Maxima and Minima
1. Find all critical points of f (x, y) = x2 + y 2 .
Solution. We want fx = fy = 0. We calculate fx = 2x, fy = 2y , and the only way that this can be
true is for x and y to both be 0. Thus, the only critical point is (x, y) = (0, 0) .
2. Find all critical points of f (x, y) = 5 − x3 + y 3 + 3xy.
Solution. We want fx = fy = 0, and we can calculate fx = −3x2 + 3y, fy = 3y 2 + 3x. So, we want
−3x2 + 3y = 0 and 3y 2 + 3x = 0. The first equation tells us that x2 = y; plugging this into the second,
we get that x4 + x = 0 = x(x3 + 1), so x = 0 or −1 . Plugging this back into the first equation gives
y = 0 or 1, so (0, 0) and (−1, 1) are the only two critical point.
3. Find all critical points of f (x, y) = xy 2 − x2 − 2y 2 and determine whether each is a local minimum,
local maximum, or saddle point.
Solution. We want ∇f = ~0, and we can calculate ∇f = hy 2 − 2x, 2xy − 4yi. So, we want y 2 − 2x = 0
2
and 2xy − 4y = 0. The first equation tells us that x = y2 ; plugging this into the second equation gives
y 3 − 4y = 0. We can factor this as y(y − 2)(y + 2) = 0, so y = 0, 2, or −2. Since x =
three points, (0, 0) and (2, ±2).
y2
2 ,
this gives us
2
To classify our three critical points, we need to calculate the discriminant fxx fyy − fxy
at each point.
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Let’s first just calculate it in terms of x and y: fxx = −2, fxy = 2y, and fyy = 2x − 4, so fxx fyy − fxy
=
2
−4(x − 2) − 4y . So:
• At (0, 0), the discriminant is 8, which tells us that (0, 0) is either a local minimum or a local maximum. To decide which, we can use the fact that fxx = −2 < 0, which tells us that
(0, 0) is a local maximum .
• At (2, 2), the discriminant is −16, so (2, 2) is a saddle point .
• At (2, −2), the discriminant is −16, so (2, −2) is a saddle point .
4. Find the points on the cone z 2 = x2 + y 2 that are closest to the point (4, 2, 0).
p
Solution. The distance from a point (x, y, z) to (4, 2, 0) is (x − 4)2 + (y − 2)2 + z 2 . Finding minimal
values of distance is the same as finding minimal values of the square of the distance. So we let
f = (x − 4)2 + (y − 2)2 + z 2 and search for the minimum of f .
Before finding the critical points, notice that if (x, y, z) is on the cone, z is dependent on x, y, so we
substitute z 2 = x2 + y 2 in f then f = f (x, y) = (x − 4)2 + (y − 2)2 + x2 + y 2 is a function of two free
variables. Next, fx = 2(x − 4) + 2x = 4x − 8, fy = 2(y − 2) + 2y = 4y − 4. Therefore fx = fy = 0
2
implies x = 2, y = 1. D = fxx fyy − fxy
= 4 · 4 − 02 = 16. The point x = 2, y = 1 is a minimum.
√
√
z 2 = 22 + 12 = 5, z = 5 or − 5.
√
√
Conclusion: (2, 1, 5) and (2, 1, − 5) are the closest point to (4, 2, 0) on the cone.
5. Find the absolute maximum and minimum values of f (x, y) = y 2 − x2 on the square |x| ≤ 1, |y| ≤ 1.
Solution. Since |x| ≤ 1, |y| ≤ 1 is a closed bounded region in R2 , we know that f (x, y) must attain
its maximum and minimum values on this region. Remember that the basic strategy is to check the
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critical points and boundary separately. That is, we’ll find all the critical points, and then we’ll find all
the points on the boundary where the absolute minimum or maximum might occur; after we’ve done
that, we’ll plug each point into f to see which gives the highest value and which gives the lowest.
The critical points are where ∇ = ~0. Since ∇f = h−2x, 2yi, the only critical point is (0, 0).(1)
Now, we look at the boundary. The boundary is composed of four separate pieces, and we’ll look at
each one individually:
• Let’s look at the piece where x = 1. Here, f (x, y) = f (1, y) = y 2 − 1, and we are concerned with
y in the interval [−1, 1]. So, we want to maximize and minimize y 2 − 1 on the interval [−1, 1].
You can do this using calculus or just by looking at the graph of y 2 − 1; either way, you should
find that it’s biggest when y = ±1 and smallest when y = 0. This gives us three candidate points,
(1, 1), (1, −1), and (1, 0).
• On the piece where x = −1, f (x, y) = f (−1, y) = y 2 − 1, and we again want to maximize and
minimize y 2 − 1 on the interval [−1, 1]. The maximum is when y = ±1 and the minimum is when
y = 0. This gives us three more candidate points, (−1, 1), (−1, −1), and (−1, 0).
• On the piece y = 1, f (x, y) = f (x, 1) = 1 − x2 , and we want to maximize and minimize this on
the interval [−1, 1]. The maximum occurs when x = 0, and the minimum occurs when x = ±1,
so we get three more candidates, (0, 1), (−1, 1), and (1, 1).
• On the piece y = −1, f (x, y) = f (x, −1) = 1 − x2 , and we get three more candidates, (0, −1),
(−1, −1), and (1, −1).
So, we now have a big list of points where the absolute minimum and maximum might occur: (0, 0),
(1, 1), (1, −1), (1, 0), (−1, 1), (−1, −1), (−1, 0), (0, 1), (−1, 1), (1, 1), (0, −1), (−1, −1), and (1, −1).
We evaluate f at each of these points, and we find that the absolute maximum occurs at (0, ±1), where
the function is equal to 1. The absolute minimum occurs at (±1, 0), where the function is equal to
−1.
6. Find the absolute maximum and minimum values of 2x3 +y 4 on the unit disk D = {(x, y) | x2 +y 2 ≤ 1}.
Solution. Let f (x, y) = 2x3 + y 4 , Since D is a closed and bounded region and f is continuous, we
know f must have attain global maximum and minimum in D.
Step 1: Find critical points: fx = 6x2 , fy = 4y 3 , and hence the only critical point is (0, 0) and it’s
indeed in D. f (0, 0) = 0.
Step 2: On the boundary x2 + y 2 = 1,
f (x, y) = 2x3 + y 4 = 2x3 + (1 − x2 )2 = x4 + 2x3 − 2x2 + 1.
Thus we regard it as a function of only one variable x and find extremum. This can be done by taking
derivative f 0 = 4x3 + 6x2 = 4x and set it equal to zero. f 0 = 2x(2x − 1)(x + 2) therefore
x = 0 or 1/2
√
(x = −2 is not in D). The correspoding points and values are f (0, ±1) = 1, f ( 21 , ± 23 ) = 13
16 . Don’t
forget about the endpoint x = −1 and x = 1, they are f (−1, 0) = −2, f (1, 0) = 2.
Step 3: After comparing all candidates, we find f (−1, 0) = −2 is the global minimum and f (1, 0) = 2
is the global maximum.
(1) You can use the Second Derivative Test to see what type of critical point (0, 0) is, but it’s not necessary to do so since we
are really just compiling a list of candidate points that we will check later.
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