find max, min and saddle points of the give function f(x,y)=sin(x)+sin(y)+sin(x+y)

0<=x<=pi/4
0<=y<=pi/4

i have that
dz/dx=cos(x)+cos(x+y)
dz/dy=cos(y)+cos(x+y)

and i set them equal to zero but im kinda confused on how to really solve that. i mean i got an answer but that was just from plugging things in and seeing if they worked but i want to know if i can actually solve this without doing that. do i use a trig identity? cause i did that and it came out pretty ugly.

Your going to need fxx, fyy, and fxy too. For your critical values of the dz's, you will need to change the cos(x+y) to their proper trig identity because you cant solve for the CV with what you have

d=fxx*fyy-fxy^2
if d=0, then its a saddle point
if d>0 and so is fxx, then it is a min value
if D>0, but fxx is <0, then its a max value

To find the critical points of the given function, you correctly calculated the partial derivatives with respect to x and y:

dz/dx = cos(x) + cos(x+y)
dz/dy = cos(y) + cos(x+y)

To find the critical points, you need to set these derivatives equal to zero and solve for x and y.

Setting dz/dx = 0:
cos(x) + cos(x+y) = 0

Setting dz/dy = 0:
cos(y) + cos(x+y) = 0

To simplify this, you can use the trigonometric identity relating cosine values:

cos(a) + cos(b) = 2*cos((a+b)/2)*cos((a-b)/2)

Using this identity, you can rewrite the first equation as:

2*cos((2x+y)/2)*cos((2x-y)/2) = 0

Since cos((2x+y)/2) cannot equal zero, the first term must be zero for this equation to hold. Therefore, you have:

cos((2x-y)/2) = 0

Similarly, using the identity on the second equation, you get:

2*cos((x+2y)/2)*cos((x-2y)/2) = 0

Again, cos((x-2y)/2) cannot equal zero, so you have:

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

Now, you can solve these equations separately.

For the first equation, cos((2x-y)/2) = 0, you know that the cosine function is zero when the argument is (2n + 1) * pi/2 for any integer n. Therefore, you can set:

(2x-y)/2 = (2n + 1) * pi/2

Simplifying this equation gives you:

2x - y = (2n + 1) * pi

For the second equation, cos((x+2y)/2) = 0, the argument must satisfy:

(x+2y)/2 = (2m + 1) * pi/2

Simplifying this equation yields:

x + 2y = (2m + 1) * pi

Now, you have to consider the given domain for x and y:

0 <= x <= pi/4
0 <= y <= pi/4

For both equations, you can substitute the minimum and maximum values for both x and y into the equations and solve for n and m to find the specific values of x and y that satisfy the conditions.

This process would involve substituting 0 and pi/4 for x and y, and then solving for n and m. These values of n and m will determine the critical points of the function within the given domain.

Once you have determined the critical points, you can further classify them as maxima, minima, or saddle points by using the second partial derivative test or evaluating the function at these points.