104/184 Quiz 1
Thursday September 24
First Name:
Last Name:
Student-No:
Section:
Grade:
Very short answer questions
1. 2 marks Each part is worth 1 mark. Please write your answers in the boxes.
Marking scheme: 1 for each correct, 0 otherwise
(a) Solve 2x =
1
5
for x.
Answer: log2
log 5
5
− log
= − ln
.
2
ln 2
1
5
= − log2 5 =
Solution: We have to allow several forms of the solution.
3
q − q2
(b) Compute lim
.
q→2
q+1
Answer:
4
3
Solution: Direct substitution.
Short answer questions — you must show your work
2. 4 marks Each part is worth 2 marks.
(a) Given that ln a = 3, ln b = 4, and ln c = −2, find the value of ln √abc .
Answer: 2
Solution:
√
a
1
ln √ = ln(a) − ln( bc) = ln a − ln(bc)
2
bc
1
= ln a − (ln b + ln c)
2
1
= 3 − (4 − 2) = 2 .
2
Marking scheme: If correct answer and some work, then 2 marks. If some work in
the right direction and wrong answer, or correct answer and rubbish, then 1 mark. If
no solution, then 0.
x2 + 3x − 4
x→1 4x2 − 7x + 3
(b) Compute the limit lim
Answer: 5
Solution: Direct substitution yields 0/0, so we simplify first:
x2 + 3x − 4
(x − 1)(x + 4)
x+4
=
=
2
4x − 7x + 3
(x − 1)(4x − 3)
4x − 3
x+4
= 5.
x→1 4x − 3
Marking scheme: 1 for any correct factoring, 1 for answer.
Hence the limit is lim
Long answer question — you must show your work
3. 4 marks Pea Sea Inc. sells 200 laptops a month at the price of $640 each. For each $5 increase
in price, the sales would drop by 2 laptops. Their factory costs $250,000 per month to operate
and each laptop costs an additional $225 to make.
Note: in this problem you are ONLY setting up the equations. You do NOT have to solve for
break even values or any optimal production values.
Marking scheme: 1 for any correct answer. Only part a requires work. Answers do not need
to be simplied. If students correctly use the wrong demand curve in (c) & (d), they still get
those 2 points.
(a) Find the linear demand equation for the laptops. Use the notation p for the unit price
and q for the monthly demand.
Answer: q = − 25 p + 456, or p = 1140 − 52 q
Solution: A data point is (p, q) = (price, demand). So two points on the linear
∆q
demand curve are (640, 200) and (645, 198). Its slope is m = ∆p
= 198−200
= − 52 .
645−640
Thus, q = − 25 p + b, b being a constant. Substitute in (640, 200) to get b = 456.
Therefore, q = − 25 p + 456. or equivalently: p = 1140 − 25 q.
(b) Find the cost function, C = C(q), for producing q laptops per month.
Answer: C(q) = 250000 + 225q
Solution: No work is needed
(c) Find the monthly revenue function, R = R(q).
Answer: R(q) = p · q = 1140q − 52 q 2
Solution: Either expression is acceptable.
(d) Find the monthly profit (or net income) function P (q).
Answer: P (q) = − 52 q 2 + 915q − 250000
Solution: P (q) = R(q)−C(q) = 1140q − 52 q 2 −(250000+225q) = 915q − 52 q 2 −250000.
104/184 Quiz 1
Thursday September 24
First Name:
Last Name:
Student-No:
Section:
Grade:
Very short answer questions
1. 2 marks Each part is worth 1 mark. Please write your answers in the boxes.
Marking scheme: 1 for each correct, 0 otherwise
(a) Solve 3y+1 = 2 for y.
Answer: y = log3 2 − 1
Solution: 3y+1 = 2 ⇒ y + 1 = log3 2 ⇒ y = log3 2 − 1
(b) Compute lim
p→(−3)
p2 − p − 12
.
p−3
Answer: 0
Solution:
lim
p→(−3)
p2 − p − 12
p−3
lim (p2 − p − 12)
=
p→(−3)
lim (p − 3)
=
0
= 0.
