Calculus Problems: Minima, Maxima, and Integration, Assignments of Calculus

Calculus problems focusing on finding minima and maxima of functions, as well as integrating expressions. The problems involve finding the x and y values that result in the minimum and maximum values of the functions, and integrating expressions using maple.

Typology: Assignments

Pre 2010

Uploaded on 08/09/2009

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1. Section 12.8, Page 912, question 54.
0 f(x,2x) are Minima Absolute
f(2,0)at 16 is Maximum Absolute
0)1)(( 16)1)2(()1())1(2()(),(
20 ,1 line On the
0)4)1(4(16)4)2(4()44())42(2()(),(
21 ,42 line On the
0)11(1)10()1())1(2()(),(
10 ,1 line On the
0))2(2()2,(
26124248
042),(048),(
42)2(2),(482)2(2),(
)2(),(
2
5
2
2
5
2
2
5
2
2
5
2
2
1
2
1
2222
2222
2
2
=
=====+==
+=
=====+==
+=
=====+==
+=
==
===
====
====
=
MinimumMaximumxxxxfyxf
xxy
MinMaxxxxxfyxf
xxy
MinimumMaximumxxxxfyxf
xxy
xxxxf
xyyxxyyx
xyyxfyxyxf
xyyxyxfyxyxyxf
yxyxf
yx
yx
2. Section 12.9, Page 918, question 22.
32 32
|
y. real allfor defined is
)342(Re3232
08123416168424840
1
8424442
)2()02()2()02()0()00(
3
2
3
2
3
2
3
2
6
4812
6
)3)(8(414412
2222
84
4284
84
42
22
222222
2
2
2
=
><
+
+±=
===++=+=+=
=+=
++=+++=
=+++++=
±
±
+
++
+
yy
ject
yyyyyyyyyy
yyyyyyS
yyyS
y
S
y
S
y
S
yy
yyy
yy
y
y
S
3. Section 12.10, Page 928, question 40.
150500100)0,25,0(
25500:50
00:0
00:21:20
02222:122
22
),,(0),,(
222),,(50),,(
22),,(100),,(
2
2
222
222
22
=++=
=====++
====
=====
=+=+===
+=
===
++==++=
+=++=
T
yzxSincezyx
zxzSincezx
zSincezSincez
xxxxSinceyy
xx
zyxhzxzyxh
zyxzyxgzyxzyxg
yxzyxTyxzyxT
µµλµλ
µµµλλλ
µλ
µµµ
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ki
kji
ji
pf2

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Page 1 of 2

1. Section 12.8, Page 912, question 54.

AbsoluteMinimaaref(x,2x) 0

AbsoluteMaximumis 16 atf(2,0)

On theline 1 , 0 2

On theline 2 4 , 1 2

On theline 1 , 0 1

2 5

2 2

(^25) 2

(^25) 2

(^25) 2

1

2

1

2 2 2 2

2 2 2 2

2

2

f x y f x x x x Maximum Minimum

y x x

f x y f x x x x Max Min

y x x

f x y f x x x x Maximum Minimum

y x x

f x x x x

x y y x x y y x

f xy x y f x y y x

f xy x y x y f x y x y y x

f x y x y

x y

x y

2. Section 12.9, Page 918, question 22.

isdefinedforallrealy.

2 3 2 3 (Re 2 4 3 )

3

2 3

2

3

2 3

2 6

12 48

6

2 2 2 2 12 1444 (^8 )(^3 )

4 8

4 82 4 4 8

2 4

2 2

2 2 2 2 2 2

2

2 2

∂ ∂

±

± − ∂

− +

− ++ − − +

− ∂

y y

ject

y y y y y y y y y y

S y y y y y y

S y y y

y S

y

S

y

S

y y

y y y y y

y y

S

3. Section 12.10, Page 928, question 40.

2 2

2 2 2

2 2 2

2 2

T

x y z Since x z y

x z Since z x z

z Since z Since z

y y Since x x x x

x x

hxyz x z hxyz

gxyz x y z gxyz x y z

Txyz x y Txyz x y

i k

i j k

i j

Page 2 of 2

4. Section 13.1, Page 942, question 36.

9 2

9 3

3 2

9

2

(^99) 3

3 2 0

9

3

9 3

0

9

3

9 /

0

3

0 0

ln ln 3

1 1 ( 0 ) ( 0 ) ] [ 9 ln| |] 9 ln| 9 | 9 ln| 3 |

9

18

2

Area dydx dydx x dx dx x x x

x x

5. (MAPLE) Section 13.1, Page 943, question 68.

4

3

164

0 2 2

3

2

2

0 2 2

2

0

2

4 2 2

4

2 2

2

0

4 / 4

4 2 2

2

2

2

2

dx dy x y

xy dxdy x y

xy dxdy x y

xy

y x y y

y x x y

dydx x y

xy

y

y

x

x

x

> plot([sqrt(4-x^2), 4-x^2/4], x=-10..10, color=[red,blue]);

> evalf( int(int((xy)/(x^2+y^2+1),y=sqrt(4-x^2)..4-x^2/4),x=0..2));*

> evalf(int(int((xy)/(x^2+y^2+1),x=sqrt(4-*

y^2)..2),y=0..2)+int(int((xy)/(x^2+y^2+1),x=0..2),y=2..3)+int(int((xy**

)/(x^2+y^2+1),x=0..sqrt(16-4y)),y=3..4));*