Partial Derivatives

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[SUP]-1[SUP]x^-1y^-2

y = [tex]\sqrt{2x^2}[/tex]

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

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

Plug x and z into the original

([tex]\sqrt{(y^2)/2}[/tex])(y²)([tex]\sqrt{(3y^2)/2}[/tex])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