# Maximum and minimum of a function of 2 variables (Lagrange multipliers)

• Jun 4th 2009, 05:13 PM
arbolis
Maximum and minimum of a function of 2 variables (Lagrange multipliers)
My question is : Can I solve the problem without passing by Lagrange multipliers? Because I have some difficulties by doing so. If not, then I don't want any tip and I'll work harder.
I must find the minimum and the maximum of the function $\displaystyle f(x,y)=x^2+y^2+\frac{2xy\sqrt{2}}{3}$ in the ellipse $\displaystyle x^2+2y^2 \leq 1$.
So it seems a problem that can be solved via Lagrange multipliers, but I didn't find a way to find $\displaystyle \lambda$, $\displaystyle x$ and $\displaystyle y$. (I even had to divide by $\displaystyle 0$ if $\displaystyle \lambda \neq 1$ but then I found another condition ( if $\displaystyle \lambda \neq 0$ then ...) and even with it I couldn't isolate $\displaystyle \lambda$ nor $\displaystyle x$ nor $\displaystyle y$).
• Jun 4th 2009, 06:35 PM
Spec
The requirement that the two gradients are parallel can also be expressed as

$\displaystyle \left|\begin{array}{cc}f'_x&f'_y\\g'_x&g'_y\end{ar ray} \right|=0$

This will produce exactly the same result as using the Lagrange multipliers method (once you've eliminated the multiplier).

Both these methods are a bit impractical here though. Since we're dealing with an ellipse, we can express the main function in terms of one variable.

$\displaystyle f(x,y)=f(\varphi)=\left\{x=\cos\varphi,\ y=\frac{\sin \varphi}{\sqrt{2}} \right\}=\cos^2 \varphi+\frac{\sin^2 \varphi}{2}+\frac{2\cos \varphi\sin \varphi}{3}$

You will of course also need to find the inner points where $\displaystyle f'_x=f'_y=0$
• Jun 4th 2009, 06:57 PM
arbolis
Quote:

Originally Posted by Spec
The requirement that the two gradients are parallel can also be expressed as

$\displaystyle \left|\begin{array}{cc}f'_x&f'_y\\g'_x&g'_y\end{ar ray} \right|=0$

This will produce exactly the same result as using the Lagrange multipliers method (once you've eliminated the multiplier).

Both these methods are a bit impractical here though. Since we're dealing with an ellipse, we can express the main function in terms of one variable.

$\displaystyle f(x,y)=f(\varphi)=\left\{x=\cos\varphi,\ y=\frac{\sin \varphi}{\sqrt{2}} \right\}=\cos^2 \varphi+\frac{\sin^2 \varphi}{2}+\frac{2\cos \varphi\sin \varphi}{3}$

You will of course also need to find the inner points where $\displaystyle f'_x=f'_y=0$

Thank you VERY much. I'll work on that! (I might ask for further help in a few days)