a4 solutions

# a4 solutions - problem 1 First compute fx = y(x(x2 y 2 1...

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problem 1 . First compute: f x = - y ( x ) ( x 2 + y 2 + 1) 2 = 0 and f y = ( x 2 + y 2 + 1) - y (2 y ) ( x 2 + y + 2 + 1) 2 = x 2 - y 2 + 1 ( x 2 + y 2 + 1) 2 Thus, at a critical point, xy = 0 and x 2 - y 2 + 1 = 0 . The ﬁrst equation says that x = 0 or y = 0. If x = 0, then by solving the second equation, we ﬁnd that y = ± 1. If y = 0, then the second equation has no solutions. Thus the only critical points are (0 , 1) and (0 , - 1). It is easy to check that f (0 , 1) = 1 2 and f (0 , - 1) = - 1 2 . Also, note that f ( x, y ) 0 as r = p x 2 + y 2 → ∞ . This is true because | y | = p y 2 p x 2 + y 2 + 1 = r 2 + 1, so that | f ( x, y ) | ≤ r 2 + 1 r 2 + 1 = 1 r 2 + 1 . Then it is easy to check that the limit as r → ∞ = 0, as we asserted above. It follows that f (0 , 1) = 1 2 is the absolute maximum, and f (0 , - 1) = - 1 2 is the absolute minimum. problem 2 . Let x, y and z denote the lengths of the three sides of the rectangle. Since they are lengths, none of them can be negative.

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a4 solutions - problem 1 First compute fx = y(x(x2 y 2 1...

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