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Unformatted text preview: MATH 2513, Fall 2007 Simon Fraser University Assignment 8: Solutions Additional Question Due: 4:30pm, Monday 19 November 2007 1. Find the point(s) on the surface x 2 y 4 z = 8 that is/are closest to the origin. Solution: We should immediately begin by formulating the problem mathematically: The distance of a point ( x, y, z ) to the origin is p x 2 + y 2 + z 2 , so the problem becomes: Minimize p x 2 + y 2 + z 2 for points ( x, y, z ) satisfying x 2 y 4 z = 8 . It is equivalent, and simpler (strongly recommended!) to minimize f ( x, y, z ) = x 2 + y 2 + z 2 . Note that the numbers in this question turn out to be somewhat complicated and un wieldy — though quite appropriate for a homework problem; as suggested in the Notes at the end, they are fairly straightforward to deal with if one just considers appropriate powers of 2. Problems on a timed exam will have “simpler” numbers. You should solve the problem twice, using the following two methods: (a) Reduce the problem to an unconstrained problem in two variables; that is, eliminate one of the variables in terms of the other two, and find the critical point(s) [use the methods of Section 14.7]. Solution: We can solve, for example, for z ; I choose to solve for x since both the constraint and the function f ( x, y, z ) contain x 2 . We have x 2 = 8 y 4 z 1 , and substituting into f , the problem becomes to find the minimum of...
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 Spring '09
 jcliu
 Addition, Optimization, Power of two, Constraint, lagrange multipliers, Joseph Louis Lagrange, surface x2

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