−6
p→(−3)
Short answer questions — you must show your work
2. 4 marks Each part is worth 2 marks.
(a) Given that log2 x = 5, log2 y = −3, find the value of log2 (4x3 y −2 ).
Answer: 23
Solution:
log2 4(x3 y −2 ) = log2 4 + log2 x3 − log2 y 2 = 2 + 3 log2 x − 2 log2 y
= 2 + 3(5) − 3(−2) = 23 .
Marking scheme: If correct answer and some work, then 2 marks. If some work in
the right direction and wrong answer, or correct answer and rubbish, then 1 mark. If
no solution, then 0.
3x2 − x − 10
x→2
x2 − 4
(b) Compute the limit lim
Answer: 11/4
Solution: Direct substitution yields 0/0, so we simplify first:
3x2 − x − 10
(x − 2)(3x + 5)
3x + 5
=
=
2
x −4
(x − 2)(x + 2)
x+2
3x + 5
= 11/4 .
x→2 x + 2
Marking scheme: 1 for any correct factoring, 1 for answer.
Hence the limit is lim
Long answer question — you must show your work
3. 4 marks A carver sells 80 carvings a month at the price of $160 each. For each $12 decrease
in price, he can sell 4 more carvings. The workshop costs $20,000 per month to operate and
each carving costs $40 to make.
Note: in this problem you are ONLY setting up the equations. You do NOT have to solve for
break even values or any optimal production values.
Marking scheme: 1 for any correct answer. Only part a requires work. Answers do not need
to be simplied. If students correctly use the wrong demand curve in (c) & (d), they still get
those 2 points.
(a) Find the linear demand equation for the carvings. Use the notation p for the unit price
and q for the monthly demand.
Answer: p=-3q+400
Solution: A data point is (q, p) = (quantity, price). So two points are (80, 160) and
(84, 148). Since the rate of change is constant, the demand curve must be a line with
12
= 160−148
= −4
= −3.
slope m = ∆p
∆q
80−84
Thus, p = −3q + K when K is some constant. Substitute in (80, 160) to get K = 400.
Therefore, p = −3q + 400.
(b) Find the cost function, C = C(q), for producing q carvings per month.
Answer: C(q) = 20, 000 + 40q
Solution: No work is needed
(c) Find the monthly revenue function, R = R(q).
Answer: R(q) = q · (−3q + 400)
Solution: R = p · q = q · (−3q + 400) = −3q 2 + 400q
(d) Find the monthly profit (or Net Income) function P (q).
Answer: P (q) = −3q 2 + 360q − 20, 000
Solution: P (q) = R(q)−C(q) = −3q 2 +400q −(20, 000+40q) = −3q 2 +360q −20, 000
104/184 Quiz 1
Thursday September 24
First Name:
Last Name:
Student-No:
Section:
Grade:
Very short answer questions
1. 2 marks Each part is worth 1 mark. Please write your answers in the boxes.
Marking scheme: 1 for each correct, 0 otherwise
(a) Solve 5x =
1
3
for x.
Answer: log5
Solution: 5x =
(b) Compute lim
q→2
1
3
⇒ log5 5x = log5
1
3
⇒ x = log5
1
3
1
3
3
= − log
log 5
= log5 (3−1 ) = − log5 3
q3 − 6
.
q+4
Answer:
Solution: limq→2
q 3 −6
q+4
1
3
lim(q 3 − 6)
=
q→2
lim(q + 4)
=
2
6
= 13 .
q→2
Short answer questions — you must show your work
2. 4 marks Each part is worth 2 marks.
