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- #26

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?

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- #26

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?

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- #27

- Feb 21, 2013

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Unfortently I dont know. Do you also mean there is another point?

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?

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- #28

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- #29

- Feb 21, 2013

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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 \(\displaystyle f(1,\sin(2\pi))\)Yes, we need to check the endpoints, or boundaries for $t$, which we gave as:

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

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- #30

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

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

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.

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- #31

- Feb 21, 2013

- 739

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

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

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

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.

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- #32

- Feb 21, 2013

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- #33

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?

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- #34

- Feb 21, 2013

- 739

- - - Updated - - -

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

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- #35

- Feb 21, 2013

- 739

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)

Edit: (I did think wrong.. here is the intervall.We get this intervall. 3<_x<_-3. (Equal or less)

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- #36

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- #37

- Feb 21, 2013

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I do not know? I Should check all critical point so yes?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?

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- #38

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- #39

- Feb 21, 2013

- 739

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..?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?

\(\displaystyle f(3,11)\),\(\displaystyle f(3,23)\)

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- #40

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...

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- #41

- Feb 21, 2013

- 739

Yeah my bad I forgot that we did subsitute for y. One question what did you mean with interior (1,2)

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...

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- #42

- Feb 21, 2013

- 739

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?

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- #43

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

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:

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?

\(\displaystyle -1=\lambda x\)

\(\displaystyle -2=\lambda y\)

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.

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- #44

- Feb 21, 2013

- 739

I get x=9 and x=7Recall 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.

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- #45

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- #46

- Feb 21, 2013

- 739

\(\displaystyle \lambda=-\frac{1}{x}\) and \(\displaystyle -\lambda=\frac{2}{y}\)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.

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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- #47

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