# Extremum of f

• Mar 17th 2011, 12:40 AM
problem
Extremum of f
If $f(x,y) = sin x+ siny + sin (x+y)$, $0\le x\le2\pi$, $0\le y \le 2\pi$, find the extremum of $f$.

I find $f_x$and $f_y$ and equate them to zero. Then from these two equations, I get $cos x= cos y$.

Can anyone help me to proceed from here to get the extremum?
• Mar 17th 2011, 01:11 AM
CaptainBlack
Quote:

Originally Posted by problem
If $f(x,y) = sin x+ siny + sin (x+y)$, $0\le x\le2\pi$, $0\le y \le 2\pi$, find the extremum of $f$.

I find $f_x$and $f_y$ and equate them to zero. Then from these two equations, I get $cos x= cos y$.

Can anyone help me to proceed from here to get the extremum?

With both $x$ and $y$ in $[0,2\pi]$, $\cos(x)=\cos(y)$ implies that $x=y$, which will imply something else ...

CB
• Mar 17th 2011, 04:18 AM
HallsofIvy
Not quite. x, y in $[0, 2\pi]$ and cos(x)= cos(y) implies either x= y or $x= 2\pi- y$. Put those into the two equations cos(x)+ cos(x+y)= 0 and cos(y)+ cos(x+y)= 0.

(Although I don't believe that second case gives a new solution.)
• Mar 17th 2011, 09:08 AM
problem
By getting $x=y$ or $x=2\pi-y$, we still can not imply anything for a critical value as we have $x$ and $y$ in between $0$ and $2\pi$.
Is it that we still have to create another equation to get the critical value?

By the way, the answer given is $x=y=\frac{5\pi}{3}$ is the minimum and $x=y=\frac{\pi}{3}$ is the maximum value.
• Mar 17th 2011, 11:16 AM
HallsofIvy
Quote:

Originally Posted by problem
By getting $x=y$ or $x=2\pi-y$, we still can not imply anything for a critical value as we have $x$ and $y$ in between $0$ and $2\pi$.
Is it that we still have to create another equation to get the critical value?

If x= y, then sin(x)+ sin(y)+ sin(x+ y)= sin(x)+ sin(x)+ sin(x+x)= 2sin(x)+ sin(2x).

Quote:

By the way, the answer given is $x=y=\frac{5\pi}{3}$ is the minimum and $x=y=\frac{\pi}{3}$ is the maximum value.
I suggest you check that again. The values of x and y are NOT values of the function and cannot be maximum and minimum values for the function.
• Mar 17th 2011, 08:04 PM
CaptainBlack
Quote:

Originally Posted by problem
By getting $x=y$ or $x=2\pi-y$, we still can not imply anything for a critical value as we have $x$ and $y$ in between $0$ and $2\pi$.
Is it that we still have to create another equation to get the critical value?

By the way, the answer given is $x=y=\frac{5\pi}{3}$ is the minimum and $x=y=\frac{\pi}{3}$ is the maximum value.

What is $f_x(x,y)$ (and/or $f_y(x,y)$), now that you know that $x=y$ or $x=2\pi-y$ what do the equations $f_x(x,y)=0$ and $f_y(x,y)=0$ become in each of these cases?

CB