MCV4U1 Grade 12 Calculus and Vectors
APPLICATION OF DERIVATIVES
REVIEW
1
1. Determine f' (x) and f'' (x) if f (x) = x4 – 4.
𝑥
Answer:
4
f' (x) = 4x3 + 5.
𝑥
f'' (x) = 12x2 –
20
𝑥6
.
𝑑2 𝑦
2. For y = x9 – 7x3 + 2, find 2 .
𝑑𝑥
Answer:
𝑑𝑦
= 8x8 – 21x2
𝑑𝑥
𝑑2 𝑦
𝑑𝑥 2
= 72x7 – 42x
3. Calculate each of the following:
a) f'' (2) if f (x) = 5x3 – x
Answer:
f' (x) = 15x2 – 1
f'' (x) = 30x
f'' (2) = 60.
b) f'' (–1) if f (x) = –2x –3 + x2
Answer:
f' (x) = 6x –4 + 2x
f'' (x) = –24 x –5 + 2
f'' (–1) = –24 (–1)–5 + 2 = 24 + 2 = 26.
c) f'' (0) if f (x) = (4x – 1)4
Answer:
f' (x) = 4(4x – 1)3(4x – 1)' = 16(4x – 1)3
f'' (x) = 48(4x – 1)2(4x – 1)' = 192(4x – 1)2
f'' (0) = 192(0 – 1)2 = 192.
2𝑥
d) f'' (1) if f (x) =
𝑥−5
Answer:
f' (x) =
(2𝑥)′ (𝑥 − 5)− (2𝑥)(𝑥 − 5)′
(𝑥 − 5)2
20
f'' (x) = (𝑥
f'' (1) = (1
− 5)3
20
= – 5/16
− 5)3
e) f'' (4) if f (x) = √𝑥 + 5
Answer:
1
f' (x) = (x + 5) –0.5
2
1
f'' (x) = – (x + 5) –1.5
4
1
f'' (4) = – (9) –1.5 = –1/108
4
3
f) f'' (8) if f (x) = √𝑥 2
Answer:
f (x) = 𝑥
2
3
1
2
f' (x) = 𝑥 −3
3
2
f'' (x) = – 𝑥
9
2
4
−3
−
f'' (8) = – (8)
9
4
3
= –1/72
=
2(𝑥 − 5) − 2𝑥
(𝑥 − 5)2
= –(𝑥
10
− 5)2
4. Determine the velocity and acceleration of an object that moves along a
straight line in such a way that its position is s (t) = t2 + (2t – 3)0.5.
Answer:
1
−2
3
−2
v = 2t + (2𝑡 − 3) , a = 2 – (2𝑡 − 3)
5. Determine the velocity and acceleration as functions of time, t, for, s (t)
5
= t – 7 + , t ≠ 0.
𝑡
Answer:
v (t) = 1 – 5t – 2, a (t) = 10 t – 3
6. The position of an object moving along a straight line is described by
the function s (t) = – t3 + 4t2 – 10 for t ≥ 0. Is the object moving toward or
away from its starting position when t = 3?
Answer:
s (0) = – 10
s (3) = – (3)3 + 4(3)2 – 10 = –1
v (t) = s' (t) = –3t2 + 8t
v (3) = –3 < 0
s (0) < s (3) and v (3) < 0 ∴ the object is moving away from the origin and
towards its starting position
7. An object moves along a straight line. The object's position at time t is
given by s (t). Find the position, velocity, and acceleration at the specified
time.
2𝑡
a) s (t) =
, t = 3.
𝑡+3
Answer:
2(3)
s (3) =
=1
3+3
v (t) = s' (t) =
v (3) = (3
6
+ 3)2
(2𝑡)′ (𝑡 + 3)− (2𝑡)(𝑡 + 3)′
(𝑡
= 1/ 6
+ 3)2
=
2(𝑡 + 3) − 2𝑡
(𝑡 + 3)2
= (𝑡
6
+ 3)2
a (t) = v' (t) = – (𝑡
a (3) = – (3
12
+ 3)3
12
+ 3)3
= –1/18
5
b) s (t) = t +
, t = 1.
