4 = 1

then taking the gradient
\nabla g(x,y,z) = \nabla(x^2 + y^2 + z ^2) = 2(x,y,z)
note is teh direction from the origin to the point as you say

\nablaf(x,y,z) = \nabla(xy^2z^4) = (y^2z^4 ,2xyz^4 ,4xy^2z^3 )

so i think your 3rd equation is missing a y^2
 
Since the value of the lagrange multiplier, \beta, is not necessary to the solution, I often find it simplest to start by dividing equations to eliminate the multiplier.

Dividing \beta x= y^2z^4 by \beta y= 2xyz^4 gives
\frac{x}{y}= \frac{y^2z^4}{2xyz^4}= \frac{y}{2x}
so that 2x^2= y^2 and y= \sqrt{2}x.

Dividing \beta y= 2xyz^4 by \beta z= 4x^2z^3 gives
\frac{y}{z}= \frac{2xyz^4}{4x^2z^3}= \frac{yz}{2x}
so that 2x= z^2 and x= z^2/2 and then y= \sqrt{2}{x}= z^2/\sqrt{z}.

Putting those into the equation of the surface, xy^2z^4= 1 gives (z^2/2)(z^4/2)(z^4)= z^10/4= 1 so that z= (1/4)^{1/10}.

Check my arithmetic.
 
i don't think equation 3 is correct - shouldn't the equation set be
\beta x= y^2z^4
\beta y = 2xyz^4
\beta z= 4xy^2z^3
1= xy^2z^4
 
also as the surface is symmetric about the xy and xz axes, i'd think there should be at least 2 points... if not 4...
 
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1 div 2
2x^2= y^2

1 div 3
\frac{x}{z}= \frac{y^2z^4}{4xy^2z^3 }= \frac{z}{4x}
4x^2 = z^2
 
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