Final_02(spring)sol

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

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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 0 , y 0 , z 0 . Then, we can get a two variables function f ( x, y ) = 5 xy 1 - x 2 3 2 - y 2 4 2 , x 0 , y 0 ∂f ∂x = 5 y (1 - 3 x 2 3 2 - y 2 4 2 ) 1 - x 2 3 2 - y 2 4 2 ∂f ∂y = 5 x (1 - x 2 3 2 - 3 y 2 4 2 ) 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 . 3 0 1 y/ 3 sin( x 2 ) dxdy = 1 0 3 x 0 sin( x 2 ) dydx = 1 0 3 x sin( x 2 ) dx = 3 2 1 0 sin( x 2 ) d ( x 2 ) = - 3 2 cos( x 2 ) | 1 0 = 3 2 (1 - cos(1)) 1

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2 (3) V = 1 0 x 0 1 - x 2 x - 1 dzdydx = 1 0 x 0 1 - x 2 - ( x - 1) dydx = 1 0 x 1 - x 2 - x ( x - 1) dx = 1 0 - 1 2 1 - x 2 d (1 - x 2
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