√
(a) Given that log3 x = 7, log3 y = 5, and log3 z = 4, find the value of log3
xy
.
z
Answer: 2
Solution:
√
xy
(xy)1/2
1
log3
= log3
= log3 (xy) − log3 z
z
z
2
1
= [log3 x + log3 y] − log3 z
2
1
= [7 + 5] − 4 = 2 .
2
Marking scheme: If correct answer and some work, then 2 marks. If some work in
the right direction and wrong answer, or correct answer and rubbish, then 1 mark. If
no solution, then 0.
x2 − 2x − 3
x→3 2x2 − 3x − 9
(b) Compute the limit lim
Answer: 4/9
Solution: Direct substitution yields 0/0, so we simplify first:
x2 − 2x − 3
(x − 3)(x + 1)
x+1
=
=
2
2x − 3x − 9
(x − 3)(2x + 3)
2x + 3
x+1
= 4/9 .
x→3 2x + 3
Marking scheme: 1 for any correct factoring, 1 for answer.
Hence the limit is lim
Long answer question — you must show your work
3. 4 marks A manufacturer sells 100 tables per month at the price of $450 each. For each $10
decrease in price, he can sell 5 more tables per month. Their factory costs $150,000 per month
to operate and each table costs an additional $325 to make.
Note: in this problem you are ONLY setting up the equations. You do NOT have to solve for
break even values or any optimal production values.
Marking scheme: 1 for any correct answer. Only part a requires work. Answers do not need
to be simplied. If students correctly use the wrong demand curve in (c) & (d), they still get
those 2 points.
(a) Find the linear demand equation for the tables. Use the notation p for the unit price and
q for the monthly demand.
Answer: p=-2q+650
Solution: A data point is (q, p) = (quantity, price). So two points are (100, 450) and
(105, 440). Since the rate of change is constant, the demand curve must be a line with
10
= 450−440
= −5
= −2. Thus, p = −2q + K when K is some constant.
slope m = ∆p
∆q
100−105
Substitute in (100, 450) to get K = 650.
Therefore, p = −2q + 650.
(b) Find the cost function, C = C(q), for producing q tables per month.
Answer: C(q) = 150, 000 + 325q
Solution: No work is needed
(c) Find the monthly revenue function, R = R(q).
Answer: R(q) = q · (−2q + 650)
Solution: R = p · q = q · (−2q + 650) = −2q 2 + 650q
(d) Find the monthly profit (or Net Income) function P (q).
Answer: P (q) = −2q 2 + 325q − 150, 000
Solution: N I(q) = R(q) − C(q) = −2q 2 + 650q − (150, 000 + 325q) = −2q 2 + 325q −
150000
104/184 Quiz 1
Thursday September 24
First Name:
Last Name:
Student-No:
Section:
Grade:
Very short answer questions
1. 2 marks Each part is worth 1 mark. Please write your answers in the boxes.
Marking scheme: 1 for each correct, 0 otherwise
(a) Solve 6x =
2
5
for x.
Answer: log6
Solution: 6x =
(b) Compute lim
q→4
2
5
⇒ x = log6
2
5
2
5
=
log 2/5
log 6
q2 + 2
.
q−5
Answer: −18
Solution: Direct substituiton.
Short answer questions — you must show your work
2. 4 marks Each part is worth 2 marks.
(a) Given that log5 x = 4, log5 y = 3, and log5 z = −2, find the value of log5
yz
√ .
5 x
Answer: -2
Solution:
√
√
yz
log5 √ = log5 (yz) − log5 (5 x) = log5 y + log5 z − (log5 5 + log5 x)
5 x
= 3 + (−2) − (1 + 1/2 log5 x)
= 1 − (1 + 1/2 × 4)
= −2 .
