8/13/2019 Lec7[1]Valores Maximos y Minimos
1/22
MATH 209
Calculus,III
Volker Runde
Maximum andmimimumvalues MATH 209Calculus, III
Volker Runde
University of Alberta
Edmonton, Fall 2011
8/13/2019 Lec7[1]Valores Maximos y Minimos
2/22
MATH 209
Calculus,III
Volker Runde
Maximum andmimimumvalues
Local and absolute maxima and minima
Definition
The function f(x, y) has alocal maximumat (a, b) iff(x, y) f(a, b)for all (x, y) near (a, b). In this case, we call
f(a, b) a local maximum valueoff. Iff(x, y)f(a, b)for all(x, y)Df, we say that f has anabsolute maximumat (a, b).
Definition
The function f(x, y) has alocal minimumat (a, b) iff(x, y) f(a, b)for all (x, y) near (a, b). In this case, we callf(a, b) a local minimum valueoff. Iff(x, y)f(a, b)for all(x, y)Df, we say that f has anabsolute minimumat (a, b).
8/13/2019 Lec7[1]Valores Maximos y Minimos
3/22
MATH 209
Calculus,III
Volker Runde
Maximum andmimimumvalues
First derivative test
Theorem (first derivative test)
If f has a local maximum or minimum at(a, b) and if f x(a, b)
and f y(a, b) exist, then
fx(a, b) =fy(a, b) = 0.
IdeaIffhas a local maximum or minimum at (a, b), then thetangent plane to the graph off must be horizontal.
8/13/2019 Lec7[1]Valores Maximos y Minimos
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MATH 209
Calculus,III
Volker Runde
Maximum andmimimumvalues
Examples, I
Example
Letf(x, y) =x2 +y2 2x 6y+ 14,
so that
fx(x, y) = 2x 2 and fy(x, y) = 2y 6.
It follows that
fx(a, b) =fy(a, b) = 0 (a, b) = (1, 3).
Hence, a local maximum or minimum off canonlyoccur at(1, 3).
8/13/2019 Lec7[1]Valores Maximos y Minimos
5/22
MATH 209
Calculus,III
Volker Runde
Maximum andmimimumvalues
Examples, II
Example (continued)
For any (x, y) R2:
f(x, y) =x2 +y2 2x 6y+ 14
= (x 1)2 + (y 3)2 + 4
4
=f(1, 3)
Hence, fhas an absolute minimum at (1, 3).
8/13/2019 Lec7[1]Valores Maximos y Minimos
6/22
MATH 209
Calculus,III
Volker Runde
Maximum andmimimumvalues
Stationary points, I
Definition
A point (a, b) is calledstationaryorcriticalforf iffx(a, b) =fy(a, b) = 0.
Refomulation of the first derivative test
Iffhas a local maximum or minimum at (a, b), then (a, b)
must be a critical point for f.
8/13/2019 Lec7[1]Valores Maximos y Minimos
7/22
MATH 209
Calculus,III
Volker Runde
Maximum andmimimumvalues
Stationary points, II
ExampleLet
f(x, y) =x2 +y3.
Then
fx(x, y) = 2x and fy(x, y) = 3y2,
so that (0, 0) is the only stationary point for f.Since f(0, y) =y3 for all y, we get both positive and negativevalues offno matter how close (x, y) gets to (0, 0).
Thus, there is no local maximum or minimum of f at (0, 0).
Definition
Let (a, b) a stationary point for f such that f has neither alocal maximum or minimum at (a, b). Then we call (a, b) a
saddleforf.