𝑡+2
Answer:
5
s (1) = 1 +
= 8/ 3
1+2
v (t) = s' (t) = 1 – (𝑡
v (1) = 1 – (1
a (t) = v' (t)
a (1) = (1
10
5
5
+ 2)2
= 4/9
+ 2)2
10
= (𝑡
+ 2)3
+ 2)3
= 10/27
8. A particle moving along a straight line will be s centimetres from a
16
fixed point at time t seconds, where t > 0 and s = 27t3 + + 10.
𝑡
a) Determine when the velocity will be zero.
Solution
16
v (t) = s' (t) = 81t2 – 2
v (t) = 0 ⇒ 81t –
2
16
𝑡2
𝑡
= 0 ⇒ t4 =
16
81
⇒t=
b) Is the particle accelerating? Explain.
32
a (t) = v' (t) = 162t + 3 > 0 for all t > 0.
𝑡
∴ the particle is accelerating
2
3
9. Sam applies the brakes steadily to stop his car, which is travelling at 20
m/s. The position of the car, s, in metres at t seconds, is given by s (t) =
20t – 0.3t3. Determine:
a) the stopping time
s (t) = 20t – 0.3t3
v (t) = s' (t) = 20 – 0.9t2
200
10√2
v (t) = 0 ⇒ t2 =
⇒t=
≈ 4.7 s
9
3
b) the stopping distance
Solution
s = 20(4.7) – 0.3(4.7)3 ≈ 62.9 m
c) the deceleration at 2 s
Solution
a (t) = v' (t) = – 1.8t
a (2) = –3.6 m/s2
The deceleration at 2 s is 3.6 m/s2
10. A motorist starts braking for a stop sign. After t seconds, the distance,
in metres, from the front of the car to the sign is s (t) = 62 – 16t + t2
a) How far was the front of the car from the sign when the driver started
braking?
Solution
s (0) = 62 m ∴ the car was 62 m from the sign when the driver started
braking.
b) Does the car go beyond the stop sign before stopping?
Solution
v (t) = s' (t) = 2t – 16
v (t) = 0 ⇒ 2t – 16 = 0 ⇒ t = 8 s.
s (2) = 62 – 16(8) + (8)2 = –2 m.
Yes, 2 m beyond the stop sign
c) Explain why it is unlikely that the car would hit another vehicle that is
travelling perpendicular to the motorist's road when the car first comes to a
stop at the intersection.
Answer:
Stop signs are located two or more metres from an intersection. Since the
car only went 2 m beyond the stop sign, it is unlikely the car would hit
another vehicle travelling perpendicular.
11. A pellet is shot into the air. Its position above the ground at any time, t,
is defined by s (t) = 45t – 5t2 m. For what values of t, t ≥ 0, is the upward
velocity of the pellet positive? For what values of t is the upward velocity
zero and negative? Draw a graph to represent the velocity of the pellet.
Answer:
v (t) = 45 – 10t
The upward velocity is positive for 0 ≤ t ≤ 4.5 s, zero for t = 4.5 s, and
negative for t > 4.5 s.
2
12. The function s (t) = (𝑡 2 + 𝑡)3 , t ≥ 0, represents the displacement s, in
metres, of a particle moving along a straight line after t seconds.
a) Determine v(t) and a(t).
Solution
2
1
v (t) = s' (t) = (𝑡 2 + 𝑡)−3 (t2 + t)'
3
2
1
−3
2
= (𝑡 + 𝑡)
3
2
−
(2t + 1)
1
3
′
2
1
3
4
1
a (t) = v' (t) = ((𝑡 + 𝑡) ) (2t + 1) + (𝑡 2 + 𝑡)−3 (2t + 1)'
2
3
4
−3
2
= – (𝑡 2 + 𝑡)
9
2
=
4
−3
(𝑡 2 + 𝑡)
9
(2t + 1)2 + (𝑡 2 + 𝑡)−3
3
[– (2t + 1)2 + 6(t2 + t)]
4
2
= (𝑡 2 + 𝑡)−3 (2t2 + 2t – 1)
9
b) Find the average velocity during the first 5 s.