Marking scheme: If correct answer and some work, then 2 marks. If some work in
the right direction and wrong answer, or correct answer and rubbish, then 1 mark. If
no solution, then 0.
x2 + x − 12
x→3 2x2 − 3x − 9
(b) Compute the limit lim
Answer: 7/9
Solution: Direct substitution yields 0/0, so we simplify first:
x2 + x − 12
(x − 3)(x + 4)
x+4
=
=
2
2x − 3x − 9
(x − 3)(2x + 3)
2x + 3
x+4
= 7/9 .
x→3 2x + 3
Marking scheme: 1 for any correct factoring, 1 for answer.
Hence the limit is lim
Long answer question — you must show your work
3. 4 marks A manufacturer sells 120 tables per month at the price of $400 each. For each $12
decrease in price, he can sell 6 more tables per month. Their factory costs $125,000 per month
to operate and each table costs an additional $250 to make.
Note: in this problem you are ONLY setting up the equations. You do NOT have to solve for
break even values or any optimal production values.
Marking scheme: 1 for any correct answer. Only part a requires work. Answers do not need
to be simplied. If students correctly use the wrong demand curve in (c) & (d), they still get
those 2 points.
(a) Find the linear demand equation for the tables. Use the notation p for the unit price and
q for the monthly demand.
Answer: p=-2q+640
Solution: A data point is (q, p) = (quantity, price). So two points are (120, 400) and
(126, 388). Since the rate of change is constant, the demand curve must be a line with
12
= −6
== −2.
slope m = ∆p
∆q
Thus, p = −2q +K when K is some constant. Substitute in (120, 400) to get K = 640.
Therefore, p = −2q + 640.
(b) Find the cost function, C = C(q), for producing q tables per month.
Answer: C(q) = 125, 000 + 250q
Solution: No work is needed
(c) Find the monthly revenue function, R = R(q).
Answer: R(q) = p · q = q · (−2q + 640)
Solution: R = p · q = q · (−2q + 640) = −2q 2 + 640q
(d) Find the monthly profit (or Net Income) function P (q).
Answer: P (q) = −2q 2 + 390q − 125, 000
Solution: N I(q) = R(q) − C(q) = −2q 2 + 650q − (125, 000 + 250q) = −2q 2 + 400q −
125, 000
104/184 Quiz 1
Friday September 24
First Name:
Last Name:
Student-No:
Section:
Grade:
Very short answer questions
1. 2 marks Each part is worth 1 mark. Please write your answers in the boxes.
Marking scheme: 1 for each correct, 0 otherwise
(a) Solve 7x =
1
5
for x.
Answer: log7
Solution: 7x =
(b) Compute lim
q→3
1
5
⇒ x = log7
1
5
1
5
5
= − log
log 7
= log7 (5−x ) = − log7 5
q3 − q + 1
.
q+2
Answer: 5
Solution: lim
q→3
q3 − q + 1
q+2
lim(q 3 − q + 1)
=
q→3
lim(q + 2)
q→3
=
25
= 5.
5
Short answer questions — you must show your work
2. 4 marks Each part is worth 2 marks.
√
(a) Given that log3 x = 3, log3 y = 5, and log3 z = 1, find the value of log3
xy
.
z
Answer: 3
Solution:
√
xy
(xy)1/2
log3
= log3
= log3 (xy)1/2 − log3 z
z
z
1
= [log3 x + log3 y] − log3 z
2
1
= [3 + 5] − 1 = 3 .
2
Marking scheme: If correct answer and some work, then 2 marks. If some work in
the right direction and wrong answer, or correct answer and rubbish, then 1 mark. If
no solution, then 0.
3x2 − 7x + 2
x→2 x2 − 5x + 6
(b) Compute the limit lim
Answer: -5
Solution: Direct substitution yields 0/0, so we simplify first:
3x2 − 7x + 2
(x − 2)(3x − 1)
3x − 1
=
=
2
x − 5x + 6
(x − 2)(x − 3)
x−3
3x − 1
= −5 .
x→2 x − 3
Marking scheme: 1 for any correct factoring, 1 for answer.
