15911 - correct Thus we can divide out the 2 and get b 2...

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Solution to problem 11 in section 15.9 The problem asks you to maximize the volume of a rectangular solid which is inscribed in an ellipsoid with equation x 2 a 2 + y 2 b 2 + z 2 c 2 = 1 which is equivalent to the equation g ( x, y, z ) = b 2 c 2 x 2 + a 2 c 2 y 2 + a 2 b 2 z 2 - a 2 b 2 c 2 = 0 This is our constraint. The formula we want to maximize is f ( x, y, z ) = xyz the volume of this Fgure. We use Lagrange multipliers to get < yz, xz, xy > = λ < 2 b 2 c 2 x, 2 a 2 c 2 y, 2 b 2 c 2 z > (1) We need to solve this along with g ( x, y, z ) = 0. Now, we get three equations from (1): yz = 2 λ 2 b 2 c 2 x (2) xz = 2 λ 2 a 2 c 2 y (3) xy = 2 λ 2 a 2 b 2 z (4) Multiplying (2) by x , (3) by y , and (4) by z , we get that xyz = 2 λ 2 b 2 c 2 x 2 = 2 λ 2 b 2 c 2 y 2 = 2 λ 2 a 2 b 2 z 2 We can certainly assume λ n = 0, since if it were, one of x, y, z would have to equal 0, which would give us a Fgure with no volume, which is clearly not
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Unformatted text preview: correct. Thus, we can divide out the 2 λ and get b 2 c 2 x 2 = a 2 c 2 y 2 = a 2 b 2 z 2 Since these are all equal, we can reduce g ( x, y, z ) = 0 to 3 b 2 c 2 x 2-a 2 b 2 c 2 = 0 from which we quickly get that x = a √ 3 . Similarly, we can get that y = b √ 3 and z = c √ 3 . Now, the problem asks us for the maximum volume, so we need to use these to make a volume. But note that if we just plugged these values into f , we would not get the whole volume, but rather we would get one-eighth of it (Fgure out why). Thus, the total volume is 8 f ( a √ 3 , b √ 3 , c √ 3 ) = 8 abc 3 √ 3 1...
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This note was uploaded on 08/11/2008 for the course MATH 234 taught by Professor Dickey during the Fall '08 term at University of Wisconsin.

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