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Finding the local extrema of $f(x, y) = sin(x) + sin(y) + sin(x+y)$ on the domain $(0, 2 pi) times (0, 2 pi)$

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?

enter image description here

2 Answers

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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