# Local & global extrema

#### mathmari

##### Well-known member
MHB Site Helper
Hey !!

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?

#### Klaas van Aarsen

##### MHB Seeker
Staff member
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 $f$.
Hey mathmari !!

At this stage we don't know yet if $1$ is an actual maximum of the function.
We only know that $1$ is an upper bound of the function.

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.
We now have a point where we achieve the upper bound, which implies that $1$ is indeed a maximum of the function.
However, at this stage we don't know yet, if it is a strictly local maximum.

Last edited:

#### mathmari

##### Well-known member
MHB Site Helper
However, at this stage we don't know yet, if it is a strictly local maximum.
Do we have to show that in the neighbourhood of the point the function achieves only at this point the value $1$ ? Or how do we show that?

#### Klaas van Aarsen

##### MHB Seeker
Staff member
Do we have to show that in the neighbourhood of the point the function achieves only at this point the value $1$ ? Or how do we show that?
Yes. That is what it means that a local maximum is strict.

The function $\sin x$ has a strict local maximum at $x=\frac\pi 2$, doesn't it?

#### mathmari

##### Well-known member
MHB Site Helper
The function $\sin x$ has a strict local maximum at $x=\frac\pi 2$, doesn't it?
So we have that for each other $x$ it holds that $\sin(x)<1$ and so we get that $\sin^2 (x)\cdot \cos^2 (y)<1$ for $x\neq \frac{\pi}{2}$, right?

So can we say that this means that at $\left (\frac{\pi}{2}, 0\right )$ the function has a strictly local maximum?

#### Klaas van Aarsen

##### MHB Seeker
Staff member
So we have that for each other $x$ it holds that $\sin(x)<1$ and so we get that $\sin^2 (x)\cdot \cos^2 (y)<1$ for $x\neq \frac{\pi}{2}$, right?
Only in a small enough neighborhood of $(\frac\pi 2,0)$.

So can we say that this means that at $\left (\frac{\pi}{2}, 0\right )$ the function has a strictly local maximum?
We still have the case that we are in a small enough neighborhood with $x= \frac{\pi}{2}$.
Can we deduce in that case as well that the function value is strictly less than $1$?

#### mathmari

##### Well-known member
MHB Site Helper
Only in a small enough neighborhood of $(\frac\pi 2,0)$.

We still have the case that we are in a small enough neighborhood with $x= \frac{\pi}{2}$.
Can we deduce in that case as well that the function value is strictly less than $1$?
In a small neighborhood of $0$ the function $\cos^2(y)$ is only $1$ at $y=0$ and in a small neighborhood of $\frac{\pi}{2}$ the function $\sin^2(x)$ is only $1$ at $x=\frac{\pi}{2}$, that means that in a small neighborhood of $(\frac\pi 2,0)$ the function $f$ is less than $1$.

Ia that correct so far?

#### Klaas van Aarsen

##### MHB Seeker
Staff member
In a small neighborhood of $0$ the function $\cos^2(y)$ is only $1$ at $y=0$ and in a small neighborhood of $\frac{\pi}{2}$ the function $\sin^2(x)$ is only $1$ at $x=\frac{\pi}{2}$, that means that in a small neighborhood of $(\frac\pi 2,0)$ the function $f$ is less than $1$.

Is that correct so far?
Yep.

#### mathmari

##### Well-known member
MHB Site Helper
So we get that $f$ has at this point a strictly local maximum , right?

To show that this is also a global maximum do we say that it cannot be greater than $1$ ?

Also can we just mention the parts with the small neighborhoods at #7 or do we have to prove that?

#### Klaas van Aarsen

##### MHB Seeker
Staff member
So we get that $f$ has at this point a strictly local maximum , right?

To show that this is also a global maximum do we say that it cannot be greater than $1$ ?
Indeed.
Suppose it was not a global maximum. Then there would have to be another point where $f$ is greater than $1$.
But we already established that $1$ is an upper bound, leading to a contradiction.
So it is indeed also a global maximum.

Also can we just mention the parts with the small neighborhoods at #7 or do we have to prove that?
The functions $\sin$ and $\cos$ are well known to have a strict local maximum at $\frac\pi 2$ respectively $0$.
So they are less than $1$ in a sufficiently small neighborhood around those points.
Consequently their squares and product are also less than $1$ in a sufficiently small neighborhood.
I believe that is sufficient to serve as proof.

#### mathmari

##### Well-known member
MHB Site Helper
Indeed.
Suppose it was not a global maximum. Then there would have to be another point where $f$ is greater than $1$.
But we already established that $1$ is an upper bound, leading to a contradiction.
So it is indeed also a global maximum.

The functions $\sin$ and $\cos$ are well known to have a strict local maximum at $\frac\pi 2$ respectively $0$.
So they are less than $1$ in a sufficiently small neighborhood around those points.
Consequently their squares and product are also less than $1$ in a sufficiently small neighborhood.
I believe that is sufficient to serve as proof.
Ok! I see!!

At the second part,where we have to determine all points at which $f$ gets its global minimum, do we have to do the same as in the first part just for minimum instead of maximum ?

#### Klaas van Aarsen

##### MHB Seeker
Staff member
At the second part,where we have to determine all points at which $f$ gets its global minimum, do we have to do the same as in the first part just for minimum instead of maximum ?
Yep.

#### mathmari

##### Well-known member
MHB Site Helper
Ah the minimum in this case is $0$ because of the square, right?

#### Klaas van Aarsen

##### MHB Seeker
Staff member
Ah the minimum in this case is $0$ because of the square, right?
Indeed.

#### mathmari

##### Well-known member
MHB Site Helper
So we have that $\sin^2(x)\cdot \cos^2(y)\geq0$ and at all points with $x=n\pi$ and all points with $y=(2k-1)\frac{\pi}{2}$ the function achieves that minimum, which is a global minimum because there is no other lower value of the function.

Is that correct and complete? Or do we have to prove something more?

#### Klaas van Aarsen

##### MHB Seeker
Staff member
So we have that $\sin^2(x)\cdot \cos^2(y)\geq0$ and at all points with $x=n\pi$ and all points with $y=(2k-1)\frac{\pi}{2}$ the function achieves that minimum, which is a global minimum because there is no other lower value of the function.

Is that correct and complete? Or do we have to prove something more?
Correct and complete.

MHB Site Helper