Solution
s (0) = 0
2
3
s (5) = (52 + 5)3 = √900 ≈ 9.655
𝑠 (5)− 𝑠 (0)
9.655 − 0
vav =
≈
= 1.931 m/s
5
5
c) Determine the velocity at exactly 5 s.
Solution
2
1
−3
v (5) = (30)
3
(11) ≈ 2.36 m/s
d) Find the average acceleration during the first 5 s.
Solution
aav =
𝑣 (5)− 𝑣 (0)
5
2
1
−3
2
v (0) = (0 + 0)
3
(0 + 1) =
2
3
3 √0
= undefined
∴ aav is undefined
e) Determine the acceleration at exactly 5 s.
Solution
2
a (5) = (30)
9
4
−3
(59) ≈ 0.141 m/s
13. Determine the maximum and minimum of each function on the given
interval.
a) f (x) = 2x3 – 9x2, – 2 ≤ x ≤ 4
Solution
f' (x) = 6x2 – 18x = 6x (x – 3)
f' (x) = 0 ⇒ x = 0 ∈ [– 2, 4] or x = 3 ∈ [– 2, 4]
f (– 2) = 2(– 2)3 – 9(– 2)2 = –16 – 36 = –52
f (0) = 0
f (3) = 2(3)3 – 9(3)2 = 54 – 81 = –27
f (4) = 2(4)3 – 9(4)2 = 128 – 144 = –16
∴ min f (x) = f (– 2) = –52, max f (x) = f (0) = 0
–2≤𝑥≤4
–2≤𝑥≤4
b) f (x) = 12x – x3, x ∈ [–3, 5]
Solution
f' (x) = 12 – 3x2 = 3(4 – x2) = 3(2 – x)(2 + x)
f' (x) = 0 ⇒ x = –2 ∈ [–3, 5] or x = 2 ∈ [–3, 5]
f (–3) = 12(–3) – (–3)3 = –36 + 27 = –9
f (–2) = 12 (–2) – (–2)3 = –24 + 8 = –16
f (2) = 12(2) – (2)3 = 24 – 8 = 16
f (5) = 12(5) – (5)3 = 60 – 125 = –65
∴ min f (x) = f (5) = –65, max f (x) = f (2) = 16
–3≤𝑥≤5
–3≤𝑥≤5
c) f (x) = 2x +
Solution
18
𝑥
,1≤x≤5
2(𝑥 2 − 9)
18
2(𝑥 − 3)(𝑥 + 3)
f' (x) = 2 – 2 =
=
𝑥
𝑥2
𝑥2
f' (x) = 0 ⇒ x = –3 ∉ [1, 5] or x = 3 ∈ [1, 5]
18
f (1) = 2(1) + = 20
f (3) = 2(3) +
1
18
3
18
= 12
f (5) = 2(5) + = 13.6
5
min f (x) = f (3) = 12,
1≤𝑥≤5
max f (x) = f (1) = 20
1≤𝑥≤5
14. The position function of an object that moves in a straight line is s (t) =
8
1 + 2t – 2 , 0 ≤ t ≤ 2. Calculate the maximum and minimum velocities
𝑡 +1
of the object over the given time interval.
Solution
8
16 𝑡
v (t) = s' (t) = 2 + (𝑡 2 2 ∙ (t2 + 1)' = 2 + (𝑡 2 2 > 0 for all t > 0
+ 1)
+ 1)
∴ v (t) is increasing on [0, 2]
∴ min v (t) = v (0) = 2, max v (t) = v (2) = 3.28
0≤𝑡≤2
0≤𝑡≤2
15. Suppose that the cost, in dollars, of manufacturing x items is
approximated by C (x) = 625 + 15x + 0.01x2 for 1 ≤ x ≤ 500. The unit cost
𝐶 (𝑥)
(the cost of manufacturing one item) would then be U (x) =
. How
𝑥
many items should be manufactured to ensure that the unit cost is
minimized?
Solution
625
U (x) =
+ 15 + 0.01x
𝑥
U' (x) = –
625
𝑥2
U' (x) = 0 ⇒ –
U (1) =
625
1
U (250) =
+ 0.01
625
𝑥2
+ 0.01 = 0 ⇒ x2 = 62500 ⇒ x = 250
+ 15 + 0.01(1) = 640.01
625
250
625
+ 15 + 0.01(250) = 2.5 + 15 + 2.5 = 20
U (500) =
+ 15 + 0.01(500) = 1.25 + 15 + 5 = 21.25
500
∴ min U (x) = U (250) = 20
1 ≤ 𝑥 ≤ 500
250 items should be manufactured to ensure that the unit cost is
minimized.
