Multivariable Calc Absolute Extrema Problem

AI Thread Summary
To find the absolute extrema of the function F(x,y) = sin(x)sin(y)sin(x+y) over the square defined by 0 ≤ x ≤ π and 0 ≤ y ≤ π, critical points must be identified by calculating the partial derivatives with respect to x and y and setting them to zero. The partial derivative with respect to x is sin(y)[cos(x)sin(x+y) + sin(x)cos(x+y)], leading to critical points at y = 0 and y = π. Similarly, the partial derivative with respect to y is sin(x)[cos(y)sin(x+y) + sin(y)cos(x+y)], yielding critical points at x = 0 and x = π. The discussion highlights the challenge of solving for additional critical points beyond these boundaries. Further exploration is needed to determine the complete set of critical points and evaluate the function at these locations for absolute extrema.
joemabloe
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Homework Statement


F(x,y)= sin(x)sin(y)sin(x+y) over the square 0\underline{}<x\underline{}<pi and 0\underline{}<y\underline{}<pi


(The values for x and y should be from 0 to pi INCLUSIVE)

Homework Equations





The Attempt at a Solution



I know I need to do the partial derivatives in terms of x and y and set them equal to 0 to find the critical points, but I am having some trouble with that.
 
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Do you know \frac{\partial}{\partial x}\sin(x), \frac{\partial}{\partial x}\sin(y) and \frac{\partial}{\partial x}\sin(x+y)? You can use the product rule to put them together- it works the same with partial derivatives. Then do the same but with \frac{\partial}{\partial y}.
 
Tomsk said:
Do you know \frac{\partial}{\partial x}\sin(x), \frac{\partial}{\partial x}\sin(y) and \frac{\partial}{\partial x}\sin(x+y)? You can use the product rule to put them together- it works the same with partial derivatives. Then do the same but with \frac{\partial}{\partial y}.


I already found that:

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 solutions


partial 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 and am stuck
 
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