# Extremum of f

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

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

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 $\displaystyle f(x,y) = sin x+ siny + sin (x+y)$, $\displaystyle 0\le x\le2\pi$, $\displaystyle 0\le y \le 2\pi$, find the extremum of $\displaystyle f$.

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

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

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

CB
• Mar 17th 2011, 04:18 AM
HallsofIvy
Not quite. x, y in $\displaystyle [0, 2\pi]$ and cos(x)= cos(y) implies either x= y or $\displaystyle x= 2\pi- y$. 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 $\displaystyle x=y$ or $\displaystyle x=2\pi-y$, we still can not imply anything for a critical value as we have $\displaystyle x$ and $\displaystyle y$ in between $\displaystyle 0$ and $\displaystyle 2\pi$.
Is it that we still have to create another equation to get the critical value?

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

Originally Posted by problem
By getting $\displaystyle x=y$ or $\displaystyle x=2\pi-y$, we still can not imply anything for a critical value as we have $\displaystyle x$ and $\displaystyle y$ in between $\displaystyle 0$ and $\displaystyle 2\pi$.
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 $\displaystyle x=y=\frac{5\pi}{3}$ is the minimum and $\displaystyle x=y=\frac{\pi}{3}$ 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 $\displaystyle x=y$ or $\displaystyle x=2\pi-y$, we still can not imply anything for a critical value as we have $\displaystyle x$ and $\displaystyle y$ in between $\displaystyle 0$ and $\displaystyle 2\pi$.
Is it that we still have to create another equation to get the critical value?

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

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

CB