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

[Jamin2112]

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 2^nd and 3^rd 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?

 

[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 $x,y,z \neq 0$

 

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

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

 

[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.

 

[lanedance]

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

 

[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...

 

[lanedance]

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

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