# Thread: Lagrange Multipiers

1. ## Lagrange Multipiers

The teacher gave us some problems on the board for us to jot down. He said it would prepare us with the final.

Problem:

Optimize the function $\displaystyle f(x,y,z)=8x-4z$ subject to the constraint $\displaystyle x^2 + 10y^2 + z^2 = 5$

I am not 100% sure what optimize means and if Lagrange multiplier applies but this is what i have so far.

$\displaystyle f_{x}= 8$
$\displaystyle f_{y}= 0$
$\displaystyle f_{z}= -4$

$\displaystyle g_{x}=2x$
$\displaystyle g_{y}= 20y$
$\displaystyle g_{z}= 2z$

Combining we have:

(1) $\displaystyle 8 = 2x\lambda$
(2) $\displaystyle 0 = 20y\lambda$
(3) $\displaystyle -4 = 2z\lambda$

Solving (1) for x:

(4) $\displaystyle x = \frac{4}{\lambda}$
(5) $\displaystyle y = \lambda$
(6) $\displaystyle z = \frac{-2}{\lambda}$

Now, add in (4),(5), and (6) for the constraint

$\displaystyle \left(\frac{4}{\lambda}\right)^2 + 10\lambda^2+\left(\frac{-2}{\lambda}\right)^2 = 5$

then

$\displaystyle \frac{16}{\lambda^2} + 10\lambda^2+\frac{4}{\lambda^2} = 5$

So, at this point i am stuck. I don't know how to finish.
I don't even know if i was suppose to use Lagrange multiplier or if its correct up to that point?

If someone could let me know if i did the problem correctly and help me finish?
Thank you so much.

2. ## Re: Lagrange Multipiers

Hey icelated.

Hint: Multiply both sides by lambda^2 and you should get an equation after re-arranging: a*lambda^4 + b*lambda^2 + c = 0.

Make a substitution u = lambda^2 and solve the quadratic a*u^2 + b*u + c = 0 and then solve for lambda.

3. ## Re: Lagrange Multipiers

Originally Posted by icelated
The teacher gave us some problems on the board for us to jot down. He said it would prepare us with the final.

Problem:

Optimize the function $\displaystyle f(x,y,z)=8x-4z$ subject to the constraint $\displaystyle x^2 + 10y^2 + z^2 = 5$

I am not 100% sure what optimize means and if Lagrange multiplier applies but this is what i have so far.

$\displaystyle f_{x}= 8$
$\displaystyle f_{y}= 0$
$\displaystyle f_{z}= -4$

$\displaystyle g_{x}=2x$
$\displaystyle g_{y}= 20y$
$\displaystyle g_{z}= 2z$

Combining we have:

(1) $\displaystyle 8 = 2x\lambda$
(2) $\displaystyle 0 = 20y\lambda$
(3) $\displaystyle -4 = 2z\lambda$

Solving (1) for x:

(4) $\displaystyle x = \frac{4}{\lambda}$
(5) $\displaystyle y = \lambda$
No, $\displaystyle 0= 20y\lambda$ gives $\displaystyle y= 0$ NOT "$\displaystyle y= \lambda$"

(6) $\displaystyle z = \frac{-2}{\lambda}$

Now, add in (4),(5), and (6) for the constraint

$\displaystyle \left(\frac{4}{\lambda}\right)^2 + 10\lambda^2+\left(\frac{-2}{\lambda}\right)^2 = 5$

then

$\displaystyle \frac{16}{\lambda^2} + 10\lambda^2+\frac{4}{\lambda^2} = 5$

So, at this point i am stuck. I don't know how to finish.
I don't even know if i was suppose to use Lagrange multiplier or if its correct up to that point?

If someone could let me know if i did the problem correctly and help me finish?
Thank you so much.