# Thread: [SOLVED] F(x,y) maximum problem

1. ## [SOLVED] F(x,y) maximum problem

When I try and solve for critical points i end up with x=y. Where the domain for the function is inclusivly between $\displaystyle 0$ and$\displaystyle 2\pi$.

If I have infinitely many critical points is there another method to solving this?

$\displaystyle f(x,y)=sin(x)+sin(y)+sin(x+y)$

$\displaystyle f_x(x,y)=cos(x) + cos(x+y)$
$\displaystyle f_y(x,y)=cos(y) + cos(x+y)$

Now $\displaystyle f_x=0=f_y$
So when i set them equal-->
$\displaystyle cos(x) + cos(x+y)=cos(y) + cos(x+y)$
$\displaystyle cos(x)=cos(y)$
$\displaystyle x=y$

Now I am unsure what to do next, normally i get some critical points, but instead i got infinately many critical points...

EDIT: Here is a link to the graph I think it'll work: http://www.wolframalpha.com/input/?i...sin%28x%2By%29
if not paste this in: graph z=sin(x)+sin(y)+sin(x+y) to get an idea of the graph.

EDIT:Going to bed now, will check again around 12hrs from now...for what it is worth

2. since cos(x) + cos(x+y) = 0

and since x = y

cos(x) + cos(2x) = 0

2cos^2(x) +cos(x) -1 = 0

(2cos(x) -1)(cos(x)+1) = 0

cos(x) = 1/2 cos(x) = -1

I'm assuming the domain is the square [0,2pi] x[0,2pi] ?

In which case x = pi/3, 5pi/3, pi

Again since x = y

(pi/3,pi/3) , (5pi/3,5pi/3) (pi,pi) are the critical points

3. Originally Posted by snaes
When I try and solve for critical points i end up with x=y. Where the domain for the function is inclusivly between $\displaystyle 0$ and$\displaystyle 2\pi$.

If I have infinitely many critical points is there another method to solving this?

$\displaystyle f(x,y)=sin(x)+sin(y)+sin(x+y)$

$\displaystyle f_x(x,y)=cos(x) + cos(x+y)$
$\displaystyle f_y(x,y)=cos(y) + cos(x+y)$

Now $\displaystyle f_x=0=f_y$
So when i set them equal-->
$\displaystyle cos(x) + cos(x+y)=cos(y) + cos(x+y)$
$\displaystyle cos(x)=cos(y)$
$\displaystyle x=y$

Now I am unsure what to do next, normally i get some critical points, but instead i got infinately many critical points...
From this is follows that if $\displaystyle f_x= f_y$ then cos(x)= cos(y). It does NOT follow that x= y. You might also have $\displaystyle y= 2\pi- x$. So you actually have twice as many solutions as you thought!
Those are the two diagonals of the square $\displaystyle [0,2\pi]\times[0,2\pi]$.

And, it does not follow that the two derivatives are equal to 0 for all y= x, only that they are equal two each other. Caculus26 showed that, using y= x in one of the equations, $\displaystyle f_x= 0$ or $\displaystyle f_y= 0$ you get $\displaystyle (\pi/3,\pi/3)$ , $\displaystyle (5\pi/3,5\pi/3)$, and $\displaystyle (\pi,\pi)$.

Do the same with $\displaystyle y= 2\pi- x$. Since $\displaystyle x+y= x+ (2\pi- x)= 2\pi$ you have $\displaystyle cos(x)+ cos(x+y)= cos(x)+ cos(2\pi)= cos(x)+ 1= 0$ and that is true only for [tex]x= \pi[/itex]. Then $\displaystyle y= 2\pi- \pi= \pi$ so that just gives the point $\displaystyle (\pi, \pi)$ that Caculus26 got before- it is the center point of the square and lies on both diagonals.

4. Ahh, I see. Thank you!