APPM 2350 Midterm: appm2350spring2014exam2_sol_0

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31 Jan 2019
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Spring 2014: f (x, y) = 10 + x2 + y2. + x2(1 y) (a) fx = 2x + 2x(1 y) = 2x(2 y) = 0 fy = y x2 = 0 (1) (2) Equation (1) requires x = 0 or y = 2. If x = 0, then eq. (2) yields y = 0 so one critical point is (0, 0). D(0, 0) = 4 > 0 and with fyy = 1 > 0 implies (0, 0) is a local minimum. D( 2, 2) = 4 2(2) 4(2) = 8 < 0 implies ( 2, 2) is a saddle point. In the case here we have l(x, y) = f (0, 0) = 10. Y (cid:29) (cid:28) dx dt dy dt(cid:29). When ying straight north, v = 5j = 2 5 = When ying due east, v = 5i = 5 5 =

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