1. ## lagrange multiplier

I need to know if I'm solving for Lagrange multiplier correctly.

$\displaystyle f(x,y,z)= 8x-4y; \ x^2+10y^2+z^2=5$

so far I have:

$\displaystyle \begin{array}{l}f_x = 8, \ f_y = -4 \ f_z = 0 \\ \\ f_x = 2x, \ f_y = 20y \ f_z = 2z \\ \\ 8=\lambda 2x \rightarrow \lambda = \frac{4}{x} \\ \\ -4 = \lambda 2y \rightarrow -1 = \lambda 5y = \left( \frac{4}{x} \right) 5y \ \rightarrow y = -\frac{x}{20}\\ \\ 0 = \lambda z \rightarrow z= 0 \end{array}$

putting the above values in the constraint I get:

$\displaystyle x^2+10 \left( -\frac{x}{20} \right) ^2+(0)^2=5 = \frac{41}{40} x^2 \rightarrow x = \pm \sqrt{ \frac{200}{41}}$

then just substitute for y...

is this correct?

2. Originally Posted by lllll
I need to know if I'm solving for Lagrange multiplier correctly.

$\displaystyle f(x,y,z)= 8x-4y; \ x^2+10y^2+z^2=5$

so far I have:

$\displaystyle \begin{array}{l}f_x = 8, \ f_y = -4 \ f_z = 0 \\ \\ f_x = 2x, \ f_y = 20y \ f_z = 2z \\ \\ 8=\lambda 2x \rightarrow \lambda = \frac{4}{x} \\ \\ -4 = \lambda 2y \rightarrow -1 = \lambda 5y = \left( \frac{4}{x} \right) 5y \ \rightarrow y = -\frac{x}{20}\\ \\ 0 = \lambda z \rightarrow z= 0 \end{array}$

putting the above values in the constraint I get:

$\displaystyle x^2+10 \left( -\frac{x}{20} \right) ^2+(0)^2=5 = \frac{41}{40} x^2 \rightarrow x = \pm \sqrt{ \frac{200}{41}}$

then just substitute for y...

is this correct?
I did it a little differently and came up with

$\displaystyle x=\frac{10\sqrt{82}}{41}$

It is the same as yours

Nice work.!!