Final_02(spring)sol

Final_02(spring)sol - 2002 (spring) Final solutions of math...

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Unformatted text preview: 2002 (spring) Final solutions of math 100 Solutions: (1) We only consider the first octant instead of the whole ellipsoid. Let V ( x, y, z ) = xyz be the volume of rectangular on the first octant. Since the maximum volume must occur in the boundary, we can only consider in the region x 2 3 2 + y 2 4 2 + z 2 5 2 = 1 . x ≥ , y ≥ , z ≥ . Then, we can get a two variables function f ( x, y ) = 5 xy r 1- x 2 3 2- y 2 4 2 , x ≥ , y ≥ ∂f ∂x = 5 y (1- 3 x 2 3 2- y 2 4 2 ) r 1- x 2 3 2- y 2 4 2 ∂f ∂y = 5 x (1- x 2 3 2- 3 y 2 4 2 ) r 1- x 2 3 2- y 2 4 2 By solving ∂f ∂x = 0 and ∂f ∂y = 0, we have x = 3 2 , y = 4 2 or x = 0 , y = 0. You need to check x = 3 2 , y = 4 2 which is the maximum of f ( x, y ) by second partial derivatives. So the maximum volume of rectangular box is V = 8 × 3 2 × 4 2 × 5 2 = 60 in the ellipsoid. (2) We cannot integrate it directly. We should change dxdy to dydx ....
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This note was uploaded on 09/30/2010 for the course MATH 100 taught by Professor Qt during the Fall '09 term at HKUST.

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Final_02(spring)sol - 2002 (spring) Final solutions of math...

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