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 September 16th, 2012, 09:20 AM #1 Newbie   Joined: Sep 2012 Posts: 4 Thanks: 0 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
 September 16th, 2012, 10:17 AM #2 Senior Member     Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,211 Thanks: 521 Math Focus: Calculus/ODEs 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.
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

 September 16th, 2012, 02:45 PM #4 Member   Joined: May 2012 Posts: 78 Thanks: 0 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 : $f(x,y)=x^2+y^2-4x+4y+5 \right f(y)=(-y+5)^2+y^2-4(-y+5)+4y+5=2y^2-2y+10$. Now we can derive and get $f'(y)=4y-2$ 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).
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

 September 16th, 2012, 06:16 PM #6 Senior Member     Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,211 Thanks: 521 Math Focus: Calculus/ODEs Re: critical points of a function Using Lagrange multipliers, we define: $f(x,y)=x^2+y^2-4x+4y+5$ subject to the constraint: $g(x,y)=x+y-5=0$ Next we use the system: $f_x(x,y)=\lambda g_x(x,y)$ $f_y(x,y)=\lambda g_y(x,y)$ in our case becomes: $2x-4=\lambda$ $2y+4=\lambda$ which implies: $2x-4=2y+4$ $x=y+4$ Substitute into the constraint: $y+4+y-5=0$ $2y=1$ $y=\frac{1}{2}$ $x=\frac{9}{2}$ Hence, we have the critical point: $$$x,y$$=$$\frac{9}{2},\frac{1}{2}$$$
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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