### Solutions for 3/11

```Solutions to problems from 3/11
1. Find all critical points of the following functions.
a. f (x, y) = x3 + 3xy − y 2 (there are exactly 2)
2
b. f (x, y) = ex − ey + y (there is exactly one)
p
c. f (x, y) = cos x2 + y 2 (there are infinitely many)
d. f (x, y) = ex + ey
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e. f (x, y) = 5x2 y − 6x3 y 2
f. f (x, y) = x2 + y + 2xy + y 2
g. f (x, y) = x3 − 3x + y 3 − 12y + 7
Solution: The critical points ofa function
f are those points at which 5f is either zero or undefined.
0
So for each problem, we set 5f =
and solve for x and y. In most cases, there are a ton of ways to
0
do this; I’ll just use one of them.
(a): fx = 3x2 + 3y, fy = 3x − 2y. So the critical points are when 3x2 + 3y = 0 and 3x − 2y = 0. Solving
the first equation, we get y = −x2 . Plugging this into the second one, we get 3x + 2x2 = 0. This has two
solutions: x = 0 and x = − 32 . If x = 0, then y = −x2 = 0. If x = − 32 , then y = −x2 = − 94 . So our
critical points are (0, 0) and (− 32 , − 94 ). Side note: Make sure you understand why (0, − 94 ) and (− 32 , 0) are
not critical points.
2
2
(b): fx = 2xex , fy = −ey + 1. So the critical points are when 2xex = 0 and −ey + 1 = 0. In order for
2
2
2xex to be zero, it must be that either 2x = 0 or ex = 0. But e to any power is positive, so that second
option can’t happen; so 2x = 0, and hence x = 0. Solving the second equation for y, we have ey = 1, so
y = 0. Therefore the only point that makes both equations true is (0, 0).
(c): fx = − √
x
x2 +y 2
p
x2 + y 2 , fy = − √
y
x2 +y 2
p
x2 + y 2 . The critical points are where both of these
p
2
2
are zero. Denominators
can’t make things be zero,
p
p so we’ll clear them out; now we have −x sin x + y = 0
2
2
2
2
and −y sin x + y = 0. This is true if sin x + y = 0;pif not, then x and y have to both be zero. So
our critical points are (0, 0) and every point such that sin x2 + y 2 = p
0. When does that happen? Well,
sin θ = 0 exactly when θ = nπ for some whole number n. So we need x2 + y 2 = nπ for integer n. But
that’s the equation for a circle of radius nπ. So our final answer is: the critical points are (0, 0) and every
point on a circle of radius nπ for each integer n.
sin
sin
3
3
(d): fx = ex , fy = 3y 2 ey . So the critical points are when ex = 0 and 3y 2 ey = 0. But ex 6= 0, so no such
points exist; f has no critical points.
1
(e): fx = 10xy − 18x2 y 2 , fy = 5x2 − 12x3 y. So the critical points are when 10xy − 18x2 y 2 = 0 and
5x − 12x3 y = 0. Here’s where factoring comes in handy. Factoring each equation, we get xy(10 − 18xy) = 0
and x2 (5 − 12xy) = 0. If x = 0, both of these equations are true regardless of y; so that means every point
(0, y) for any y is a critical point.
If x 6= 0, then we can divide both equations by x as many times as we like; this gives us y(10 − 18xy) = 0
and 5 − 12xy = 0. Using the second equation, xy = 5/12. So 10 − 18xy = 10 − 18 · 5/12 = 10 − 7.5 = 2.5,
which is not zero; so in order for y(10 − 18xy) = 0, it must be that y = 0. But then there’s no x that can
make xy = 5/12. So there’s no such (x, y). That means that x = 0 captured all of the critical points. So
our final answer is: the critical points are all points (0, a) for a a real number.
2
(f): fx = 2x + 2y, fy = 1 + 2x + 2y. So the critical points are when 2x + 2y = 0 and 1 + 2x + 2y = 0.
But, substituting the first equation into the second, we get 1 + 0 = 0. This is clearly false no matter what x
and y are, so there are no critical points.
(g): fx = 3x2 − 3, fy = 3y 2 − 12. So the critical points are when 3x2 − 3 = 0 and 3y 2 − 12 = 0. Solving
the first equation for x, we get x2 = 1 so x = ±1. Solving the second equation for y, we get y 2 = 4 so
y = ±2. Notice that these were independent of each other: we didn’t say “if x is 1 then y is 2” or anything
like that. So every combination of these options will work. That gives us four critical points: (1, 2), (−1, 2),
(1, −2), and (−1, −2).