15. Find the production level that minimizes the average cost per unit for
the cost function C (x) = 0.004x2 + 40x + 16000.
Solution
The unit cost (the cost of manufacturing one item) would then be
𝐶 (𝑥)
16000
U (x) =
= 0.004x + 40 +
𝑥
U' (x) = 0.004 –
𝑥
16000
𝑥2
16000
U' (x) = 0 ⇒ 0.004 –
x
0 < x < 2000
–
U' (x)
U (x)
↘
𝑥2
= 0 ⇒ x2 = 4000000 ⇒ x = 2000
2000
x > 2000
0
+
min
↗
The production level that minimizes the average cost per unit is 2000.
16. A box with a square base and no top must have a volume of l0 000
cm3. If the smallest dimension is 5 cm, determine the dimensions of the
box that minimize the amount of material used.
Solution
Let x cm represent the side length of the
square base and y cm represent the height
of the box. Then the volume of the box
would be V = x2y and the surface area
would be A = x2 + 4xy.
We get:
10 000
V = l0 000 ⇒ x2y = l0 000 ⇒ y = 2
y≥5⇒
10 000
𝑥2
𝑥
≥ 5 ⇒ x ≤ 2000 ⇒ x ≤ 20√5 ≈ 44.72 cm
2
∴ 5 ≤ x ≤ 20√5
40 000
3
A' (x) = 0 ⇒ 2x – 2 = 0 ⇒ x3 = 20 000 ⇒ x = 10 √20 ≈ 27.14 cm
𝑥
x
5 ≤ x < 27.14
–
A' (x)
A'(x)
↘
27.14
0
min
27.14 < x ≤ 44.72
+
↗
Thus, the dimensions of the box that minimize the amount of material used
10 000
are x = 27.14 cm and y =
≈ 13.75 cm
27.14 2
Answer: 27.14 cm by 27.14 cm for the base and height 13.57 cm
17. An animal breeder wishes to create five adjacent rectangular pens,
each with an area of 240 m2. Find the dimensions required for the pens to
keep the amount of fencing used to a minimum.
Solution
xm
xm
ym
xm
ym
ym
xm
xm
ym
xm
ym
y m = length of the pens, y m = width of the pens
240
xy = 240 ⇒ y =
𝑥
The length of the fencing equals:
2400
2400
L = 6x + 10y = 6x +
⇒ L (x) = 6x +
L' (x) = 6 –
𝑥
2400
𝑥2
L' (x) = 0 ⇒ 6 –
L' (10) = 6 –
2400
2400
100
𝑥2
= 0 ⇒ x2 = 400 ⇒ x = 20 m.
= –18 < 0, L' (40) = 6 –
2400
𝑥
2400
1600
= 4.5 > 0
L (20) = 6(20) +
= 120 + 120 = 240 m.
20
x
0 < x < 20
20
x > 20
–
0
+
L' (x)
L (x)
240
↘
↗
min
Therefore, the dimensions required for the pens to keep the amount of
240
fencing used to a minimum are x = 20 m and y =
= 12 m.
20
Answer: length 20 m, width 12 m
18. You are given a piece of sheet metal that is twice as long as it is wide
and has an area of 800 m2. Find the dimensions of the rectangular box that
would contain a maximum volume if it were constructed from this piece of
metal by cutting out squares of equal area at all four corners and folding
up the sides. The box will not have a lid. Give your answer correct to one
decimal place.
Solution
Determine the dimensions of the piece of sheet metal.
A = l × w, A = 800 m2, l = 2w ⇒ 800 = 2w2 ⇒ w = 20 m, l = 40 m
Let x cm be the length of the sides of squares cut out from the piece of
sheet metal. Then the volume of the box would be:
V = x (20 – 2x)(40 – 2x)
= 4x (10 – x)(20 – x)
= 4 (x3 – 30x2 + 200 x)
∴ V (x) = 4 (x3 – 30x2 + 200 x),
0 < x < 10.
