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February 3rd, 2014, 12:34 AM  #1 
Member Joined: Jan 2013 Posts: 96 Thanks: 0  Min
Find the minimum of if and . 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 , then and then, since you only need , you can take , and hence , arbitrarily close to . There is no maximum either. For example, if you fix , then and then, since you only need , taking arbitrarily close to will make arbitrarily close to . 
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: 7,385 Thanks: 844 Math Focus: Elementary mathematics and beyond  Re: Min 