# Homework Help: Theres gotta be an easier way

1. May 5, 2009

### joemama69

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

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

2. Relevant equations

3. 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.

2. May 5, 2009

### 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 $\lambda$.

Last edited by a moderator: May 5, 2009
3. May 5, 2009

### matt grime

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

4. May 5, 2009

### HallsofIvy

Thanks, matt, I have editted that.

5. May 6, 2009

### joemama69

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

6. May 7, 2009

### 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.

7. May 8, 2009

### 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