V ' (x) = 4 (3x2 – 60x + 200)
V ' (x) = 0 ⇒3x2 – 60x + 200 = 0
x=
30 ± 10√3
3
30 − 10√3
0 < x < 10 ⇒ x =
≈ 4.2 m
3
x
0 < x < 4.2
4.2
4.2 < x < 10
+
0
–
V ' (x)
V (x)
max
↗
↘
∴ max V (x) = V (4.2)
0 < 𝑥 < 10
20 – 2x ≈ 20 – 2(4.2) = 12.6 m
40 – 2x ≈ 40 – 2(4.2) = 32.6 m
Answer: the dimensions of the rectangular box are 31.6 cm by 11.6 cm by
4.2 cm
19. A cylindrical can needs to hold 500 cm3 of apple juice. The height of
the can must be between 6 cm and 15 cm, inclusive. How should the can
be constructed so that a minimum amount of material will be used in the
construction? (Assume that there will be no waste.)
Solution
Let r cm represent the radius of the can, h cm represent its height.
500
V = πr2h ⇒ 500 = πr2h ⇒ h = 2
6 ≤ h ≤ 15 ⇒ 6 ≤
⇒
100
π𝑟
1
500
π𝑟
2 ≤ 15 ⇒
≤ r2 ≤
3𝜋
250
3𝜋
15
⇒√
π𝑟 2
≤
500
100
3𝜋
≤
1
6
≤ r ≤√
250
3𝜋
⇒ 3.26 ≤ r ≤ 5.15
The surface area of the can is:
500
1000
A = 2𝜋r2 + 2πrh = 2𝜋r2 + 2πr ∙ 2 = 2𝜋r2 +
A (r) = 4𝜋r –
π𝑟
1000
r
𝑟2
A (r) = 0 ⇒ 4𝜋r –
1000
𝑟2
3
=0⇒r= √
250
𝜋
≈ 4.30 cm.
A (3.26) = 373.52 cm2
A (4.30) = 348.73 cm2
A (5.15) = 360.82 cm2
∴
min
A (r) = A (4.30) = 348.73 cm2
3.26 ≤ 𝑟 ≤ 5.15
∴ r ≈ 4.3 cm, h =
500
π(4.30)2
≈ 8.6 cm
Answer: radius 4.3 cm, height 8.6 cm
20. In oil pipeline construction, the cost of pipe to go underwater is 60%
more than the cost of pipe used in dry-land situations. A pipeline comes to
a river that is 1 km wide at point A and must be extended to a refinery, R,
on the other side, 8 km down the river. Find the best way to cross the river
(assuming it is straight) so that the total cost of the pipe is kept to a
minimum. (Give your answer correct to one decimal place.)
Solution
Let the length of pipe used in dry-land situations be x km, 0 ≤ x ≤ 8, then
the length of pipe to go underwater would be
BR = √(8 − 𝑥)2 + 1 = √𝑥 2 − 16𝑥 + 65 km.
Assume that the cost of pipe used in dry-land situations is 1 unit. Then the
cost of pipe to go underwater is 1.6 units. The total cost of the oil pipeline
construction is
C(x) = x + 1.6√𝑥 2 − 16𝑥 + 65 km
Determine the value of x so that the total cost of the pipe is kept to a
minimum.
𝑥−8
C' (x) = 1 + 1.6 ∙ 2
√𝑥 − 16𝑥 + 65
C' (x) = 0 ⇒ √𝑥 2 − 16𝑥 + 65 = 1.6 (8 – x)
5√𝑥 2 − 16𝑥 + 65 = 8 (8 – x)
25(x2 –16x + 65) = 64(64 –16x + x2)
25x2 – 400x + 1625 = 4096 –1024x + 64x2
39x2 – 624x + 2471 = 0
624 ± √389376−4(39)(2471)
624 ± 62.45
x=
≈
78
78
x ≈ 7.2 or x ≈ 7.2
0 ≤ x ≤ 8 ⇒ x ≈ 7.2 km
C (0) = 1.6√65 ≈ 12.9
C (8) = 8 + 1.6√𝑥 2 − 16𝑥 + 65 ≈ 9.6
C (7.2) = 7.2+ 1.6√7.22 − 16(7.2) + 65 ≈ 9.2
∴ the total cost of the pipe is kept to a minimum when x ≈ 7.2 km.
