Department of Mathematical Sciences
Instructor: Daiva Pucinskaite
Calculus III
July 11, 2016
Quiz 12
Find the absolute minimum and maximum values of
f (x, y) = x2 + y 2 − 2y
on the set D, where D is the closed triangular region with vertices (−2, 0), (2, 0), and (0, −2).
Recall: Let a function f be continuous on a closed bounded region D in R2 . The absolute maximum as well
as the absolute maximum of f must take place at critical points (inside D) or on the boundary points, i.e.
there are points (a, b), (m, n) among the critical points, and the boundary points of D, such that
f (a, b) ≤ f (x, y) ≤ f (m, n)
The number f (a, b) is the absolute minimum, and the number f (m, n) is the absolute maximum of f .
• Critical points:
(An interior point (a, b) in D is a critical point if either fx (a, b) = fy (a, b) = 0, or fx (a, b) fails
to exist, or fy (a, b) fails to exist)
x2 + y 2 − 2y x
x2 x + y 2 x − (2y)x
fx (x, y) =
=
fy (x, y) =
and
=
x2 + y 2 − 2y y
x2 y + y 2 y − (2y)y
= 2y − 2
= 2x
show that fx (x, y) and fy (x, y) never fails to exist. Because fx (1, 0) = 2 · 0 = 0, and
fy (0, 1) = 2 · 1 − 2 = 0 the only critical point of f is (0, 1). Since (0, 1) is not inside D,
(0, 1)
1
-3
-2
-1
0
1
2
3
-1
-2
the absolute maximum and minimum value the function f takes on the boundary of D,
i.e. the points on the black, red or blue line segments.
• The set of boundary points:
(∗) The red line segment contains the points (x, 0) for −2 ≤ x ≤ 2.
(∗) The black line segment contains the points (x, −x − 2) for −2 ≤ x ≤ 0.
(∗) The blue line segment contains the points (x, x − 2) for 0 ≤ x ≤ 2.
The absolute maximum/minimum value of f is the largest/smallest number among
f(x) = f (x, 0) = x2 + 02 − 2 · 0 = x2 ,
for x ∈ [−2, 2],
f(x) = f (x, −x − 2) = x2 + (−x − 2)2 − 2(−x − 2) = 2x2 + 6x + 8,
f(x) = f (x, x − 2) = x2 + (x − 2)2 − 2(x − 2) = 2x2 − 6x + 8,
for [−2, 0],
for [0, 2].
We consider three functions f, f, f in one variable x on the given intervals. If the absolute
maximum/minimum of f, resp. f, resp. f occurs for some r, k ∈ [−2, 2], resp. r, k ∈ [−2, 0],
resp. r, k ∈ [0, 2], i.e.
f(k) ≤ f(x) ≤ f(r) ,
|{z} |{z} |{z}
f (k,0)
f(k)
|{z}
f (x,0)
≤
f (k,−k−2)
for all x ∈ [−2, 2]
f (r,0)
f(x)
|{z}
f (x,−x−2)
≤
f(r) ,
|{z}
f (r,−r−2)
f(k) ≤ f(x) ≤ f(r) ,
|{z}
|{z}
|{z}
f (k,k−2)
f (x,x−2)
for all x ∈ [−2, 0]
for all x ∈ [0, 2]
f (r,r−2)
then the absolute maximum of f (x, y) is the largest number among
f(r) = f (r, 0),
f(r) = f (r, −r − 2),
f(r) = f (r, r − 2)
and the absolute minimum of f (x, y) is the smallest number among
f(k) = f (k, 0),
f(k) = f (k, −k − 2),
f(k) = f (k, k − 2)
We have to determine the absolute maximum/minimum of f, f, and f
Recall: Let f be a continuous function in one variable defined on the interval [a, b], and c is the only critical
point of f, then
∗ the absolute maximum of f is the largest number among f(a), f(c), f(b), and
∗ the absolute minimum of f is the smallest number among f(a), f(c), f(b).
◮ f′ (x) = (x2 )′ = 2x implies, that c = 0 is the only critical point of f on [−2, 2]. Since
f(−2) = (−2)2 = 4, f(0) = (0)2 = 0, f(2) = (2)2 = 4, we have
f(0) ≤ f(x) ≤ f(2) for all
|{z} |{z} |{z}
0
4
x2
2
x ∈ [−2, 2].
Thus r = 2 (or −2), and k = 0.
◮ f′ (x) = (2x2 + 6x + 8)′ = 4x + 6 implies, that c = − 32 is the only critical point of f on
[−2, 0]. Since f(−2) = 2(−2)2 + 6(−2) + 8 = 4, f(− 32 ) = 2(− 23 )2 + 6(− 32 ) + 8 = 27 ,
f(0) = 2(0)2 + 6(0) + 8 = 8, we have
3
≤ f(x) ≤ f(0) for all x ∈ [−2, 0].
f −
|{z}
|{z}
2
| {z } 2x2 +6x+8
8
7
2
Thus r = 0, and k = − 32 .
◮ f′ (x) = (2x2 − 6x + 8)′ = 4x − 6 implies, that c = 23 is the only critical point of
f on [0, 2]. Since f(0) = 2(0)2 − 6(0) + 8 = 8,
f( 23 ) = 2( 32 )2 − 6( 23 ) + 8 = 27 ,
2
f(2) = 2(2) − 6(2) + 8 = 4, we have
3
≤ f(x) ≤ f(0) for all x ∈ [0, 2].
f
|{z}
|{z}
2
| {z } 2x2 −6x+8
8
7
2
Thus r = 0, and k = 32 .
3
The maximal value among f(2) , f(0) , f(0) is 8, and the minimal value among f(0) , f − ,
|{z} |{z} |{z}
|{z}
2
| {z }
8
4
8
0
7
2
3
f
is 0.
2
| {z }
7
2
The absolute maximum of the function f (x, y) = x2 +y 2 −2y is 8 at (0, −2), and the absolute
minimum is 0 at (0, 0); in other words:
f (0, 0) ≤ f (x, y) ≤ f (0, −2)
| {z }
| {z }
| {z }
0
x2 +y 2 −2y
8
3
for all (x, y) in D.