Question

Calculus II please help

[see attachment]

Answer 1

i'm up to where 2y+2z+2hy +2hz=0 , 2y+2z+2hx+2hz= 0, 2x+2y+2hx+2hy=0

Answer 2

Is there even enough information given here?

Answer 3

It's for Lagrange Multipliers

Answer 4

\[h= \frac{ -2y-2z }{ 2y+2z } h= \frac{ -2y-2z }{ 2x+2z } h=\frac{ -2x-2y }{ 2x+2y}\]

Answer 5

\[\frac{ -2y-2z }{ 2y+2z } = \frac{ -2y-2z }{ 2z+2z } \]

Answer 6

looking for what y=

Answer 7

then looking for what z = from \[\frac{ -2y-2z }{ 2y+2z } = \frac{ -2x-2y }{ 2x+27 }\]

Answer 8

not 27 .... 2y

Answer 9

f(x,y,z)=2xy+2xz+2yz
g(x,y,z)=xyz=216

\(\frac{d f}{dx}+h\frac{dg}{dx}=0\)
\(\frac{d f}{dy}+h\frac{dg}{dy}=0\)
\(\frac{d f}{dz}+h\frac{dg}{dz}=0\)
\(g(x,y,z)=216\)
substitute values for the derivatives and solve this system

Answer 10

got it?

Answer 11

yes ty... just did the constraint equation wrong.

Answer 12

:)

Answer 13

so for the first one would be be 2hy+2hz?