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QUIZ 3, Version A : MATH 251, Section 506
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Write up your result, detail your calculations if necessary and BOX your final answer
1. Let f (x, y) = 4x3 + y 3 − 12x − 3y
(a) [50pts] Find the critical points of this function.
(b) [50pts] Classify these points ( say if these are maximum, minimum points or saddle points by
justifying).
1. (a) Critical points of f are the couple (x, y) such that ∇f (x, y) =< 12x2 − 12, 3y 2 − 3 >=< 0, 0 >.
The first equality gives us 12(x2 −1) = 0 equivalent to x = ±1. The second equality 3(y 2 −1) = 0
equivalent to y = ±1.. So, (±1, ±1) are critical points. The function f has 4 critical points :
(1, 1), (−1, −1), (1, −1), (−1, 1) .
(b) Classify the critical points (local maximum, local minimum or saddle points).
We need to compute the partial derivatives of second order of f : fxx (x, y) = 24x, fyy (x, y) = 6y,
fxy (x, y) = fyx (x, y) = 0 by Clairaut’s Theorem.
For (1, 1), Det(1,1) f = fxx (1, 1).fyy (1, 1) − (fxy (1, 1))2 = 6.24 > 0 and fxx (1, 1) = 24 > 0, so
(1, 1) is a local minimum point.
For (−1, 1), Det(−1,1) f = −24.6 < 0, so (−1, 1) is a saddle point.
For (1, −1), Det((1,−1) f = −24.6 < 0, so (1, −1) is also a saddle point.
For (−1, −1), Det(−1,−1) f = 24.6 > 0 and fxx (−1, −1) = −24 < 0, so (−1, −1) is a local maximum point.