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2. Classify all the critical points you found in 1 as either local minima, local maxima, or saddle points.
Solution: There are two basic strategies. The second derivative test is usually the best: for each critical
2
2
point, we compute fxx fyy − fxy
and fxx , and use their signs to classify the point. If fxx fyy − fxy
> 0 and
2
fxx > 0, then we have a local minimum. If fxx fyy − fxy > 0 and fxx < 0, then we have a local maximum. If
2
f xxfyy − fxy
< 0, then it doesn’t matter what fxx is, we always have a saddle point. If none of these work,
we have to move on to the other option: we compute the level set of f , and ask whether f goes down or up
in each of the sections the level set creates.
I’ll divide up the points by sub-problem.
2
=
(a): The critical points were (0, 0) and (− 32 , − 94 ). fxx = 6x, fyy = −2, and fxy = 3. So fxx fyy − fxy
2
−12x − 9. At the first point, x = 0, so fxx fyy − fxy = −9 < 0; so (0, 0) is a saddle point. At the
2
= 18 − 9 = 9 > 0, so we either have a maximum or a minimum.
second point, x = − 23 , so fxx fyy − fxy
3
9
fxx = 6x = −9 < 0, so (− 2 , − 4 ) is a local maximum.
2
2
(b): The only critical point was (0, 0). fxx = 2ex + 4x2 ex , fyy = −ey , and fxy = 0. At the critical
2
= −2 < 0. So (0, 0) is a saddle point.
point, fxx = 2, fyy = −1, and fxy = 0. So fxx fyy − fxy
(c): This is one of the very few cases in which I recommend skipping
the second derivative test; the
p
2
reason is because that involves taking the second derivatives of cos x + y 2 , which looks like no fun at
all. Instead, we’ll try something else: remember that the critical points were the points on the circles of
radius nπ for each n. What values does f take on these points? Well, if n = 0, 2, 4, . . ., then f (x, y) = 1.
If n = 1, 3, 5, . . ., then f (x, y) = −1. These are already the minimum and maximum values of cos! So the
points on circles of radius nπ for even n are maxima, and the points on circles of radius nπ for odd n are
minima.
(d): This one didn’t have any critical points, so there’s nothing to do.
(e): The only critical point was (0, 0). fxx = 10y − 36xy 2 , fyy = −12x3 , and fxy = 10x − 36x2 y. At
2
= 0. Unfortunately, that puts us in none of the cases!
(0, 0), fxx = 0, fyy = 0, and fxy = 0. So fxx fyy − fxy
So the second derivative test was inconclusive, and we’ll have to try something else.
f (0, 0) = 0. The z = 0 level set of f is where 0 = 5x2 y − 6x3 y 2 . Factoring, we get 0 = x2 y(5 − 6xy). So
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either x = 0, y = 0, or 5 − 6xy = 0. In that third case, y = 6x
. So the z = 0 level set consists of the x-axis,
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the y-axis, and the curve y = 6x , which doesn’t touch the origin.
(−1, 1) is in the upper left-hand quadrant. f (−1, 1) = 11 > 0, so the function goes up as we move
north-west. (1, −1) is in the lower right-hand quadrant. f (1, −1) = −11 < 0, so the function goes down
as we move south-east. Since the function goes up in one direction and down in another, (0, 0) must be a
Note: There’s no particular reason why I started by choosing points in those two quadrants. I just didn’t
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want to deal with the y = 6x
part of the level set, and I knew it didn’t touch those quadrants.
(f): This one didn’t have any critical points either, so there’s nothing to do.
(g): The critical points were (−1, −2), (1, −2), (−1, 2), and (1, 2). fxx = 6x, fyy = 6y, and fxy = 0.
2
So fxx fyy − fxy
= 36xy. At (−1, −2), 36xy = 72 > 0, and fxx = 6x = −6 < 0, so (−1, −2) is a local
3
maximum. At (−1, 2), 36xy = −72 < 0, so (−1, 2) is a saddle point. At (1, −2), 36xy = −72 < 0,
so (1, −2) is also a saddle point. At (1, 2), 36xy = 72 > 0, and fxx = 6x = 6 > 0, so (1, 2) is a local
minimum.
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```