Question

Use Lagrange multipliers to find the indicated extrema, assuming that x,y, and z are positive.

Maximize: xyz
Constraint: x+y+z=32, x-y+z=0

Answer 1

ah wait I typed in the wrong problem, let me retype it...

Answer 2

\[f(x,y,z)=x^2+y^2+z^2\]
\[constraint: g(x,y,z): x+y+z-9=0\]

Answer 3

after taking the partial derivative of each term for x,y,and z I get:

\[x=y=z=\lambda/2\]

Now here is where I'm stuck at, my book is telling me that when substituing into the constraint, I'm suppose to get 3 and I'm not getting that. For their x,y, and z values I think they get 1 but I'm not sure it just says "x=y=z".

Answer 4

Partial derivatives of f and g.
\[\Delta f(x,y,z)=<2x,2y,2z>\]
\[\ \lambda \Delta g (x,y,z)=<\lambda,\lambda,\lambda>\]

Find x and y values:
\[\ 2x =lambda\]
\[\ 2y =lambda\]
\[\ 2y =lambda\]
x,y, and z = \[\ lambda/2\]