Hence the limit is lim
Long answer question — you must show your work
3. 4 marks A manufacturer sells 90 tables a month at the price of $450 each. For each $9
decrease in price, he can sell 3 more tables. Their factory costs $115,000 per month to operate
and each table costs an additional $300 to make.
Note: in this problem you are ONLY setting up the equations. You do NOT have to solve for
break even values or any optimal production values.
Marking scheme: 1 for any correct answer. Only part a requires work. Answers do not need
to be simplied. If students correctly use the wrong demand curve in (c) & (d), they still get
those 2 points.
(a) Find the linear demand equation for the tables. Use the notation p for the unit price and
q for the monthly demand.
Answer: p=-3(q-90)+450=-3q+720
Solution: A data point is (q, p) = (quantity, price). So two points are (90, 450) and
(93, 441). Since the rate of change is constant, the demand curve must be a line with
9
= 450−441
= −3
= −3.
slope m = ∆p
∆q
90−93
Thus, p = −3q + K when K is some constant. Substitute in (90, 450) to get K = 720.
Therefore, p = −3q + 720.
(b) Find the cost function, C = C(q), for producing q tables per month.
Answer: C(q) = 115, 000 + 300q
Solution: No work is needed
(c) Find the monthly revenue function, R = R(q).
Answer: R(q) = q · (−3q + 720)
Solution: R = p · q = q · (−3q + 720) = −3q 2 + 720q
(d) Find the monthly profit (or Net Income) function P (q).
Answer: P (q) = −3q 2 + 420q − 115, 000
Solution: P = R − C = −3q 2 + 720q − 300q − 115, 000 = −3q 2 + 420q − 115, 000.
104/184 Quiz #1
Friday September 24
First Name:
Last Name:
Student-No:
Section:
Grade:
Very short answer questions
1. 2 marks Each part is worth 1 mark. Please write your answers in the boxes.
scheme: 1 for each correct, 0 otherwise
Marking
(a) Solve 2x = 25 for x.
Answer: 2 log2 (5)
Solution: 2x = 25 =⇒ x = log2 (2x ) = log2 (25) = log2 (52 ) = 2 log2 (5)
(b) Compute lim
r→−1
r2 −5r+1
r3 +4
.
Answer:
Solution: lim
r→−1
r2 −5r+1
r3 +4
lim
=
r→−1
(r2 −5r+1)
lim (r3 +4)
r→−1
=
7
3
(−1)2 −5(−1)+1
(−1)3 +4
=
7
3
Short answer questions — you must show your work
2. 4 marks Each part is worth 2 marks.
(a) Given that log5 x = 6, log5 y = 1, and log5 z = 3, find the value of log5
y
√
3 xz
.
Answer: −2
Solution:
log5
y
√
3
xz
= log5
y
(xz)1/3
= log5 (y) − log5 (xz)1/3
1
= log5 (y) − (log5 (x) + log5 (z))
3
1
= 1 − (6 + 3) = −2
3
Marking scheme: If correct answer and some work, then 2 marks. If some work in
the right direction and wrong answer, or correct answer and rubbish, then 1 mark. If
no solution, then 0.
2
(b) Compute the limit lim 2xx2 −2x−3
.
−2x−12
x→3
Answer: 2/5
Solution: Direct substitution yields 0/0, so we simplify first:
x2 − 2x − 3
(x − 3)(x + 1)
x+1
=
=
.
2
2x − 2x − 12
(x − 3)(2x + 4
2x + 4
2
x+1
Hence the limit is lim 2xx2 −2x−3
= lim 2x+4
= 4/10 = 2/5
−2x−12
x→3
x→3
Marking scheme: 1 for any correct factoring, 1 for answer.
Long answer question — you must show your work
3. 4 marks A manufacturer sells 200 tables a month at the price of $350 each. For each $15
decrease in price, he can sell 10 more tables. There factory costs $50,000 per month to operate
and each table costs an additional $125 to make.
