# Theres gotta be an easier way

## Homework Statement

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

## 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
Homework Helper
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 $\lambda$.

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

HallsofIvy
Homework Helper
Thanks, matt, I have editted that.

ok i got a little stuck, heres 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 im lost

matt grime
y = $$\sqrt{2x^2}$$
x = $$\sqrt{(y^2)/2}$$
z = $$\sqrt{(3y^2)/2}$$