Extremum of f

**problem** · March 17th 2011, 12:40 AM

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?

**CaptainBlack** · March 17th 2011, 01:11 AM

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

**HallsofIvy** · March 17th 2011, 04:18 AM

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.)

**problem** · March 17th 2011, 09:08 AM

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.

**HallsofIvy** · March 17th 2011, 11:16 AM

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).

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.

**CaptainBlack** · March 17th 2011, 08:04 PM

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

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