difference quotient homework€¦ · find the difference quotient. 1. f(x) = 5 2. f(x) = 2x – 3...
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Joe Vasta
Difference Quotient – Homework
( ) ( )f x h f xDQ
h
Find the difference quotient.
1. f(x) = 5
2. f(x) = 2x – 3
3. f(x) = –4x + 1
4. f(x) = 3x2 + 2x + 1
5. f(x) = –x2 + 3x – 2
6. 3
( )5
xf x
x
7. 2
( )1
xf x
x
8. f(x) = x3 – x
2
Joe Vasta
Difference Quotient – Homework Answers
1. DQ = 0
2. DQ = 2
3. DQ = –4
4. DQ = 6x + 3h + 2
5. DQ = –2x – h + 3
6. DQ = 3
5 ( )x x h
7. DQ = 2
( 1)( 1)x x h
8. DQ = 3x2 + 3hx + h
2 – 2x – h
Joe Vasta
Modeling with Functions – Homework
1. A rectangle has perimeter 100 feet. Find a function that models its area A in
terms of its length l .
2. A rectangle has area 20 square feet. Find a function that models its perimeter
P in terms of its width w.
3. A rectangle has width 4 feet. Find a function that models its area A in terms
of its perimeter P.
4. A rectangle has length 5 feet. Find a function that models its perimeter P in
terms of its area A.
5. The length of a rectangle is 5 feet longer than the width. Find a function that
models its area A in terms of its perimeter P.
6. The length of a rectangle is 5 times the width. Find a function that models its
perimeter P in terms of its area A.
7. The short leg of a right triangle is 4 feet. Find a function that models its area
A in terms of its hypotenuse z.
8. The long leg of a right triangle is 5 feet. Find a function that models its
perimeter P in terms of its short leg x.
9. The hypotenuse of a right triangle is 10 feet. Find a function that models its
area A in terms of its short leg x.
10. The hypotenuse of a right triangle is 6 feet. Find a function that models its
perimeter P in terms of its long leg y.
11. The long leg of a right triangle is 5 times the short leg. Find a function that
models its area A in terms of its hypotenuse z.
12. The hypotenuse of a right triangle is 6 inches longer than the short leg. Find a
function that models its perimeter P in terms of its short leg x.
13. Find a function that models the area A of a circle in terms of the
circumference C.
14. Find a function that models the circumference C of a circle in terms of the
area A.
15. Find a function that models the area A of a square in terms of the diagonal d.
16. Find a function that models the diagonal d of a square in terms of the
perimeter P.
Joe Vasta
Modeling with Functions – Homework Answers
1. 250)( lllA
2. ww
wP 240
)(
3. 162)( PPA
4. 5
210)(
AAP
5. 16
100)(
2
pPA
6. 5
12)(
AAP
7. 162)( 2 zzA
8. 255)( 2 xxxP
9. 2
100)(
2xxxA
10. 2366)( yyyP
11. 52
5)(
2zzA
12. 93262)( xxxP
13. 4
)(2C
CA
14. AAC 2)(
15. 2
)(2d
dA
16. 4
2)(
PPd
Joe Vasta
Related Rates & Circles – Homework
A circle is growing in size. The area, circumference, and radius are denoted by A, C, and r,
respectively. Time is denoted by t.
1. If dr
dt= 5 ft/sec and r = 4 ft, then find
dA
dt.
2. If dA
dt= 6 ft
2/sec and r = 3 ft, then find
dr
dt.
3. If dr
dt= 7 ft/sec and A = 10 ft
2, then find
dA
dt.
4. If dA
dt= 3 ft
2/sec and A = 13 ft
2, then find
dr
dt.
5. If dr
dt= 3 ft/sec and r = 4 ft, then find
dC
dt.
6. If dC
dt= 9 ft/sec and r = 1 ft, then find
dr
dt.
7. If dC
dt= 7 ft/sec and C = 6 ft, then find
dA
dt.
8. If dA
dt= 2 ft
2/sec and C = 10 ft, then find
dC
dt.
9. If dC
dt= 4 ft/sec and r = 8 ft, then find
dA
dt.
10. If dA
dt= 6 ft
2/sec and r = 9 ft, then find
dC
dt.
