Max and min value, multi variable (open sets)

[Petrus]

Calculate max and min value of the function \(\displaystyle f(x,y)=x^2+y^2-2x-4y+8\)
in the range defined by the \(\displaystyle x^2+y^2≤9\)
Progress:
\(\displaystyle f_x(x,y)=2x-2\)
\(\displaystyle f_y(x.y)=2y-4\)
So I get \(\displaystyle x=1\) and \(\displaystyle y=2\) We got one end point that I don't know what to do with \(\displaystyle x^2+y^2≤9\)
If I got this right it should be a elips that x can max be 3,-3 and y 3,-3

 

[MarkFL]

You are correct to first find any critical points in the interior of the region, which as you found is (1,2).

In order to examine $f$ on the boundary of the region, we may represent the circle $x^2+y^2=3^2$ by means of the parametric equations $x=3\cos(t),\,y=3\sin(t),\,0\le t\le 2\pi$. Thus, on the boundary we can write $f$ as a function of a single variable $t$:

\(\displaystyle f(\cos(t),\sin(t))=3^2\cos^2(t)+3^2\sin^2(t)-2\cos(t)-4\sin(t)+8=17-2\cos(t)-4\sin(t)\)

Can you proceed now, differentiating $f(t)$ and equating to zero to find the critical point(s)? And don't neglect the endpoints...

 

[Petrus]

You are correct to first find any critical points in the interior of the region, which as you found is (1,2).

In order to examine $f$ on the boundary of the region, we may represent the circle $x^2+y^2=3^2$ by means of the parametric equations $x=3\cos(t),\,y=3\sin(t),\,0\le t\le 2\pi$. Thus, on the boundary we can write $f$ as a function of a single variable $t$:

\(\displaystyle f(\cos(t),\sin(t))=3^2\cos^2(t)+3^2\sin^2(t)-2\cos(t)-4\sin(t)+8=17-2\cos(t)-4\sin(t)\)

Can you proceed now, differentiating $f(t)$ and equating to zero to find the critical point(s)? And don't neglect the endpoints...

I got to this \(\displaystyle 2(\sin(t)-2\cos(t))=0\) how do I solve that
Can't I also solve this two way? The way I try to use early?

 

[MarkFL]

You cannot use the extrema of the variables, you want the extrema of the function on the boundary.

Now, to solve the equation you found, first divide through by 2 and rearrange to get:

\(\displaystyle \sin(t)=2\cos(t)\)

What do you think you should do next?

Also, you may find the critical points through use of Lagrange multipliers rather than parametrization, using the boundary of the region as the constraint. I would highly recommend doing both just for practice and insight.

 

[Petrus]

You cannot use the extrema of the variables, you want the extrema of the function on the boundary.

Now, to solve the equation you found, first divide through by 2 and rearrange to get:

\(\displaystyle \sin(t)=2\cos(t)\)

What do you think you should do next?

Also, you may find the critical points through use of Lagrange multipliers rather than parametrization, using the boundary of the region as the constraint. I would highly recommend doing both just for practice and insight.

I would love to solve this as lagrange multipliers as well:)
rewrite sin(t) as 2cos(t) in our orginal function before we derivate?

 

[MarkFL]

We'll deal with the parametrization method first.

You want to find the value(s) of $t$ which satisfy the equation that resulted from equating the derivative to zero. Now, we could observe that $\sin(t)=2\cos(t)$ implies $y=2x$ (which is what you will find with Lagrange) and plug this into the equation of the circle, but let's solve for $t$ first. What happens if we divide through by $\cos(t)$?

 

[Petrus]

We'll deal with the parametrization method first.

You want to find the value(s) of $t$ which satisfy the equation that resulted from equating the derivative to zero. Now, we could observe that $\sin(t)=2\cos(t)$ implies $y=2x$ (which is what you will find with Lagrange) and plug this into the equation of the circle, but let's solve for $t$ first. What happens if we divide through by $\cos(t)$?

\(\displaystyle \tan(t)=2\)

 

[MarkFL]

Correct. We should be aware that we needn't be worried about division by zero as there is no real value of $t$ such that $\sin(t)=\cos(t)=0$.

Now, in which quadrants will we find solutions for $\tan(t)=2$?

 

[Petrus]

Correct. We should be aware that we needn't be worried about division by zero as there is no real value of $t$ such that $\sin(t)=\cos(t)=0$.

Now, in which quadrants will we find solutions for $\tan(t)=2$?

first quadrants and I did forget about that when I divided

 

[MarkFL]

Yes, there is a first quadrant solution, but there is another...think of the periodicity of the tangent function.

 

[Petrus]

and second, so first and second :)?

