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 February 3rd, 2014, 12:34 AM #1 Member   Joined: Jan 2013 Posts: 96 Thanks: 0 Min Find the minimum of $x^2-2x+y$ if $x,y \ge 0$ and $x+y \le 2$. I need a graphic interpretation of the problem (in the plane xOy). Thank you!
 February 21st, 2014, 10:45 AM #2 Member   Joined: Mar 2013 Posts: 90 Thanks: 0 Re: Min There is no minimum. For example, if you fix $x=0$, then $k=x^2-2x+y=y$ and then, since you only need $y\leq2$, you can take $y$, and hence $k$, arbitrarily close to $-\infty$. There is no maximum either. For example, if you fix $y=0$, then $k=x^2-2x+y=x^2-2x$ and then, since you only need $x\leq2$, taking $x$ arbitrarily close to $-\infty$ will make $k$ arbitrarily close to $+\infty$.
 February 26th, 2014, 02:46 AM #3 Member   Joined: Jan 2013 Posts: 96 Thanks: 0 Re: Min Really? I sugest you to read again the statement!
 February 26th, 2014, 03:16 AM #4 Senior Member   Joined: Jun 2013 From: London, England Posts: 1,312 Thanks: 115 Re: Min The min must be at (1, 0). For every value of y, x^2 - 2x will have the same minimum value of -1 (at x = 1). So, the minimum will be -1 at (1, 0).
 March 16th, 2014, 05:54 PM #5 Global Moderator     Joined: Oct 2008 From: London, Ontario, Canada - The Forest City Posts: 7,362 Thanks: 826 Math Focus: Elementary mathematics and beyond Re: Min $f(x,\,y)\,=\,x^2\,-\,2x\,+\,y\,=\,(x\,-\,1)^2\,+\,y\,-\,1\,\Rightarrow\,\min$$f(x,\,y)$$\,=\,-1\text{ at }(x,\,y)\,=\,(1,\,0)$

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