Problems: Second Derivative Test | Massachusetts Institute of Technology - Edubirdie

University:
Massachusetts Institute of Technology
Course:
18.02SC | Multivariable Calculus
Academic year:

2022
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441

Pages:

1
Author:

ekspektoe0

Second derivative test 1. Find and classify all the critical points of f (x, y) = x6 + y 3 + 6x − 12y + 7. Answer: Taking the ﬁrst partials and setting them to 0: ∂z = 6x5 + 6 = 0 ∂x and ∂z = 3y 2 − 12 = 0. ∂y The ﬁrst equation implies x = −1 and the second implies y = ±2. Thus, the critical points are (−1, 2) and (−1, −2). Taking second partials: ∂2z = 30x4 , ∂x2 ∂2z = 0, ∂xy ∂2z = 6y. ∂y 2 We analyze each critical point in turn. At (-1,-2): A = zxx (−1, −2) = 30, B = zxy (−1, −2) = 0, C = zyy (−1, −2) = −12. Therefore AC − B 2 = −360 < 0, which implies the critical point is a saddle. At (-1,2): A = zxx (−1, 2) = 30, B = zxy (−1, 2) = 0, C = zyy (−1, 2) = 12. Therefore AC − B 2 = 360 > 0 and A > 0, which implies the critical point is a minimum.

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