Simplifying the Distance Formula for Points on the Surface xy^2z^3 = 2

[joemama69]

Homework Statement

Determine the points on the surface xy^2z^3 = 2 that are closest to the origin

Homework Equations

The Attempt at a Solution

is there an easier way to do this than to plug it into the distance formula and taking the derivative set to 0.

 

[HallsofIvy]

Use the square of the distance formula.

Also, and this may be what you are looking for, instead of using xy²z³= 2 to replace one variable with the other two, use "Lagrange multipliers". If we write f(x,y,z)= x²+ y²+ z², the square of the distance to the origin, and g(x,y,z)= xy²z³= 2, then max or min values of f, for points that satisfy g(x,y,z)= 2 must have \nabla f parallel to \nabla g- one must be a multiple of the other. Setting \nabla f= \lambda g and comparing the components, together with g(x,y,z)= 2, gives 4 equations to solve for x, y, z, and \lambda.

Tip: since you are not interested in the value of \lambda, and \lambda is simply multiplied by the functions of x, y, and z, often a best first thing to do is to divide one equation by another to eliminate \lambda.

 

[matt grime]

I case you need to look that up, Halls meant Lagrange multipliers, not Laplace.

 

[HallsofIvy]

Thanks, matt, I have editted that.

 

[joemama69]

ok i got a little stuck, here's my work, BTW k = lagrange multiplier

f(x,y,z) = x² + y² + z²
g(x,y,z) = xy²z³ = 2

grad f = 2xi + 2yj + 2zk
k grad g = k(y²z³i + 2xyz³j + 3z²xy²k

k = 2x/y²z³
k = 1/xz³
k = 2/3zxy²

and I am lost

 

[matt grime]

So you know that 2x/y^2z^3 = 1/xz^3 ie. 2x^2=y^2, and you can subs that in and see what happens.

 

[joemama69]

ok so i can sove for each of the variables

y = \sqrt{2x^2}

x = \sqrt{(y^2)/2}

z = \sqrt{(3y^2)/2}

how does that help