Mathematics Asked on January 5, 2022
I am trying to find the relative extrema of
$$f(x, y) = sin(x) + sin(y) + sin(x+y), text{ where } (x, y) in (0, 2 pi) times (0, 2 pi)$$
Setting the partial derivatives equal to zero gives
$$frac{partial f}{partial x}(x, y) = cos(x) + cos(x+ y) = 0$$
$$frac{partial f}{partial y}(x, y) = cos(y) + cos(x+ y) = 0$$
Subtracting the equations gives $cos(x) = cos (y)$, and since $0 < x, y < 2 pi,$ we can see from the unit circle that this equation holds $iff y = 2pi – x iff x+y = 2pi$. Now using this information in the two equations above, we get
$$cos(x) + cos(2 pi) = 0 implies x = pi$$
$$cos(y) + cos(2 pi) = 0 implies y = pi$$
However, I graphed $f$, and this seems incorrect. In the pictures, we can see that there appears to be a local maximum around $(1, 1)$ and around $(5.5, 5.5)$. Could someone please tell me my mistake?
hint
The equation $$cos(x)=cos(y)$$ has a solution, other than the one you gave, which is simply $ x=y$.
after replacing this in the two equations, one gets $$cos(x)+cos(2x)=0$$ or $$2cos^2(x)+cos(x)-1=0$$ thus $$cos(x)=frac{-1pm sqrt{9}}{4}$$ $$=frac 12 text{ or } -1$$
Answered by hamam_Abdallah on January 5, 2022
We have that
$$cos(x) = cos (y) implies x=y :lor: x=2pi-y$$
it seems you only have considered the second condition.
Answered by user on January 5, 2022
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