Joe Vasta
Related Rates & Circles – Homework Answers
1. 40π ft2/sec
2. 1
ft/sec
3. 14 10 ft2/sec
4. 3
2 13 ft/sec
5. 6π ft/sec
6. 9
2 ft/sec
7. 21
ft
2/sec
8. 2
5
ft/sec
9. 32 ft2/sec
10. 2
3 ft/sec
Joe Vasta
Related Rates & Right Triangles – Homework
A right triangle is changing in size. The area is denoted by A, the perimeter is denoted by P, and
time is denoted by t.
1. If y is fixed at 4 ft, dz
dt= 21 ft/sec, and x = 3 ft, then find
dx
dt,
dA
dt,
d
dt
,
dP
dt.
2. If x is fixed at 6 ft, dA
dt= –12 ft
2/sec, and z = 7 ft, then find
dy
dt,
dz
dt,
d
dt
,
dP
dt.
3. If x is fixed at 3 ft, d
dt
= 6 rad/sec, and y = 5 ft, then find
dy
dt,
dz
dt,
dA
dt,
dP
dt.
4. If z is fixed at 13 ft, dy
dt= –24 ft/sec, and y = 5 ft, then find
dx
dt,
dA
dt,
d
dt
,
dP
dt.
5. If z is fixed at 4 ft, dx
dt= –8 ft/sec, and x = 2 ft, then find
dy
dt,
dA
dt,
d
dt
,
dP
dt.
6. If z is fixed at 6 ft, d
dt
= 2 rad/sec, and x = 5, then find
dx
dt,
dy
dt,
dA
dt,
dP
dt.
7. If dx
dt= –1 ft/sec,
dz
dt= 4 ft/sec, x = 4 ft, and z = 5 ft, then find
dy
dt,
dA
dt,
d
dt
,
dP
dt.
8. If dA
dt=
23
2 ft
2/sec,
dx
dt= 2 ft/sec, x = 5 ft, and y = 4 ft, then find
dy
dt,
dz
dt,
d
dt
,
dP
dt.
9. If d
dt
= –3 rad/sec,
dy
dt= –4 ft/sec, x = 1 ft, and y = 2 ft, then find
dx
dt,
dz
dt,
dA
dt,
dP
dt.
y
x θ
z
Joe Vasta
Related Rates & Right Triangles – Homework Answers
1. dx
dt= 35 ft/sec
dA
dt= 70 ft
2/sec
28
5
d
dt
rad/sec
dP
dt= 56 ft/sec
2. dy
dt= –4 ft/sec
4 13
7
dz
dt ft/sec
24
49
d
dt
rad/sec
4 134
7
dP
dt ft/sec
3. dy
dt= 68 ft/sec 10 34
dz
dt ft/sec
dA
dt= 102 ft
2/sec 68 10 34
dP
dt ft/sec
4. dx
dt= 10 ft/sec
dA
dt= –119 ft
2/sec
d
dt
= –2 rad/sec
dP
dt= –14 ft/sec
5. 8
3
dy
dt ft/sec
16
3
dA
dt ft
2/sec
4
3
d
dt
rad/sec
88
3
dP
dt ft/sec
6. 2 11dx
dt ft/sec
dy
dt= 10 ft/sec
dA
dt= 14 ft
2/sec 10 2 11
dP
dt ft/sec
7. dy
dt= 8 ft/sec
29
2
dA
dt ft
2/sec
7
5
d
dt
rad/sec
dP
dt= 11 ft/sec
8. dy
dt= 3 ft/sec
22
41
dz
dt ft/sec
7
41
d
dt
rad/sec
225
41
dP
dt ft/sec
9. 11
2
dx
dt ft/sec
5
2
dz
dt ft/sec
7
2
dA
dt ft
2/sec
3 5
2 2
dP
dt ft/sec
Joe Vasta
Optimization – Homework
Round answers to two decimal places if necessary.
1. A farmer with 100 feet of fencing wants to enclose a rectangular field and then
divide it into 8 equal pens. See the diagram below. What are the field
dimensions that maximize the area? What is the largest area?
2. A farmer with 200 feet of fencing wants to enclose a rectangular field and then
divide it into 6 equal pens. See the diagram below. What are the field
dimensions that maximize the area? What is the largest area?
3. A farmer wants to fence an area of 300 square feet in a rectangular field and then
divide it into 12 equal pens. See the diagram below. What are the field
dimensions that minimize the amount of fencing? What is the minimal amount of
fencing?
4. A farmer wants to fence an area of 400 square feet in a rectangular field and then
divide it into 24 equal pens. See the diagram below. What are the field
dimensions that minimize the amount of fencing? What is the minimal amount of
fencing?
