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5) (From Hardcover Book, Marsden/Tromba, Vector Calculus, 6th ed., Exercises for Section 3.3, #34) Let f (x, y) еба — . = 5ye

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5 e

a) First we find the critical points.

41 5r 5r fx 05ye - 5e 0ye = e ye fy05e 16re e15 1x 0 4 5y0y e .y 1

Hence (0, 1 is the critical point .

Loca maxima is F(0, 1) 5-1-1 = 3

b) Condition for existence of extremum at (0, 1 is

f(0, 1)fy(0, 1) - (fy(0, 1))> 0

fr(0,1)5 fyy (0, 1)20 fry(0, 1) 5 . fr:(0, 1)fpy (0, 1) - (fay(0,1)) 5 x (-20) 25 125 <0

Hence f does not have a global maxima.

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