MA 261
Quiz 6
23 Φβρoυ άριoς 2016
Instructions: Write your name and section number on your quiz. Show
all work, with clear logical steps. No work or hard-to-follow work will lose
points.
Problem 1. (5 points) The point (0, 0) is a critical point of the function
f (x, y) = x3 − 3xy 2 .
What conclusions can you make about this critical point?
Solution. fx = 3x2 − 3y 2 , fy = −6xy, so fxx = 6x, fxy = −6y, fyy = −6x.
So
fxx fxy 6x −6y =
= −36x2 − 36y 2 .
D = fxy fyy −6y −6x
At the point (0, 0), we have D(0, 0) = 0. So the Second derivative test is
inconclusive. As it turns out, this is the monkey saddle.
,
Problem 2. (7 points) Find the absolute maximum and minimum values of
f on D, where
f (x, y) = 2x3 + y 4 ,
D = {(x, y) ∈ R2 | x2 + y 2 ≤ 1}
Solution. Recall that a continuous function on a closed and bounded region
in Rn always attains it’s maximum and minimum values. So we know the
absolute maximum and minimum exist. They occur at one of two places: a
critical point of f or the boundary of the region we are interested in.
We begin by checking whether f has any critical points on the interior of
D. fx = 6x2 , fy = 4y 3 . Setting each of these equal to 0, we conclude that
the only critical point for f is (0, 0), and f (0, 0) = 0.
To check the boundary of D, we have that x2 +y 2 = 1. So y 2 = 1−x2 , and
let g(x) = f (x, y) = 2x3 + (1 − x2 )2 = x4 + 2x3 − 2x2 + 1, with −1 ≤ x ≤ 1.
Then g 0 (x) = 4x3 + 6x2 − 4x = 0√⇒ x = 0, −2, or 21 .
f (0, ±1) = g(0) = 1, f ( 21 , ± 23 ) = g( 12 ) = 13
, and (−2, −3) is not in D.
16
Just as in Calc I, we need to check the endpoints: we get f (−1, 0) = g(−1) =
−2, and f (1, 0) = g(1) = 2. Thus the absolute maximum and minimum of f
on D are f (1, 0) = 2 and f (−1, 0) = −2.
Another way we could have checked for extrema on the boundary of D
would be to write x = cos θ, y = sin θ so that f (cos θ, sin θ) = 2 cos3 θ + sin4 θ
for 0 ≤ θ ≤ 2π.
Notice that we could have also adopted the approach of Lagrange multipliers. The procedure for checking for critical points of f remains the same.
On the boundary of D, however, we let g(x, y) = x2 + y 2 = 1. Then we have
∇f = λ∇g.
That is,
2 λ2x
6x
=
.
4y 3
λ2y
If x, y 6= 0, then this gives 3x = λ = 2y 2 , and substituting this back into
our equation g(x, y) = x2 + y 2 = 1, we get x2 + 23 x = 1, i.e., x2 + 23 x − 1 =
(x − 21 )(x + 2) = 0. In addition to the possibility that x = 0, this brings us
back to the case that we began with.
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Problem 3. (0 points) What is your favorite mathematical theorem, fact or
result?
Solution. It’s always fun to tell people about the Hairy Ball Theorem.
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