# Lagrange multiplier question

1. May 4, 2010

### Jamin2112

1. The problem statement, all variables and given/known data

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

2. Relevant equations

Lagrange's method for finding extrema

3. 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?

2. May 4, 2010

### lanedance

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 $x,y,z \neq 0$

Last edited: May 5, 2010
3. May 4, 2010

### lanedance

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 $f(x,y,z) = xy^2z^4 = 1$

$$\nabla g(x,y,z) = \nabla(x^2 + y^2 + z ^2) = 2(x,y,z)$$
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

4. May 5, 2010

### HallsofIvy

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.

5. May 5, 2010

### lanedance

i don't think equation 3 is correct - shouldn't the equation set be
$$\beta x= y^2z^4$$
$$\beta y = 2xyz^4$$
$$\beta z= 4xy^2z^3$$
$$1= xy^2z^4$$

6. May 5, 2010

### lanedance

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: May 5, 2010
7. May 5, 2010

### lanedance

1 div 2
$2x^2= y^2$

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