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math237_a7

# math237_a7 - SOLUTIONS FOR ASSIGNMENT 7 Problem Set 5 A6 a...

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SOLUTIONS FOR ASSIGNMENT 7 Problem Set 5: A6: a). Use Lagrange multipliers to find the greatest and least distance of the curve 6 x 2 + 4 xy + 3 y 2 = 14 from the origin. b). Illustrate the result graphically by drawing the constraint curve g ( x, y ) = 0 , the level curves f ( x, y ) = C , and the gradient vectors f and g . Clearly indicate the relation between the level curves of f and the contraint curve at the maximum and minimum. a). We look for the maximum and minimum values of the square of the distance of a point ( x, y ) on the curve from the origin, f ( x, y ) = x 2 + y 2 , subject to the constraint g ( x, y ) = 14 , where g ( x, y ) = 6 x 2 + 4 xy + 3 y 2 . We solve the system: f = λ g, g ( x, y ) = 14 , which is equivalent to: (2 x, 2 y ) = λ (12 x + 4 y, 4 x + 6 y ) g ( x, y ) = 14 , or: x = λ (6 x + 2 y ) y = λ (2 x + 3 y ) g ( x, y ) = 14 . (1) Since λ negationslash = 0 ( λ = 0 implies from the first two equations of the system (1) that x = y = 0 which is not a solution of the system since g (0 , 0 = 0 negationslash = 14 ), we can elliminate λ between the first two equations of the system (1) and get: x (2 x + 3 y ) = y (6 x + 2 y ) ( y + 2 x )(2 y x ) = 0 y = 2 x or y = x/ 2 . If y = 2 x then the constraint equation g ( x, 2 x ) = 14 gives x 2 = 7 / 5 , and hence ( ± radicalBig 7 5 , 2 radicalBig 7 5 ) are two solutions of the system (1). If y = x/ 2 then the constraint equation g ( x, x/ 2) = 14 gives x 2 = 8 / 5 , and thus ( ± 2 radicalBig 2 5 , ± radicalBig 2 5 ) are other two solutions of the system (1).

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