# Max and min value, multi variable (open sets)

#### MarkFL

##### Administrator
Staff member
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

##### Well-known member
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 dont know. Do you also mean there is another point?

#### MarkFL

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

$$\displaystyle 0\le t\le 2\pi$$.

#### Petrus

##### Well-known member
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

##### Administrator
Staff member
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. #### Petrus

##### Well-known member
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. My bad here I forgot i had 3sin(t),3cos(t) i Was using sin(t),cos(t) this problem Was à Hood one #### Petrus

##### Well-known member
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 dont that method work?

#### MarkFL

##### Administrator
Staff member
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 dont 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

##### Well-known member
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?

- - - Updated - - -

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

#### Petrus

##### Well-known member
I maybe respond too fast x cant 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)

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#### MarkFL

##### Administrator
Staff member
Yes, we actually gain a critical point to check that we did not get from the parametrization of the boundary...but does this affect the absolute extrema values?

#### Petrus

##### Well-known member
Yes, we actually gain a critical point to check that we did not get from the parametrization of the boundary...but does this affect the absolute extrema values?
I do not know? I Should check all critical point so yes?

#### MarkFL

##### Administrator
Staff member
As a recap, what are all of the critical points you have collected so far, and what is the function's value at these points?

#### Petrus

##### Well-known member
As a recap, what are all of the critical points you have collected so far, and what is the function's value at these points?
The crit point when we subsitate x. But the point is you can't divide by zero so when we did calculate that 3,-3 we say that the bottom of division is zero..?
$$\displaystyle f(3,11)$$,$$\displaystyle f(3,23)$$

#### MarkFL

##### Administrator
Staff member
The derivative is undefined for $x^2=3^2$, but the original function is not. Recall that critical values come from places where the derivative is zero or undefined.

You should have the critical point in the interior (1,2), and the points on the boundary:

$$\displaystyle \left(\pm\frac{3}{\sqrt{5}},\pm\frac{6}{\sqrt{5}} \right),\,(\pm3,0)$$

So, you want to evaluate the original function at these 5 points...

#### Petrus

##### Well-known member
The derivative is undefined for $x^2=3^2$, but the original function is not. Recall that critical values come from places where the derivative is zero or undefined.

You should have the critical point in the interior (1,2), and the points on the boundary:

$$\displaystyle \left(\pm\frac{3}{\sqrt{5}},\pm\frac{6}{\sqrt{5}} \right),\,(\pm3,0)$$

So, you want to evaluate the original function at these 5 points...
Yeah my bad I forgot that we did subsitute for y. One question what did you mean with interior (1,2)

#### Petrus

##### Well-known member
Ok for Lama way.
We got our $$\displaystyle g(x,y)=x^2+y^2-9$$
so..
$$\displaystyle f_x(x,y)=\lambda*g_x(x,y)$$
$$\displaystyle f_y(x,y)-=\lambda*g_y(x,y)$$
And our...
$$\displaystyle g_x(x,y)=2x$$
$$\displaystyle g_y(x,y)=2y$$
so we now got..
$$\displaystyle f_x(x,y)=\lambda*2x$$
$$\displaystyle f_y(x,y)-=\lambda*2y$$
Now I dont know to do but I guess ima subsitute again $$\displaystyle y=\sqrt{9-x^2}$$ and then equal each to same so I got
$$\displaystyle 2x=2\sqrt(9-x^2}*$$
I am doing correct?

#### MarkFL

##### Administrator
Staff member
...One question what did you mean with interior (1,2)
Recall this is the critical point you found in the interior of the circle by equating the first partials to zero.

Ok for Lama way.
We got our $$\displaystyle g(x,y)=x^2+y^2-9$$
so..
$$\displaystyle f_x(x,y)=\lambda*g_x(x,y)$$
$$\displaystyle f_y(x,y)-=\lambda*g_y(x,y)$$
And our...
$$\displaystyle g_x(x,y)=2x$$
$$\displaystyle g_y(x,y)=2y$$
so we now got..
$$\displaystyle f_x(x,y)=\lambda*2x$$
$$\displaystyle f_y(x,y)-=\lambda*2y$$
Now I dont know to do but I guess ima subsitute again $$\displaystyle y=\sqrt{9-x^2}$$ and then equal each to same so I got
$$\displaystyle 2x=2\sqrt(9-x^2}*$$
I am doing correct?
Using Lagrange multipliers, you could state $f(x,y)=17-2x-4y$ as the objective function and $g(x,y)=x^2+y^2-9=0$ as the constraint. This gives you the system:

$$\displaystyle -1=\lambda x$$

$$\displaystyle -2=\lambda y$$

What does this imply about the relationship between $x$ and $y$? Solve both for $\lambda$, then equate and simplify. Once you find this, substitute for $y$ in the constraint, and solve for $x$. This will give you two critical points on the boundary.

#### Petrus

##### Well-known member
Recall this is the critical point you found in the interior of the circle by equating the first partials to zero.

Using Lagrange multipliers, you could state $f(x,y)=17-2x-4y$ as the objective function and $g(x,y)=x^2+y^2-9=0$ as the constraint. This gives you the system:

$$\displaystyle -1=\lambda x$$

$$\displaystyle -2=\lambda y$$

What does this imply about the relationship between $x$ and $y$? Solve both for $\lambda$, then equate and simplify. Once you find this, substitute for $y$ in the constraint, and solve for $x$. This will give you two critical points on the boundary.
I get x=9 and x=7

#### MarkFL

##### Administrator
Staff member
How did you get those values? Without knowing what you did, I don't know what to address.

#### Petrus

##### Well-known member
Recall this is the critical point you found in the interior of the circle by equating the first partials to zero.

Using Lagrange multipliers, you could state $f(x,y)=17-2x-4y$ as the objective function and $g(x,y)=x^2+y^2-9=0$ as the constraint. This gives you the system:

$$\displaystyle -1=\lambda x$$

$$\displaystyle -2=\lambda y$$

What does this imply about the relationship between $x$ and $y$? Solve both for $\lambda$, then equate and simplify. Once you find this, substitute for $y$ in the constraint, and solve for $x$. This will give you two critical points on the boundary.
$$\displaystyle \lambda=-\frac{1}{x}$$ and $$\displaystyle -\lambda=\frac{2}{y}$$
So we got $$\displaystyle -\frac{1}{x}=\frac{2}{y}$$ that means $$\displaystyle y=2x$$
then we got $$\displaystyle x^2+4x^2=9$$ and that means $$\displaystyle x_1=\frac{3}{\sqrt{5}}$$, $$\displaystyle x_2=-\frac{3}{\sqrt{5}}$$
(I wanna apologize I did a misstake... I forgot I had y^2 in the constraint... I did calculate with y) Thanks Mark! You are a really great person! I got better thanks too you!!

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#### MarkFL

##### Administrator
Staff member
Yes, that is correct. Notice you have found the same critical points as with the parametrization, and with the substitution for $y$ without the unneeded end-points, and critical points from where the derivative was undefined.

So, what have you found are the absolute extrema? Where do they occur and what values do they have?