Joe Vasta
5. A farmer wants to fence an area of 500 square feet in a rectangular field and then
divide it into 6 equal pens. The field is adjacent to a road. See the diagram
below. The fencing next to the road must be sturdier and costs $70 per foot,
whereas the other fencing costs just $50 per foot. What are the field dimensions
that minimize the cost? What is the minimal cost?
6. A farmer wants to fence an area of 600 square feet in a rectangular field and then
divide it into 8 equal pens. The field is adjacent to a road. See the diagram
below. The fencing next to the road must be sturdier and costs $60 per foot,
whereas the other fencing costs just $20 per foot. What are the field dimensions
that minimize the cost? What is the minimal cost?
7. A farmer wants to fence in a rectangular field and then divide it into 8 equal pens.
The field is adjacent to a road. See the diagram below. The fencing next to the
road must be sturdier and costs $50 per foot, whereas the other fencing costs just
$40 per foot. The farmer wants to spend exactly $7000 on this project. What are
the field dimensions that maximize the area? What is the maximum area?
8. A farmer wants to fence in a rectangular field and then divide it into 18 equal
pens. The field is adjacent to a road. See the diagram below. The fencing next
to the road must be sturdier and costs $90 per foot, whereas the other fencing
costs just $60 per foot. The farmer wants to spend exactly $8000 on this project.
What are the field dimensions that maximize the area? What is the maximum
area?
Joe Vasta
9. An open box is to be made from a 7 foot by 10 foot rectangular piece of material
by cutting equal squares from the corners and turning up the sides. Find the side
length of the removed squares that will maximize the volume of the box. What is
the maximum volume?
10. An open box is to be made from a 8 foot by 20 foot rectangular piece of material
by cutting equal squares from the corners and turning up the sides. Find the side
length of the removed squares that will maximize the volume of the box. What is
the maximum volume?
11. An open box with a square base is to be made with a surface area of 100 square
feet. Find the box dimensions that maximize the volume. What is the maximum
volume?
12. An open box with a square base is to be made with a volume of 50 cubic feet.
Find the box dimensions that minimize the surface area. What is the minimum
surface area?
13. The top and bottom margins of a poster are each 2 inches and the side margins are
each 3 inches. If the area of the printed material on the poster is fixed at 200
square inches, find the poster dimensions with the smallest area.
14. The top and bottom margins of a poster are each 4 inches and the side margins are
each 5 inches. If the area of the printed material on the poster is fixed at 100
square inches, find the poster dimensions with the smallest area.
15. The top and bottom margins of a poster are each 1 inch and the side margins are
each 2 inches. If the area of the poster is fixed at 70 square inches, find the poster
dimensions with the largest printable area.
16. The top and bottom margins of a poster are each 2 inches and the side margins are
each 5 inches. If the area of the poster is fixed at 300 square inches, find the
poster dimensions with the largest printable area.
17. Find the point on the line y = 5x + 1 that is closest to the point (2, 3).
Write the point using fractions.
18. Find the point on the line y = –x + 3 that is closest to the point (–2, 6).
Write the point using fractions.
19. Find the point on the line y = 2x – 1 that is closest to the point (–2, –3).
Write the point using fractions.
20. Find the point on the line y = –3x that is closest to the point (1, –4).
Write the point using fractions.
Joe Vasta
Optimization – Homework Answers
1. 16.67 ft 10 ft 166.7 ft2
2. 25 ft 33.33 ft 833.25 ft2
3. 19.36 ft 15.49 ft 154.89 ft
4. 23.66 ft 16.90 ft 236.60 ft
5. 24.25 ft 20.62 ft $8246.50
6. 16.04 ft 37.42 ft $4490.80
7. 26.92 ft 17.5 ft 471.1 ft2
8. 14.81 ft 9.52 ft 140.99 ft2
9. 1.35 ft 42.38 ft3
10. 1.76 ft 129.94 ft3
11. base = 5.77 ft height = 2.89 ft volume = 96.22 ft3
12. base = 4.64 ft height = 2.32 ft surface area = 64.59 ft2
13. 23.32 in. 15.55 in.
14. 16.94 in. 21.18 in.
15. 11.83 in. 5.92 in.
16. 27.39 in. 10.95 in.
17. (6/13, 43/13)
18. (–5/2, 11/2)
19. (–6/5, –17/5)
20. (13/10, –39/10)