# Thread: Max/Min using Lagrange

1. ## Max/Min using Lagrange

Can anyone help with this problem please? I understand the concept of the langrange multiplier but this ones stumping me.

F(x,y)=2x+y subject to the ellipse g(x,y)=X^2+25y^2=1

I have f(x,y)=2i-1j. g(x,y)=2xi+50yj

When I apply the lamb though I get 2=lambda 2x. Which means lambda=1/x which doesn't make sense.

Can anyone tell me where I am wrong with is one?

2. We have

$\displaystyle (2,1)=\lambda (2x,50y)$

$\displaystyle 2=\lambda \; 2x \; \; (1)$

$\displaystyle 1=\lambda \; 50y \; \; (2)$

From (1) get $\displaystyle \lambda$ and insert to (2)

$\displaystyle x=50y$

and we can insert it to ellipse equation.

3. Because, in Lagrange multilie problems, the specific value of $\displaystyle \lambda$ is not relevant, I find it often helpful to eliminate $\displaystyle \lambda$ by dividing equations. From the two equations $\displaystyle 2= 2\lambda x$ and $\displaystyle 1= 50\lambda y$ we get $\displaystyle \frac{2}{1}= \frac{2\lambda x}{50\lambda y}$
so that
$\displaystyle 50y= x$
as zzzoak says.