Note: in this problem you are ONLY setting up the equations. You do NOT have to solve for
break even values or any optimal production values.
Marking scheme: 1 for any correct answer. Only part a requires work. Answers do not need
to be simplied. If students correctly use the wrong demand curve in (c) & (d), they still get
those 2 points.
(a) Find the linear demand equation for the tables. Use the notation p for the unit price and
q for the monthly demand.
Answer: p = − 32 q + 650
Solution: A data point is (q, p) = (quantity, price). So two points are (200, 350) and
(210, 335). Since the rate of change is constant, the demand curve must be a line with
15
= 350−335
= −10
= − 23 .
slope m = ∆p
∆q
200−210
Thus, p = − 32 q +K when K is some constant. Substitute in (200, 350) to get K = 650.
Therefore, p = − 32 q + 650.
(b) Find the cost function, C = C(q), for producing q tables per month.
Answer: C(q) = 125q + 50, 000
Solution: No work is needed.
(c) Find the monthly revenue function, R = R(q).
Answer: R(q) = − 32 q 2 + 650q
Solution: R = p · q = (− 23 q + 650) · q = − 23 q 2 + 650q.
(d) Find the monthly profit (or Net Income) function P (q) (or N I(q)).
Answer: P (q) = − 23 q 2 + 525q − 50, 000
Solution: P (q) = R(q) − C(q) = (− 32 q 2 + 650q) − (125q + 50, 000) = − 32 q 2 + 525q −
50, 000.
104/184 Quiz 1
Friday September 25
First Name:
Last Name:
Student-No:
Section:
Grade:
Very short answer questions
1. 2 marks Each part is worth 1 mark. Please write your answers in the boxes.
Marking scheme: 1 for each correct, 0 otherwise
(a) Solve 7x =
1
2
for x.
Answer: log7
Solution: 7x =
(b) Compute lim
t→1
1
2
⇒ log7 7x = log7
1
2
⇒ x = log7
1
2
1
2
2
= − log
log 7
= log7 (2−1 ) = − log7 2
3 − t2
.
1 + t3
Answer: 1
Solution: lim
t→1
3 − t2
1 + t3
lim(3 − t2 )
=
t→1
3
lim(1 + t )
t→1
=
2
= 1.
2
Short answer questions — you must show your work
2. 4 marks Each part is worth 2 marks.
(a) Given that log2 x = 5, log2 y = −1, and log2 z = 3, find the value of log2
x
.
y2z
Answer: 4
Solution:
log2
x
y2z
= log2 x − log2 y 2 − log2 z
= log2 x − 2 log2 y − log2 z
= 5 − 2(−1) − 3 = 4
Marking scheme: If correct answer and some work, then 2 marks. If some work in
the right direction and wrong answer, or correct answer and rubbish, then 1 mark. If
no solution, then 0.
2x2 − 7x − 15
x→5 x2 − 6x + 5
(b) Compute the limit lim
Answer: 13/4
Solution: Direct substitution yields 0/0, so we simplify first:
2x2 − 7x − 15
(x − 5)(2x + 3)
2x + 3
=
=
2
x − 6x + 5
(x − 1)(x − 5)
x−1
2x + 3
= 13/4 .
x→5 x − 1
Marking scheme: 1 for any correct factoring, 1 for answer.
Hence the limit is lim
Long answer question — you must show your work
3. 4 marks A manufacturer sells 100 chairs per month at the price of $520 each. For each $15
decrease in price, they can sell 5 more chairs per month. Their factory costs $125,000 per
month to operate and each chair costs an additional $370 to make.
Note: in this problem you are ONLY setting up the equations. You do NOT have to solve for
break even values or any optimal production values.
Marking scheme: 1 for any correct answer. Only part a requires work. Answers do not need
to be simplied. If students correctly use the wrong demand curve in (c) & (d), they still get
those 2 points.
(a) Find the linear demand equation for the chairs. Use the notation p for the unit price and
q for the monthly demand.
