Math 21a
Global Extrema
Fall, 2016
We want to find global extrema (absolute max and min) of f (x, y) on a region. Let’s take f (x, y) =
4x2 + y 2 .
Extreme Value Theorem
A continuous function on a closed and bounded region always attains a global
maximum value and a global minimum value at some points on the region.
1. Let f (x, y) = x2 y. Find the global extrema of f (x, y) on the region x2 + 2y 2 ≤ 6 if they exist.
2. The function f (x, y, z) = x2 − y 2 on the surface x2 + 2y 2 + 3z 2 ≤ 1 if they exist.
3. Find the absolute maximum and minimum values of f (x, y) = x2 + y 2 on the region xy 2 ≥ 2
if they exist.
4. Find the absolute maximum and minimum of the function f (x, y) = 6xy − y + 5 on the region
x2 + y 2 ≤ 1 if they exist.
Global Extrema – Answers and Solutions
Tips for Finding Global Extrema. Let’s find global extrema of f (x, y) on a region if they exist.
(0) Check your region! e.g. Is it defined by an equality or inequality? Is it closed and bounded?
Is it unbounded?
• If your region is closed and bounded, there always exist global extrema (the absolute
max and min) by the extreme value theorem!
• If not, global extrema may not exist. You have to consider the behavior of f when the
point (x, y) in the region goes away from the origin and/or approaches the boundary.
Draw the region and the level curves of f (x, y) and conclude whether f has the absolute
max and/or min on the region.
(1) Find the candidates of global extrema on the interior of the region. The critical points of f
that lie on the interior are the candidates.
(2) Find the candidates of global extrema on the boundary of the region (if the region contains
its boundary). The method of Lagrange multipliers gives the candidates.
(3) Evaluate f at the above candidates and choose the largest value and the smallest value if
you know the absolute max and min exist. In other cases, for example, if you know that the
absolute max exists but the absolute min does not exist, just choose the largest value.
Avoid unnecessary
e.g. What if you need to find the smallest distance? What if your
√
√ computation!
constraint is y x = 2? Do you always need to find the value of λ?
1. The function f (x, y) = x2 y on the region x2 + 2y 2 ≤ 6.
First of all, our region is closed and bounded. So by the extreme value theorem, there exist
the absolute maximum and minimum. We find them by dividing the region into two parts:
the interior and the boundary. Namely, our region consists of two parts: x2 + 2y 2 < 6 and
x2 + 2y 2 = 6. We put g(x, y) = x2 + 2y 2 .
2
2
2
(a) First we find the critical points of f in
√ x + 2y √< 6. As ∇f = h2xy, x i, f has the
critical point of the form (0, y) with − 3 < y < 3.
(b) Next we find the candidates of the extremal points on x2 + 2y 2 = 6 using the method of
Lagrange multipliers. Here ∇f = h2xy, x2 i and ∇g = h2x, 4yi, so we’re looking for the
solutions of the equations
2xy = λ 2x
x2 = λ 4y
x2 + 2y 2 = 6
We consider two cases suggested by the first equation:
Case 1: If x 6= 0, then λ = y. Plugging this into the second equation, we get x2 = 4y 2 .
Plugging this in turn into the constraint, we find 6y 2 = 6, so y = ±1 (and thus
x = ±2). This gives us four points: (x, y) = (±2, ±1).
√
√
Case 2: If x = 0, then the constraint gives y = ± 3. We get two points: (x, y) = (0, ± 3).
y
2
1
−2
−1
−1
1
2
x
−2
We evaluate f at all these points and get the following table:
(x, y)
f (x, y) = x2 y
(0, y)
0
(±2, 1)
4
−4
(±2, √
−1)
(0, ± 3)
0
classification
abs max
abs min
Thus f has the absolute maximum of 4 on the region, attained at (x, y) = (±2, 1), and the
absolute minimum of −4, attained at (x, y) = (±2, −1).
2. The function f (x, y, z) = x2 − y 2 on the surface x2 + 2y 2 + 3z 2 ≤ 1.
First of all, our region is closed and bounded. So by the extreme value theorem, there exist
the absolute maximum and minimum. We find them by dividing the region into two parts:
the interior and the boundary. Namely, our region consists of two parts: x2 + 2y 2 + 3z 2 < 1
and x2 + 2y 2 + 3z 2 = 1. Put g(x, y, z) = x2 + 2y 2 + 3z 2 .
(a) First we find the critical points of f in x2 + 2y 2 + 3z 2 < 1. As ∇f = h2x, −2yi, f has
critical points of the form (0, 0, z) with z 2 < 13 in x2 + 2y 2 + 3z 2 < 1.
(b) Next we find the candidates of the extremal points on x2 + 2y 2 + 3z 2 = 1 using the
method of Lagrange multipliers.
