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You have ﬁfteen minutes to complete the quiz. If anything seems unclear,
please ask.
1. The function f (x, y ) = x3 − 3x2 + y 2 has two critical points.
(a) Find both critical points. (3 points) (b) Classify the critical points (local minimum, local maximum, saddle,
or not enough information). (3 points) (c) Using this information, which of the pictures overleaf is the graph of
f ? (1 point)
2. Find the points, and values, where the function f (x, y ) = 2 − x2 − y 2
attains a global minimum and global maximum on the closed unit disk
{(x, y ) ∈ R2  x2 + y 2 ≤ 1}. (3 points) 1 (a) (b) (c) (d) 2 Solutions
If you have any questions and / or think you’ve spotted a mistake / typo, please
get in touch with me.
1. (a) The gradient vector is
f (x, y ) = 3x2 − 6x, 2y ;
setting this equal to zero gives
3x2 −6x = 0 ⇒ x(x−2) = 0 ⇒ x = 0, 2 and 2y = 0 ⇒ y = 0,
so the critical points are at (0, 0) and (2, 0).
(b) We have that
D(x, y ) = fxx (x, y )fyy (x, y ) − fxy (x, y )2 = (6x − 6)(2) = 12x − 12.
At (0, 0) this is equal to −12 < 0, so the function has a saddle at this
point. At (2, 0), this is equal to 12 > 0, and fxx (2, 0) = 6 > 0, so the
function has a local minimum at this point.
(c) The only picture that matches the answers to the earlier parts is (c).
2. First we ﬁnd its critical points and the values of f at these points: the
gradient vector is
f (x, y ) = −2x, −2y ,
so the only critical point is at (0, 0), and the function has value 2 here.
Next, we look at the boundary: on the boundary x2 + y 2 = 1, so
f (x, y ) = 2 − x2 − y 2 = 2 − (x2 + y 2 ) = 1.
Comparing these values, it must be the case that the global maximum
of f on the closed unit disk is 2, which occurs at (0, 0), and the global
minimum is 1, which occurs at all points on the boundary. 3 ...
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This note was uploaded on 09/14/2011 for the course MATH 175 taught by Professor Aldroubi during the Spring '08 term at Vanderbilt.
 Spring '08
 ALDROUBI
 Critical Point

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