Absolute Extrema F(x,y) 0<x<π, 0<y<π

[joemabloe]

Homework Statement

F(x,y)= sin(x)sin(y)sin(x+y) over the square 0< x< pi and 0< y< pi(The values for x and y should be from 0 to pi INCLUSIVE)

Homework Equations

The Attempt at a Solution

partial derivative in terms of x = siny[cosxsin(x+y)+sinxcos(x+y)]
you get y=0, pi because siny =0, but I don't know how to solve for the other solutionspartial derivative in terms of y = sinx[cosysin(x+y)+ sinycos(x+y)]
and you get x=0, pi because sinx=0

and then I have the same problem again

 

[vela]

Use a trig identity to simplify the contents in the square brackets.

 

[joemabloe]

I used the trig identities sin(x+y)= sinxcosy+cosxsiny and cos (x+y)= cosxcosy - sinxsiny

and with that, I get the equation the previous equation multiplied out which gives me
=cosxsinxcosy +cos^2(x)siny +sinxcosxcosy-sin^2(x)siny
=2cosxsinxcosy +(cos^2(x)+sin^2(x))(siny-siny)
=2cosxsinxcosy +(1)(0)= 2cosxsinxcosy
and I did the same process for the y derivative and got it to equal
= 2cosxcosysiny

that x=0,pi/2, pi and y = 0, pi/2, pi

then when I go to find the values of F(x,y) at the nine possible extrema points, the value of F(x,y) is equal to 0.

Because they all came out to 0, I doubt that they are actually the extrema and probably did something wrong in my solving.

 

[joemabloe]

anyone?

 

[vela]

I meant you should go the other direction, e.g.,

cos x sin(x+y) + sin x cos(x+y) = sin(2x+y)

 

[joemabloe]

I meant you should go the other direction, e.g.,

cos x sin(x+y) + sin x cos(x+y) = sin(2x+y)

what formula did you use to find sin(2x+y)?

 

[vela]

cos a sin b + sin a cos b = sin(a+b) with a=x and b=x+y.