What are the extrema and saddle points of #f(x, y) = 6 sin x sin y# on the interval #x,y in[-pi,pi]# ?

1 Answer
Nov 5, 2015

Answer:

#x=pi/2# and #y=pi#

#x=pi/2# and #y=-pi#

#x=-pi/2# and #y=pi#

#x=-pi/2# and #y=-pi#

#x=pi# and #y=pi/2#

#x=pi# and #y=-pi/2#

#x=-pi# and #y=pi/2#

#x=-pi# and #y=-pi/2#

Explanation:

To find the critical points of a #2#-variable function, you need to compute the gradient, which is a vector cointaining the derivatives with respect to each variable:

#(d/dx f(x,y), d/dy f(x,y))#

So, we have

#d/dx f(x,y) = 6cos(x)sin(y)#, and similarly

#d/dy f(x,y) = 6sin(x)cos(y)#.

In order to find the critical points, the gradient must be the zero vector #(0,0)#, which means solving the system

#{ (6cos(x)sin(y)=0), (6sin(x)cos(y)=0 ) :}#

which of course we can simplify getting rid of the #6#'s:

#{ (cos(x)sin(y)=0), (sin(x)cos(y)=0 ) :}#

This system is solved choosing for #x# a point which annihilates the cosine, and for #y# a point which annihilate the sine, and vice versa, so

#x=pm pi/2#, and #y=pm pi#, and vice versa #x=\pm pi# and #y=\pm pi/2#, obtaining #8# points in total.