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# yuiyi - √ 208(c z = 16-12 x 2 8 y-1 4(0 0 is a...

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MATH 241H - EXAM 2 - Solutions - October 11, 2004 1. (a) 2 / 5. (b) It doesn’t exist because the limit along the x -axis is 1, but the limit along the y -axis is -2. (c) f ( x, 0) = 1 for all x so f x (0 , 0) = 0. (d) f (0 , y ) = - 2 when y = 0 but f (0 , 0) = 1. Thus f (0 , y ) is not even continuous at y = 0 so f y (0 , 0) doesn’t exist. 2. dV/dt = ( dl/dt ) wh + ( dw/dt ) lh + ( dh/dt ) lw = - 3(40) - 2(80) - 4(50) = - 480 cubic inches/hour. 3. f x ( - 2 , 1) = - 12 and f y ( - 2 , 1) = 8. (a) u = 3 / 5 i + 4 / 5 j so D u f ( - 2 , 1) = ( - 12)(3 / 5) + 8(4 / 5) = - 4 / 5. (b) Max is length of gradient =
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Unformatted text preview: √ 208. (c) z = 16-12( x + 2) + 8( y-1). 4. (0 , 0) is a saddlepoint and (6 , 18) is a relative minimum. 5. The values of f at the critical points are f (0 , 5) =-20 , f (0 ,-5) = 20 , f ( √ 21 ,-2) = 29 , f (-√ 21 ,-2) = 29. The maximum value is 29 and it occurs at the points ( ± √ 21 ,-2). The minimum value is-20 and it occurs at (0 , 5)....
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