# Homework Help: Partial Derivatives

1. May 13, 2009

### joemama69

1. The problem statement, all variables and given/known data
Find point closest to origin xy2z3 = 2

2. Relevant equations

3. The attempt at a solution

note, k = lagrange multiplier

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

k = 2xy-2z-3 = x-1z-3 = (2/3)z-1x-1y-2

y = $$\sqrt{2x^2}$$

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

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

Plug x and z into the original

($$\sqrt{(y^2)/2}$$)(y2)($$\sqrt{(3y^2)/2}$$)3 = 2

I tried to simplify that, first i squared both sides

(y2/2)(y4)((27y6)/8) = 4

(27/16)y12 = 4

y = (4(16/27))1/12

y = (64/27)1/12 = 1.074569932

anyone agree, diagree