Here ∇f = h2x, −2y, 0i and (since g(x, y, z) = x2 + 2y 2 + 3z 2 ) ∇g = h2x, 4y, 6zi. Thus
the equations we need to solve are
2x = λ 2x
−2y = λ 4y
0 = λ 6z
2
2
x + 2y + 3z 2 = 1.
The first equation suggests we consider two cases:
Case 1: If x 6= 0, then λ = 1. The second and third equations then imply that y = z = 0.
Plugging these into the constraint gives x2 = 1, so we get the points (x, y, z) =
(±1, 0, 0).
Case 2: If x = 0, then we need to consider further cases. The second equation suggests two
more cases:
Case 2a: If y 6= 0, then λ = −1/2 and the third equation implies
z = 0. The constraint
√
equation then gives us the points (x, y, z) = (0, ±1/ 2, 0).
Case 2a: If y = 0,√ then the constraint equation then gives us the points (x, y, z) =
(0, 0, ±1/ 3).
The values of the points are then found in the following table:
(x, y, z)
(0, 0, z)
(±1, 0,
√0)
(0, ±1/ 2,
√0)
(0, 0, ±1/ 3)
f (x, y, z) = x2 − y 2
0
1
−1/2
0
classification
abs max
abs min
Thus f has the absolute
maximum of 1 at (x, y, z) = (±1, 0, 0) and the absolute minimum of
√
−1/2 at (0, ±1/ 2, 0).
3. The function f (x, y) = x2 + y 2 on the region xy 2 ≥ 2.
y
4
2
5
x
−2
Our region consists of two parts: xy 2 > 2 and xy 2 = 2. As f is the square of the distance
from the origin, we know that f does not have an absolute maximum on the region. (f can
be any big number!)
(a) First we find critical points of f in xy 2 > 2. As ∇f = h2x, 2yi, a candidate is (0, 0). But
this point is NOT in xy 2 > 2. So there is no critical point in xy 2 > 2.
(b) Next we find the candidates of the extremal point ( which we know will be the absolute
minimum ) on xy 2 = 2 using the method of Lagrange multipliers.
Solving ∇f = λ∇g is equivalent to solving h2x, 2yi = λhy 2 , 2xyi, or
2x = λy 2
2y = λ2xy
xy 2 = 2.
(1)
(2)
(3)
2y
If we solve for λ in equations (1) and (2), we find λ = 2x
= 2xy
. Multiplying this out,
y2
√
√ 2
2
2
we get 2x = y , or y = ± 2 x. Plugging this into equation (3), we get x ± 2 x = 2
√
and so x = 1. Hence y = ± 2.
√
√
As f (1, 2) = f (1, − 2) = 3, the function f has a minimum of 3 on the region xy 2 ≥ 2 and
no maximum.
4. The function f (x, y) = 6xy − y + 5 in the circular region x2 + y 2 ≤ 1.
First of all, our region is closed and bounded. So by the extreme value theorem, there exist
the absolute maximum and minimum. We find them by dividing the region into two parts:
the interior and the boundary. Namely, our region consists of two parts: x2 + y 2 < 1 and
x2 + y 2 = 1. Put g(x, y) = x2 + y 2 2.
(a) First we find the critical points of f in x2 + y 2 < 1. As ∇f = h6y, 6x − 1i, f has one
critical point ( 16 , 0) in x2 + y 2 < 1.
(b) Next we find the candidates of the extremal points on x2 + y 2 = 1 using the method of
Lagrange multipliers. The equation ∇f = λ∇g and the constraint give
6y = 2xλ
6x − 1 = 2yλ
x2 + y 2 = 1.
(4)
(5)
(6)
How to solve them? Let’s consider y × (1) and x × (2):
6y 2 = 2xyλ
(6x − 1)x = 2xyλ
So we get 6y 2 = 6x2 − x. Using equation (3), which becomes y 2 = 1 − x2 , we get
6(1 − x2 ) = 6x2 − x, hence 12x2 − x − 6 = 0. As 12x2 −√x − 6 = (4x√ − 3)(3x + 2), we
have x = 43 , − 32 . From equation (3), we get (x, y) = 34 , ± 47 , − 23 , ± 35 .
We evaluate f at all these points and get the following table:
(x, y)
( 16 ,√0)
( 34 , 4√7 )
( 34 , − √47 )
(− 23 , 3√5 )
(− 23 , − 35 )
f (x, y) = 6xy − y + 6
5
√
7 7
+5
8√
7 7
− √8 + 5
5 5
−√
+5
3
5 5
+5
3
√
classification
abs min
abs max
Thus f has an absolute maximum
of 5 3 5 + 5 on the region, attained
at (x, y) = (− 32 , −
√
√
and an absolute minimum of − 5 3 5 + 5, attained at (x, y) = (− 32 , 35 ).
√
5
),
3