8/13/2019 Lec7[1]Valores Maximos y Minimos
8/22
MATH 209
Calculus,III
Volker Runde
Maximum andmimimumvalues
Second derivative test, I
Theorem (second derivative test)
Let f have continuous second derivatives, and let(a, b) be astationary point for f . Set
D=fxx(a, b)fyy(a, b) fxy(a, b)2 =fxx(a, b) fxy(a, b)fyx(a, b) fyy(a, b)
.Then:
1 if fxx(a, b)>0 and D>0, then f has a local minimum at
(a, b);
2 if fxx(a, b)0, then f has a local maximum at(a, b);
3 if D
8/13/2019 Lec7[1]Valores Maximos y Minimos
9/22
MATH 209
Calculus,III
Volker Runde
Maximum andmimimumvalues
Second derivative test, II
Important
IfD= 0, then anything is possible:1 fmay have a local minimum at (a, b), or
2 fmax have a local maximum at (a, b), or
3 (a, b) may be a saddle for f.
Hence,ifD= 0 nothing can be said.
8/13/2019 Lec7[1]Valores Maximos y Minimos
10/22
MATH 209
Calculus,III
Volker Runde
Maximum andmimimumvalues
Examples, III
Example
Letf(
x, y) =
x4
+y4
4xy
+ 1,
so that
f
x = 4x3 4y and
f
y = 4y3 4x.
Thus, (x, y) is a stationary point for f if and only ifx3 =y andy3 =x, which implies x9 =x.
8/13/2019 Lec7[1]Valores Maximos y Minimos
11/22
MATH 209
Calculus,III
Volker Runde
Maximum andmimimumvalues
Examples, IV
Example (continued)
As
0 =x9 x
=x(x8 1)
=x(x4 1)(x4 + 1)
=x(x2 1)(x2 + 1)(x4 + 1)
=x(x 1)(x+ 1)(x2 + 1)(x4 + 1),
we have x=1, 0, 1.Hence, the critical points for f are (0, 0), (1, 1), and (1,1).
8/13/2019 Lec7[1]Valores Maximos y Minimos
12/22
MATH 209
Calculus,III
Volker Runde
Maximum andmimimumvalues
Examples, V
Example (continued)
Next, use the second derivative test. We have
2f
x2 = 12x2
,
2f
y2 = 12y2
, and
2f
xy =4,
so that
D(x, y) =2f
x2
2f
y2
2f
xy
2
= 144x2y2 16.
Therefore:
1 as D(0, 0) =16, fhas a saddle at (0, 0);
8/13/2019 Lec7[1]Valores Maximos y Minimos
13/22
MATH 209
Calculus,III
Volker Runde
Maximum andmimimumvalues
Examples, VI
Example (continued)
2 as D(1, 1) =D(1,1) = 128>0 and
2f
x2(1, 1) =
2f
x2(1,1) = 12>0,
fhas a local minimum at both (1, 1) and (1,1) with
corresponding local minimum valuesf(1, 1) =f(1,1) =1.
8/13/2019 Lec7[1]Valores Maximos y Minimos
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MATH 209
Calculus,III
Volker Runde
Maximum andmimimumvalues
Examples, VII
ExampleFind those points on the surface Sgiven by z2 =xy+ 1 thatare closest to the origin.Let P(x, y, z) be a point on S. Then:
distance from (0, 0, 0) to (x, y, z) =|x, y, z|
=
x2 +y2 +z2
=
x2 +y2 +xy+ 1.
Now:x2 +y2 +xy+ 1 is minimal
f(x, y) :=x2 +y2 +xy+ 1 is minimal.
8/13/2019 Lec7[1]Valores Maximos y Minimos
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MATH 209
Calculus,III
Volker Runde
Maximum andmimimumvalues
Examples, VIII
Example (continued)As
fx= 2x+y and fy = 2y+x,
a point (x, y) is stationary for f if and only ifx= 12 y and
x=2y, i.e., if and only ifx= 0 =y.Since
fxx= 2>0, fyy = 2, and fxy = 1,
we haveD= 4 1 = 3>0,
so that fhas a local minimum at (0, 0).As z=1 for x=y= 0, the points (0, 0, 1) and (0, 0,1) are
the points on Sclosest to (0, 0, 0).
