feb28

# feb28 - Math 10C Winter 2011 Prof Tesler Example from...

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Math 10C, Winter 2011, Prof. Tesler Example from February 28, 2011 lecture Find the global minimum and global maximum of f ( x, y ) = x 2 - y 2 on the square - 1 x 1 , - 1 y 1 . The square is a closed region (it includes the perimeter); bounded (it’s finite); and the function f ( x, y ) is continuous on it. So by the Extreme Value Theorem, there is a global minimum and a global maximum of f ( x, y ) within the square. (As opposed to f ( x, y ) tending to ±∞ , or approaching a minimum/maximum without ever reaching it, etc.) We have three categories of candidate points to examine, and then we’ll compare them them to see which gives the global minimum and global maximum. Local minima and local maxima from the second derivatives test, provided that they are within the region (square). Points where f ( x, y ) is discontinuous within the region (square); (there aren’t any in this problem). Local minima and local maxima along the perimeter.

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