- #1

Archimedess

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Hi everyone, I'm struggling with this problem:

Let ##f(x,y) =

\begin{cases}

(x-y)\ln(y-x) & \text{if } y>x \\

0 & \text{if } y\leq x

\end{cases}## and let ##C=\{(x,y)\in \mathbb{R}^2|x^2+y^2=1\}##

Then proof that ##max_Cf=1/e## and ##min_Cf=-(\ln2)/\sqrt2##

My solution:

I used Lagrange multiplier, so I have ##(x-y)\ln(y-x)-\lambda(x^2+y^2-1)=0##, then i have:

##

\begin{cases}

\ln(y-x)+1-2\lambda x=0\\

-\ln(y-x)-1-2\lambda y=0 \\

x^2+y^2-1=0

\end{cases}##

By solving the system (I also checked it with compiler) I only get the local minima which is ##-(ln2)/\sqrt2##, but by solving the system of equation I don't find the local maxima, I don't understand why. Any suggestions?

Let ##f(x,y) =

\begin{cases}

(x-y)\ln(y-x) & \text{if } y>x \\

0 & \text{if } y\leq x

\end{cases}## and let ##C=\{(x,y)\in \mathbb{R}^2|x^2+y^2=1\}##

Then proof that ##max_Cf=1/e## and ##min_Cf=-(\ln2)/\sqrt2##

My solution:

I used Lagrange multiplier, so I have ##(x-y)\ln(y-x)-\lambda(x^2+y^2-1)=0##, then i have:

##

\begin{cases}

\ln(y-x)+1-2\lambda x=0\\

-\ln(y-x)-1-2\lambda y=0 \\

x^2+y^2-1=0

\end{cases}##

By solving the system (I also checked it with compiler) I only get the local minima which is ##-(ln2)/\sqrt2##, but by solving the system of equation I don't find the local maxima, I don't understand why. Any suggestions?

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