Lagrange multiplier question

  • Thread starter Jamin2112
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  • #1
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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?
 

Answers and Replies

  • #2
lanedance
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looking at [itex] xy^2z^4=1 [/itex] can make the following observations that may help later
[itex] x =\frac{1}{y^2z^4} [/itex]
so x > 1 and [itex] x,y,z \neq 0[/itex]
 
Last edited:
  • #3
lanedance
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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 [itex] g(x,y,z) = x^2 + y^2 + z ^2 [/itex] subject to the constraint [itex] f(x,y,z) = xy^2z^4 = 1[/itex]

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

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

so i think your 3rd equation is missing a y^2
 
  • #4
HallsofIvy
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Since the value of the lagrange multiplier, [itex]\beta[/itex], is not necessary to the solution, I often find it simplest to start by dividing equations to eliminate the multiplier.

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

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

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

Check my arithmetic.
 
  • #5
lanedance
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i don't think equation 3 is correct - shouldn't the equation set be
[tex]\beta x= y^2z^4[/tex]
[tex]\beta y = 2xyz^4 [/tex]
[tex]\beta z= 4xy^2z^3 [/tex]
[tex]1= xy^2z^4 [/tex]
 
  • #6
lanedance
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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...
 
Last edited:
  • #7
lanedance
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1 div 2
[itex]2x^2= y^2[/itex]

1 div 3
[tex]\frac{x}{z}= \frac{y^2z^4}{4xy^2z^3 }= \frac{z}{4x}[/tex]
[tex]4x^2 = z^2[/tex]
 

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