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Hw2+Solutions

# Hw2+Solutions - Solutions to Homework 2 7.3.2 First we take...

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Solutions to Homework 2 7 . 3 . 2 First we take the partial derivatives and set them equal to 0 to obtain. f x = x - 3 = 0 and f y = 2 y + 2 = 0 . Solving these equations the only possible point where a relative extremum can exist is (3 , - 2). 7 . 3 . 8 First we take the partial derivatives and set them equal to 0 to obtain. f x = 4 x 3 - 8 y = 0 and f y = - 8 x + 4 y = 0 . Solving both of these equations for y we obtain y = x 3 / 2 and y = 2 x. Setting these equal to one another we obtain x 3 2 = 2 x x 3 - 4 x = 0 x ( x - 2)( x + 2) = 0 Solving for x and plugging back in for y we see the only possible points where a relative extremum can exist are ( - 2 , - 4), (0 , 0), and (2 , 4). 7 . 3 . 11 First we compute the second derivatives: f xx = 6, f xy = - 6, and f yy = 6 y . Then we define D ( x, y ) = 36 y - 36. Since D (3 , 3) = 72 > 0 and f xx (3 , 3) = 6 > 0 we know that f ( x, y ) has a relative minimum at (3 , 3). Since D ( - 1 , - 1) = - 72 < 0 we know that f ( x, y ) has a saddle point at ( - 1 , - 1). 7 . 3 . 12 First we compute the second derivatives: f xx = - 12 x , f xy = 12 x , and f yy = 12 - 36 y 2 . Hence we have D ( x, y ) = 432 xy 2 - 144 x - 144 x 2 . Since D (1 , 1) = 144 > 0 and f xx (1 , 1) = - 12 < 0 we know that f ( x, y ) has a relative maximum at (1 , 1). Since D (1 , - 1) = 144 > 0 and f xx (1 , - 1) = - 12 < 0 we know that f ( x, y ) has a relative maximum at (1 , - 1). Since D (0 ,

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Hw2+Solutions - Solutions to Homework 2 7.3.2 First we take...

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