 

[MarkFL]

No, the period of the tangent function is $\pi$ units, i.e., $\tan(\theta+k\pi)=\tan(\theta)$. To stay within the given domain of $t$, we let $k=1$, so we will add $\pi$ radians to the first quadrant solution to find we are in which quadrant?

 

[Petrus]

No, the period of the tangent function is $\pi$ units, i.e., $\tan(\theta+k\pi)=\tan(\theta)$. To stay within the given domain of $t$, we let $k=1$, so we will add $\pi$ radians to the first quadrant solution to find we are in which quadrant?

pi is half way (180 degree) and if se are in first quadant and add pi Then we get to third quadant

 

[MarkFL]

Correct! (Yes)

This means there are first and third quadrant solutions. Can you state the first quadrant solution and how the third quadrant solution relates to this in terms of the $x$ and $y$ coordinates?

 

[Petrus]

Correct! (Yes)

This means there are first and third quadrant solutions. Can you state the first quadrant solution and how the third quadrant solution relates to this in terms of the $x$ and $y$ coordinates?

What you mean:O? There is a normal tangent function and Then i draw à line in y=2 and it Will hit tan function in y=2 when x is something? I don't know what third quadant got with this problem but i guess y=-2? Hmm is this correct?

 

[MarkFL]

You have the right idea, but if $\tan(t)=2$, then you essentially want to find the points of intersection of the circle and the line $y=2x$.

There are many ways to look at this though, so first let's just work with the first quadrant solution. Can you state this solution?

 

[Petrus]

T=Arctan(2)

 

[MarkFL]

Good, yes $t=\tan^{-1}(2)$. Now, when you plug this value of $t$ into the parametric equations for $x$ and $y$, what do you find?

 

[Petrus]

Good, yes $t=\tan^{-1}(2)$. Now, when you plug this value of $t$ into the parametric equations for $x$ and $y$, what do you find?

\(\displaystyle x=3\cos(\arctan(2))\)
\(\displaystyle y=3\sin(\arctan(2))\)
I am correct?

 

[MarkFL]

Yes, now can you evaluate them, using a right triangle if necessary? Recall you want the side of the triangle opposite the angle $t$ to be 2 and the side adjacent to be 1, then use the Pythagorean theorem to compute the hypotenuse, and finally use the definitions of sine and cosine to complete the evaluations.

While there are simpler ways to go about this, which we will explore next, I think this is good practice. (Wink)

 

[Petrus]

Yes, now can you evaluate them, using a right triangle if necessary? Recall you want the side of the triangle opposite the angle $t$ to be 2 and the side adjacent to be 1, then use the Pythagorean theorem to compute the hypotenuse, and finally use the definitions of sine and cosine to complete the evaluations.

While there are simpler ways to go about this, which we will explore next, I think this is good practice. (Wink)

I get hypotenusan to \(\displaystyle \sqrt{5}\) is that correct?

 

[MarkFL]

Yes, as \(\displaystyle \sqrt{2^2+1^1}=\sqrt{5}\). Good work!

So, what do you find as the critical coordinates in the first quadrant?

 

[Petrus]

Yes, as \(\displaystyle \sqrt{2^2+1^1}=\sqrt{5}\). Good work!

So, what do you find as the critical coordinates in the first quadrant?

\(\displaystyle x=\frac{3}{\sqrt{5}}\)
\(\displaystyle y=\frac{6}{\sqrt{5}}\)

 

[MarkFL]

Excellent! Now, what do you find when you let $t=\pi+\tan^{-1}(2)$ to get the 3rd quadrant critical point? And remember to consider the endpoints of the domain for $t$.

Hint: what are $\sin(\theta+\pi)$ and $\cos(\theta+\pi)$ using the angle-sum identities?

Note: I must once again run off for about 4.5 hours, so if anyone else wants to jump in, that's fine. I will check back here first thing when I return, and we can look at some other methods for obtaining these critical points so you can see how they relate. (Nod)

 

[Petrus]

Excellent! Now, what do you find when you let $t=\pi+\tan^{-1}(2)$ to get the 3rd quadrant critical point? And remember to consider the endpoints of the domain for $t$.

Hint: what are $\sin(\theta+\pi)$ and $\cos(\theta+\pi)$ using the angle-sum identities?

Note: I must once again run off for about 4.5 hours, so if anyone else wants to jump in, that's fine. I will check back here first thing when I return, and we can look at some other methods for obtaining these critical points so you can see how they relate. (Nod)

Well in third quadrant x and y is negative so \(\displaystyle x=-\frac{3}{\sqrt{5}}\)
\(\displaystyle y=-\frac{6}{\sqrt{5}}\)? I am correct?

