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# Answer let x n y n z n d with x n y n z n xyz since z

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Answer: Let ( x n ,y n ,z n ) D with ( x n ,y n ,z n ) ( x,y,z ). Since z n = 0, z = 0. Because C is closed and ( x n ,y n ) C and ( x n ,y n ) ( x,y ), we ﬁnd ( x,y ) C , so ( x,y,z ) = ( x,y, 0) D . Thus D is closed because it contains all of its limit points. 4. Let f ( x,y,z ) = x 2 + 3 y + z 3 - 5. a ) Find an ( x 0 ,y 0 ,z 0 ) satisfying f ( x 0 ,y 0 ,z 0 ) = 0.

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MATHEMATICAL ECONOMICS MIDTERM #2, NOVEMBER 6, 2000 Page 2 Answer: The point (1 , 1 , 1) works. b ) Can x be expressed as a function g ( y,z ) in some neighborhood of ( x 0 ,y 0 ,z 0 )? Answer: Since ∂f/∂x = 2 x , ∂f/∂x = 2 at ( x 0 ,y 0 ,z 0 ). The Implicit Function Theorem yields such a function g . Alternatively, note that g ( y,z ) = (5 - 3 y - z 3 ) 1 / 2 works. c ) Compute dg . Answer: By the Implicit Function Theorem, dg = ( - 1 / 2 x )( ∂f/∂y,∂f/∂z ) = - (3 / 2)(1 , 3 z 2 ). At (1 , 1 , 1), this has the value ( - 3 / 2 , - 3 / 2). 5. Consider the function u ( x,y ) = x + y . a ) Does u attain a maximum on the set A = { ( x,y ) : x,y 0, px + y 3 } ? Why? Answer: The function u is continuous. The set A is a budget set in R 2 with strictly positive prices. We saw in class that such a set is compact. The Weierstrass Theorem says that any continuous function attains a maximum on a compact set.
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Answer Let x n y n z n D with x n y n z n xyz Since z n 0 z...

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