Wednesday

April 16, 2014

April 16, 2014

Posted by **Help Please** on Friday, March 9, 2012 at 9:30pm.

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.

- calculus -
**Anonymous**, Monday, April 2, 2012 at 2:39pmYour 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

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