4 let f x y z x 2 3 y z 3 5 a find an x y z

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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. b ) Does u attain a maximum on the set B = { ( x, y ) : y 0, px + y 3 } ? Why? Answer: This is a trickier problem. As was the case with the cost minimization problem done in class, we must simplify the problem before solving it. First, note that u is increasing in both x and y . It follows that any maximum must occur along the constraint px + y = 3. So x = (3 - y ) /p . Substituting into u
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