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- Apr 14, 2013

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Let $f:\mathbb{R}^2\rightarrow \mathbb{R}^2$, $f(x,y)=\sin^2(x)\cdot \cos^2(y)$.

- Show that $f$ has at $\left (\frac{\pi}{2}, 0\right )$ a strictly local maximum and that is also a global maximum.

- Determine all points at which $f$ gets its global minimum.

I have sone the following :

It holds that $|\sin (x)|\leq 1$ so $\sin^2(x)\leq 1$ and also that $|\cos (x)|\leq 1$ so $\cos^2(x)\leq 1$.

From that we get that $f(x,y)\leq 1$, that means that $1$ is a maximum of $f$.

At the point $\left (\frac{\pi}{2}, 0\right )$ the function gets the value $1$, that means that $f$ has at $\left (\frac{\pi}{2}, 0\right )$ a strictly local maximum.

Is that correct so far?