 

[MarkFL]

Yes, we can see this from symmetry or using the angle-sum identities.

So, you have two critical points thus far on the boundary, and one more from the end-points of the domain of $t$. Can you state this point, and why there is only one point, when there are two endpoints of $t$'s domain?

 

[Petrus]

Yes, we can see this from symmetry or using the angle-sum identities.

So, you have two critical points thus far on the boundary, and one more from the end-points of the domain of $t$. Can you state this point, and why there is only one point, when there are two endpoints of $t$'s domain?

Unfortently I don't know. Do you also mean there is another point?

 

[MarkFL]

Yes, we need to check the endpoints, or boundaries for $t$, which we gave as:

\(\displaystyle 0\le t\le 2\pi\).

 

[Petrus]

Yes, we need to check the endpoints, or boundaries for $t$, which we gave as:

\(\displaystyle 0\le t\le 2\pi\).

That's logic, So I have to check now \(\displaystyle f(\frac{3}{\sqrt{5}},\frac{6}{\sqrt{5}})\), \(\displaystyle f(-\frac{3}{\sqrt{5}},-\frac{6}{\sqrt{5}})\), \(\displaystyle f(1,0)\) and \(\displaystyle f(1,\sin(2\pi))\)

 

[MarkFL]

Your first two critical points are correct, but there is only a third because the parametric equations have the same value for the two endpoints because:

\(\displaystyle \sin(0)=\sin(2\pi)=0\)

\(\displaystyle \cos(0)=\cos(2\pi)=1\)

Hence, the third critical point on the boundary of the circle is

\(\displaystyle (x(0),y(0))=(3\cos(0),3\sin(0))=(3,0)\).

Now you have 4 values of the function to check, one in the interior of the circle, and 3 on the boundary.

When I return, we can look at some other methods for obtaining the critical values on the boundary of the circle. (Nod)

 

[Petrus]

Your first two critical points are correct, but there is only a third because the parametric equations have the same value for the two endpoints because:

\(\displaystyle \sin(0)=\sin(2\pi)=0\)

\(\displaystyle \cos(0)=\cos(2\pi)=1\)

Hence, the third critical point on the boundary of the circle is

\(\displaystyle (x(0),y(0))=(3\cos(0),3\sin(0))=(3,0)\).

Now you have 4 values of the function to check, one in the interior of the circle, and 3 on the boundary.

When I return, we can look at some other methods for obtaining the critical values on the boundary of the circle. (Nod)

My bad here I forgot i had 3sin(t),3cos(t) :P i Was using sin(t),cos(t) this problem Was à Hood one:)

 

[Petrus]

Early on I was also trying to subsitate that \(\displaystyle y=\sqrt{9-x^2}\) and then simplified and I did call my new function g(x)...so I did derivate and look for crit point and the put it on g(x) to get my y. May I ask why don't that method work?

 

[MarkFL]

Early on I was also trying to subsitate that \(\displaystyle y=\sqrt{9-x^2}\) and then simplified and I did call my new function g(x)...so I did derivate and look for crit point and the put it on g(x) to get my y. May I ask why don't that method work?

The method will work, if we take care to recognize:

\(\displaystyle x^2+y^2=9\) and \(\displaystyle y=\pm\sqrt{9-x^2}\) and so:

\(\displaystyle g(x)=17-2x\mp4\sqrt{9-x^2}\) hence:

\(\displaystyle g'(x)=-2\mp\frac{4x}{\sqrt{9-x^2}}=0\)

\(\displaystyle \frac{\sqrt{9-x^2}\pm 2x}{\sqrt{9-x^2}}=0\)

Observe that the numerator has the roots found from:

\(\displaystyle \sqrt{9-x^2}\pm 2x=0\)

\(\displaystyle 9-x^2=4x^2\)

\(\displaystyle x^2=\frac{9}{5}\)

\(\displaystyle x=\pm\frac{3}{\sqrt{5}}\) and so:

\(\displaystyle y=\pm\sqrt{9-\frac{9}{5}}=\pm\frac{6}{\sqrt{5}}\)

What critical values do we find from the denominator of the derivative?

Using the method of Lagrange multipliers, can you state the objective function and the constraint?

 

[Petrus]

I mean like this, if we use that method we lose 2 end point? \(\displaystyle f(0,0)\) and \(\displaystyle f(3,0)\) So that method would not be smart or I am wrong?

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Ohh We get \(\displaystyle x_1=3\) and \(\displaystyle x_2=-3\) so what do we do next with Them?

 

[Petrus]

I maybe respond too fast x can't be lower or equal to 3, so we got à new end point or I am wrong...
Edit: (I did think wrong.. here is the intervall.We get this intervall. 3<_x<_-3. (Equal or less)