8/13/2019 Lec7[1]Valores Maximos y Minimos
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MATH 209
Calculus,III
Volker Runde
Maximum andmimimumvalues
Examples, IX
Example
Make a rectangular box out of 12 m2 of cardboard without lidsuch that its volume is maximal.Let
x= length,
y= width
z= height,
so that
V= volume =xyz,
S= surface area = 2xz+ 2yz+xy= 12.
8/13/2019 Lec7[1]Valores Maximos y Minimos
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MATH 209
Calculus,III
Volker Runde
Maximum andmimimumvalues
Examples, X
Example (continued)
As S= 12, we obtain
z= 12 xy
2(x+y),
so that
V =xy
12 xy
2(x+y) =
12xy x2y2
2(x+y) .
8/13/2019 Lec7[1]Valores Maximos y Minimos
18/22
MATH 209
Calculus,III
Volker Runde
Maximum andmimimumvalues
Examples, XI
Example (continued)
It follows that
V
x =
1
2
(12y 2xy2)(x+y) (12xy x2y2)
(x+y)2
=1
2
12xy 2x2y2 + 12y2 2xy3 12xy+x2y2
(x+y)2
=1
2
12y2 2xy3 x2y2
(x+y)2
= y2
2
12 2xy x2
(x+y)2
and, similarly, Vy
= x2
2 122xyy2
(x+y)2 .
8/13/2019 Lec7[1]Valores Maximos y Minimos
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MATH 209
Calculus,III
Volker Runde
Maximum andmimimumvalues
Examples, XII
Example (continued)
To maximize V, we need x, y>0 such that Vx
= Vy
= 0.As
12 2xy
x2
= 0 = 12 2xy
y2
if and only ifx2 =y2 if and only ifx=y, solve
0 = 12 2x2 x2 = 12 3x2
forx. It follows that x=y= 2 (and thus z= 1). As themaximum ofVmust occur, we can skip the second derivativetest.Hence, V is maximal for x=y= 2 and z= 1 with V = 4.
8/13/2019 Lec7[1]Valores Maximos y Minimos
20/22
MATH 209
Calculus,III
Volker Runde
Maximum andmimimumvalues
Examples, XIII
Example
Find the absolute minimum and maximum values of
f(x, y) =x2 2xy+ 2y
on the rectangle
D :={(x, y) : 0x3, 0y2}.
with sides
L1 :={(x, 0) : 0 x3},
L2 :={(3, y) : 0 y2},
L3 :={(x, 2) : 0 x3},
L4
:={(0, y) : 0 y2}.
8/13/2019 Lec7[1]Valores Maximos y Minimos
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MATH 209
Calculus,III
Volker Runde
Maximum andmimimumvalues
Examples, XIV
Example (continued)
Asfx= 2x 2y and fy =2x+ 2
there is only one critical point, namely (1, 1); we havef(1, 1) = 1.On L1: y= 0, f(x, 0) =x
2, 0 x3.Hence, the minimum value of f on L1 is f(0, 0) = 0, and themaximum value is f(3, 0) = 9.On L2: x= 3, f(3, y) = 9 4y, 0y2.Hence, the minimum value of f on L2 is f(3, 2) = 1, and themaximum value is f(3, 0) = 9.
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MATH 209
Calculus,III
Volker Runde
Maximum andmimimumvalues
Examples, XV
Example (continued)
On L3: y= 2, f(x, 2) =x2 4x+ 4 = (x 2)2, 0x3.
Hence, the minimum value of f on L3 is f(2, 2) = 0, and the
maximum value is f(0, 2) = 4.On L4: x= 0, f(0, y) = 2y, 0 y2.Hence, the minimum value of f on L4 is f(0, 0) = 0, and themaximum value is f(0, 2) = 4.Hence:
the absolute minimum value off on D isf(0, 0) =f(2, 2) = 0;
the absolute maxiimum value off on D is f(3, 0) = 9.
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