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September 16th, 2012, 09:20 AM   #1
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critical points of a function

Find the critical points, if they exist, for

f:R^2 -> R
f(x,y)=x^2+y^2-4x+4y+5

WITH THE LINK x + y = 5


what is the link can somebody help me what do i do here help help help
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September 16th, 2012, 10:17 AM   #2
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Re: critical points of a function

You have two choices that I know of...use the "link" to substitute for either x or y into f giving you a function of 1 variable which you can the differentiate and equate to zero to find the critical values, or use Lagrange multipliers with the "link" as the constraint.
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September 16th, 2012, 12:01 PM   #3
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Re: critical points of a function

Quote:
Originally Posted by MarkFL
You have two choices that I know of...use the "link" to substitute for either x or y into f giving you a function of 1 variable which you can the differentiate and equate to zero to find the critical values, or use Lagrange multipliers with the "link" as the constraint.
not sure i understand here, as u may know by checking my ip im not from Amerika so thus would be nice to explain this to me in the universal langauge of math
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September 16th, 2012, 02:45 PM   #4
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Re: critical points of a function

Mark's second way is not very familiar to me, so we'll stick to the first method. The link is x+y=5.
Substitute for x and get x=-y+5.

Now we can take that link and insert it to the function :
.
Now we can derive and get and the critical point is y=0.5. Eay to prove that it is a minimum point.
According to the link we can write x=-0.5+5=4.5.

The critical point of the function is (4.5,0.5).
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September 16th, 2012, 04:20 PM   #5
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Re: critical points of a function

Quote:
Originally Posted by MarkFL
You have two choices that I know of...use the "link" to substitute for either x or y into f giving you a function of 1 variable which you can the differentiate and equate to zero to find the critical values, or use Lagrange multipliers with the "link" as the constraint.

found exactly what u just said
http://www.youtube.com/watch?v=ry9cgNx1QV8
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September 16th, 2012, 06:16 PM   #6
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Re: critical points of a function

Using Lagrange multipliers, we define:



subject to the constraint:



Next we use the system:





in our case becomes:





which implies:





Substitute into the constraint:









Hence, we have the critical point:

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September 16th, 2012, 07:41 PM   #7
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Re: critical points of a function

Quote:
Originally Posted by dean
...
as u may know by checking my ip im not from Amerika...
Unless a user behaves suspiciously, I don't check their IP for country of origin, since if a user wants it known where they are they have the option of displaying this information via the UCP. I feel this is a matter of privacy, and is really none of my business.
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