# Thread: [SOLVED] Lagrange multipliers to find max/min

1. ## [SOLVED] Lagrange multipliers to find max/min

Use Lagrange multipliers to find the max and min values of the function subject to the given constraints.

I get multiple answers for $\lambda$, which is my problem.

function:
$f(x,y,z)=x+2y$

subject to:
$x+y+z=1$ and $y^2+z^2=4$

$\nabla f <1,2,0> = \lambda<1,1,1>$
$\nabla f <1,2,0> = \lambda<0,2y,2z>$

Now I set them equal to solve for $\lambda$ using all of the possible combinations, here is where my answers get messed up. They are all different, and $1= \lambda 0$ makes no sense.
$1= \lambda$
$2= \lambda$
$0= \lambda$
$1= \lambda 0$
$2= \lambda 2y$
$0= \lambda 2z$

From here any solution i pick does not give me the solution of $1,\sqrt{2},-\sqrt{2}$ as a max. Also $1,-\sqrt{2},\sqrt{2}$ is the min.

Any help is appreciated as always.
Thanks!

2. Originally Posted by snaes
Use Lagrange multipliers to find the max and min values of the function subject to the given constraints.

I get multiple answers for $\lambda$, which is my problem.

function:
$f(x,y,z)=x+2y$

subject to:
$x+y+z=1$ and $y^2+z^2=4$

$\nabla f <1,2,0> = \lambda<1,1,1>$
$\nabla f <1,2,0> = \lambda<0,2y,2z>$

Now I set them equal to solve for $\lambda$ using all of the possible combinations, here is where my answers get messed up. They are all different, and $1= \lambda 0$ makes no sense.
$1= \lambda$
$2= \lambda$
$0= \lambda$
$1= \lambda 0$
$2= \lambda 2y$
$0= \lambda 2z$

From here any solution i pick does not give me the solution of $1,\sqrt{2},-\sqrt{2}$ as a max. Also $1,-\sqrt{2},\sqrt{2}$ is the min.

Any help is appreciated as always.
Thanks!
Lagrange multipliers with two constraints - you'll need two multipliers.

Form

$
F = f + \lambda g(x,y,z) + \mu h(x,y,z) = 0
$

Then $\nabla F = 0$, with $g(x,y,z) = 0$ and $h(x,y,z) = 0$ (the constaints).

However, looking at your problem, since $f$ has no $z$, then eliminate $z$ between your constraints and only use one multiplier.

3. Ahh thats how. Thanks alot!