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
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.
1 answer
1 answer