Answer: p = −3q + 820
Solution: A data point is (q, p) = (quantity, price). Two points are (100, 520) and
15
= 520−505
= −5
= −3.
(105, 505). The demand curve is a line with slope m = ∆p
∆q
100−105
Thus, p = −3q + K where K is some constant. Substitute (100, 520) to get K = 820.
Therefore, p = −3q + 820.
(b) Find the cost function, C = C(q), for producing q chairs per month.
Answer: C(q) = 125, 000 + 370q
Solution: No work is needed
(c) Find the monthly revenue function, R = R(q).
Answer: R(q) = q · (−3q + 820)
Solution: R = p · q = q · (−3q + 820) = −3q 2 + 820q
(d) Find the monthly profit (or Net Income) function P (q).
Answer: P (q) = −3q 2 + 450q − 125, 000
Solution:
P (q) = R(q) − C(q) = −3q 2 + 820q − (125, 000 + 370q) = −3q 2 + 450q − 125, 000
104/184 Quiz 1
Friday September 25
First Name:
Last Name:
Student-No:
Section:
Grade:
Very short answer questions
1. 2 marks Each part is worth 1 mark. Please write your answers in the boxes.
Marking scheme: 1 for each correct, 0 otherwise
(a) Solve 5x =
3
2
for x.
Answer: log5 (3/2) =
Solution: 5x = 3/2 ⇒ x = log5 (3/2) ⇒ x =
(b) Compute lim
z→3
log(3/2)
log 5
log(3/2)
.
log 5
z−5
.
z2 + 4
Answer:
Solution: lim
z→3
z−5
z2 + 4
lim (z − 5)
=
z→3
2
lim (z + 4)
z→3
=
−2
13
−2
−2
=
.
13
13
Short answer questions — you must show your work
2. 4 marks Each part is worth 2 marks.
(a) Given that log3 x = 7, log3 y = 5, and log3 z = 4, find the value of log3
xy
√
.
z
Answer: 10
Solution:
√
xy
1
log3 √ = log3 (xy) − log3 ( z) = log3 x + log3 y − log3 z
2
z
1
= 7 + 5 − ∗ 4 = 10 .
2
Marking scheme: If correct answer and some work, then 2 marks. If some work in
the right direction and wrong answer, or correct answer and rubbish, then 1 mark. If
no solution, then 0.
2x2 − 9x − 5
x→5 x2 − 4x − 5
(b) Compute the limit lim
Answer: 11/6
Solution: Direct substitution yields 0/0, so we simplify first:
2x2 − 9x − 5
(x − 5)(2x + 1)
2x + 1
=
=
2
x − 4x − 5
(x − 5)(x + 1)
x+1
2x + 1
= 11/6 .
x→5 x + 1
Marking scheme: 1 for any correct factoring, 1 for answer.
Hence the limit is lim
Long answer question — you must show your work
3. 4 marks A manufacturer sells 100 tables a month at the price of $400 each. For each $10
decrease in price, they can sell 5 more tables. Their factory costs $200,000 per month to
operate and each table costs an additional $275 to make.
Note: in this problem you are ONLY setting up the equations. You do NOT have to solve for
break even values or any optimal production values.
Marking scheme: 1 for any correct answer. Only part a requires work. Answers do not need
to be simplied. If students correctly use the wrong demand curve in (c) & (d), they still get
those 2 points.
(a) Find the linear demand equation for the tables. Use the notation p for the unit price and
q for the monthly demand.
Answer: p=-2q+600
Solution: A data point is (q, p) = (quantity, price). So two points are (100, 400) and
(105, 390). Since the rate of change is constant, the demand curve must be a line with
10
= 400−390
= −5
= −2.
slope m = ∆p
∆q
100−105
Thus, p = −2q + K where K is some constant. Substitute in (100, 400) to get
400 = −2(100) + K, implying K = 600.
