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# midterm1s - 18.02 MIDTERM#1 SOLUTIONS BJORN POONEN 1:05pm...

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Unformatted text preview: 18.02 MIDTERM #1 SOLUTIONS BJORN POONEN September 30, 2010, 1:05pm to 1:55pm (50 minutes). 1) Let f ( x, y ) = 2 x 2- 3 xy + y 2 + 6 x- 4 y . (a) (10 pts.) Find all critical points (if any) of f ( x, y ). Solution: Calculate the partial derivatives: f x = 4 x- 3 y + 6 f y =- 3 x + 2 y- 4 . Critical points are points where both of these are 0: 4 x- 3 y + 6 = 0- 3 x + 2 y- 4 = 0 . Multiply these by 2 and 3, respectively, to obtain 8 x- 6 y + 12 = 0- 9 x + 6 y- 12 = 0 . Adding the first equation to the second yields 8 x- 6 y + 12 = 0- x = 0 , which is equivalent to ( x, y ) = (0 , 2). Thus (0 , 2) is the only critical point. (b) (10 pts.) For each critical point, determine whether it is a local minimum, local maximum, saddle point, or none of these. Solution: Well use the second derivative test. First we calculate f xx = 4 , f xy =- 3 , f yy = 2 . Evaluating these at (0 , 2) yields A = 4, B =- 3, C = 2. Then AC- B 2 =- 1 &amp;lt; 0, so (0 , 2) is a saddle point....
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midterm1s - 18.02 MIDTERM#1 SOLUTIONS BJORN POONEN 1:05pm...

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