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Let $f(x,y)=\dfrac{1}{2}x^2+cos(y)$

I want to find its critical points:

• $\dfrac{\delta f}{\delta x} = x$

• $\dfrac{\delta f}{\delta y} =-sin(y)$

Now I have to solve the following system: $\left\{ \begin{array}{ll} x = 0 \\ -sin(y) = 0 \end{array} \right.$

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2 Answers

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You did the partial derivatives wrong. It's actually $\begin{cases}f_x=x=0\\f_y=-sin(y)=0\end{cases}$

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First you need to calculate the partial derivatives correctly:

$$ f_x = x, $$

and

$$ f_y = -\sin y. $$

Setting these equal to 0, we get that the $x$ coordinate must be $0$, and the $y$ coordinate is $n\pi$, where $n\in\mathbb{Z}$ is any integer. So there are infinitely many critical points, all of the form $(0, n\pi)$.

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