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February 2nd, 2014, 11:34 PM   #1
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Min

Find the minimum of if and . I need a graphic interpretation of the problem (in the plane xOy). Thank you!
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February 21st, 2014, 09:45 AM   #2
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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 .
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February 26th, 2014, 01:46 AM   #3
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Re: Min

Really? I sugest you to read again the statement!
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February 26th, 2014, 02:16 AM   #4
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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).
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March 16th, 2014, 05:54 PM   #5
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