Therefore, p = −2q + 600.
(b) Find the cost function, C = C(q), for producing q tables per month.
Answer: C(q) = 200, 000 + 275q
Solution: No work is needed
(c) Find the monthly revenue function, R = R(q).
Answer: R(q) = −2q 2 + 600q
Solution: R = p · q = q · (−2q + 600) = −2q 2 + 600q
(d) Find the monthly profit (or Net Income) function P (q).
Answer: P (q) = −2q 2 + 325q − 200, 000
104/184 Quiz 1-AD
Friday September 25
First Name:
Last Name:
Student-No:
Section:
Grade:
Very short answer questions
1. 2 marks Each part is worth 1 mark. Please write your answers in the boxes.
Marking scheme: 1 for each correct, 0 otherwise
(a) Solve 3x =
5
2
for x.
Answer: log3
Solution: 3x =
(b) Compute lim
t→2
5
2
5
2
⇒ log3 3x = log3
⇒ x = log3
5
2
5
2
3 − t2
.
1 + t3
Answer: −1/9
Solution: lim
t→2
3 − t2
1 + t3
lim(3 − t2 )
=
t→2
3
lim(1 + t )
t→2
=
−1
= 1.
9
Short answer questions — you must show your work
2. 4 marks Each part is worth 2 marks.
(a) Given that log2 x = 5, log2 y = −3, and log2 z = 3, find the value of log2
x
.
y2z
Answer: 8
Solution:
log2
x
y2z
= log2 x − log2 y 2 − log2 z
= log2 x − 2 log2 y − log2 z
= 5 − 2(−3) − 3 = 8
Marking scheme: If correct answer and some work, then 2 marks. If some work in
the right direction and wrong answer, or correct answer and rubbish, then 1 mark. If
no solution, then 0.
2x2 − 13x + 15
x→5
x2 − 6x + 5
(b) Compute the limit lim
Answer: 7/4
Solution: Direct substitution yields 0/0, so we simplify first:
2x2 − 13x + 15
(x − 5)(2x − 3)
2x − 3
=
=
2
x − 6x + 5
(x − 1)(x − 5)
x−1
2x − 3
= 7/4 .
x→5 x − 1
Marking scheme: 1 for any correct factoring, 1 for answer.
Hence the limit is lim
Long answer question — you must show your work
3. 4 marks A manufacturer sells 100 chairs per month at the price of $520 each. For each $15
decrease in price, they can sell 5 more tables per month. Their factory costs $125,000 per
month to operate and each chair costs an additional $370 to make.
Note: in this problem you are ONLY setting up the equations. You do NOT have to solve for
break even values or any optimal production values.
Marking scheme: 1 for any correct answer. Only part a requires work. Answers do not need
to be simplied. If students correctly use the wrong demand curve in (c) & (d), they still get
those 2 points.
(a) Find the linear demand equation for the chairs. Use the notation p for the unit price and
q for the monthly demand.
Answer: p = −3q + 820
Solution: A data point is (q, p) = (quantity, price). Two points are (100, 520) and
15
= 520−505
= −5
= −3.
(105, 505). The demand curve is a line with slope m = ∆p
∆q
100−105
Thus, p = −3q + K where K is some constant. Substitute (100, 520) to get K = 820.
Therefore, p = −3q + 820.
(b) Find the cost function, C = C(q), for producing q chairs per month.
Answer: C(q) = 125, 000 + 370q
Solution: No work is needed
(c) Find the monthly revenue function, R = R(q).
Answer: R(q) = q · (−3q + 820)
Solution: R = p · q = q · (−3q + 820) = −3q 2 + 820q
(d) Find the monthly profit (or Net Income) function P (q).
Answer: P (q) = −3q 2 + 450q − 125, 000
Solution:
P (q) = R(q) − C(q) = −3q 2 + 820q − (125, 000 + 370q) = −3q 2 + 450q − 125, 000