sfinal - Solutions to the Practice Problems for the Final...

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Solutions to the Practice Problems for the Final, Fall 2006 1. Find the plane that contains the lines L 1 : x = 1 + t, y = 1 t, z = t and L 2 : x = 1 + 3 t, y = 3 t, z = 3 3 t. SOLUTION: The lines are parallel. We choose two points on L 1 , say (1 , 1 , 0) with t = 0 and (2 , 0 , 1) with t = 1. We also choose a point on L 2 , say (1 , 0 , 3) with t = 0. The plane goes through the three points, so it is 2 x + 3 y z = 5. 2. Find and classify the critical points of the function f ( x, y ) = 4 x 2 e y 2 x 4 e 4 y . SOLUTION: An easy computation shows that the first order partial derivatives of f exist at every point. So the only critical points for f are the ones in which both f x and f y vanish: f x = 8 xe y 8 x 3 = 0 8 x ( e y x 2 ) = 0 x = 0 or e y = x 2 (1) f y = 4 x 2 e y 4 e 4 y = 0 4 e y ( x 2 e 3 y ) = 0 e 3 y = x 2 (2) We notice that x = 0 is not a solution of equation (2), and we solve e y = x 2 e 3 y = x 2 . This system gives: e y = e 3 y e y (1 e 2 y ) = 0 e 2 y = 1 y = 0 and x 2 = e 0 = 1, hence x = ± 1 . There are two critical points: P = (1 , 0) and Q = ( 1 , 0). To classify them as max/min/saddle, we compute the second order partial derivatives: f xx = 8 e y 24 x 2 f yy = 4 x 2 e y 16 e 4 y f xy = 8 xe y . At P = (1 , 0), we have det 16 8 8 12 = 128 > 0, and f xx = 16 < 0, so P = (1 , 0) is a local maximum. At Q = ( 1 , 0), we have det 16 8 8 12 = 128 > 0, and f xx = 16 < 0, so Q = ( 1 , 0) is also a local maximum.

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3. Find the absolute maximum of the function f ( x, y, z ) = x 2 yz on the triangle cut by the plane x + y + z = 12 and the first octant. SOLUTION: We apply the method of Lagrange multipliers, with f ( x, y, z ) = x 2 yz and g ( x, y, z ) = x + y + z 12. Since ( f ) = 2 xyz, x 2 z, x 2 y and ( g ) = 1 , 1 , 1 we need to solve the system 2 xyz = λ (1) x 2 z = λ (2) x 2 y = λ (3) x + y + z = 12 (4) . From equations (2) and (3) we get x 2 z = x 2 y , so x = 0 or z = y . First, note that the value of f is 0 at any point on the boundary of the triangle. Now assume that x = 0, y = 0, and z = 0. Hence z = y , so λ = x 2 y and equation (1) gives 2 xyz = x 2 y xy (2 z x ) = 0 x = 0 or y = 0 or x = 2 z.
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