Answer: Run the pipe 7.2 km along the river shore and then cross
diagonally to the refinery.
21. A train leaves the station at l0:00 p.m. and travels due north at a speed
of 100 km/h. Another train has been heading due west at 120 km/h and
reaches the same station at 11:00 p.m. At what time were the two trains
closest together?
Solution
Let t = 0 h at l0:00 p.m. Introduce a coordinate system with the origin at
the station, x-axis is directed due east, and y-axis is directed due north.
The first train starts at point A (0, 0) and at time t h he is in the point B(0,
100t). The second train starts at point C (120, 0) and at time t h he is in the
point D (120 – 120t, 0). The square of the distance between the two trains
at time t h equals
f (t) = (120 – 120t)2 + (100t)2 = 14400(t – 1)2 + 10000t2
Determine the value of t that minimizes f (t).
f '(t) = 28800(t – 1) + 20000t = 48800t – 28800
f '(t) = 0 ⇒ t = 288/488 ≈ 0.59 h ≈ 35 min
t
0 < t < 0.59 0.59
t > 0.59
–
0
+
f '(t)
f (t)
min
↘
↗
∴ f (t) is minimal when t ≈ 0.59 h ≈ 35 min
∴ the two trains are closest together at 10:35 p.m.
Answer: 10:35 p.m.
Try yourself:
22. A boat leaves a dock at 2:00 p.m., heading west at 15 km/h. Another
boat heads south at 12 km/ h and reaches the same dock at 3:00 p.m. When
were the boats closest to each other?
Answer: 2:23 p.m.
23. A store sells portable MP3 players for $100 each and, at this price,
sells 120 MP3 players every month. The owner of the store wishes to
increase his profit, and he estimates that, for every $2 increase in price of
MP3 prayers, one less MP3 prayer will be sold each month. If each MP3
prayer costs the store $70, at what price should the store sell the MP3
players to maximize profit?
Solution
Let n be the number of times the price of the portable MP3 players was
increased by $2. Then the new price of the portable MP3 players is $(100
+ 2n) and the number of the MP3 players sold is (120 – n). The profit
equals
P (n) = (100 + 2n) (120 – x) – 70(120 – x)
= (120 – n)(30 + 2n)
= –2n2 + 210n + 2600
Determine the value of n that maximizes the profit.
P' (n) = – 4n + 210
P' (n) = 0 ⇒ – 4n + 210 = 0 ⇒ n = 52.5
n
0 < t < 52.5 52.5
t > 52.5
–
0
+
P' (n)
P(n)
min
↘
↗
Since n is a whole number, we get: n = 52 or n = 53.
If n = 52 then the new price is 100 + 2(52) = $204
If n = 53 then the new price is 100 + 2(53) = $206
Answer: $204 or $206
24. An offshore oil well, P, is located in the ocean 5 km from the nearest
point on the shore, A. A pipeline is to be built to take oil from P to a
refinery that is 20 km along the straight shoreline from A. If it costs $100
000 per kilometre to lay pipe underwater and only $75 000 per kilometre
to lay pipe on land, what route from the well to the refinery will be the
cheapest? (Give your answer correct to one decimal place.)
Solution
Assume that the pipeline meets the shore at a point C, x km from point A.
Then the cost of the pipe line would be
C (x) = 100 000√𝑥 2 + 25 + 75 000 (20 – x)
Determine the value of x that minimizes the cost.
100 000𝑥
C' (x) = 2
– 75000
√𝑥 + 25
4𝑥
C' (x) = 0 ⇒
√𝑥 2 + 25
– 3 = 0 ⇒ 3√𝑥 2 + 25 = 4x
225
⇒ 9(x2 + 25) = 16x2 ⇒ x2 =
⇒ x ≈ 5.7 km
7
x
5.7
x > 5.7
0 ≤ x < 5.7
C' (x)
–
0
+
C (x)
min
↘
↗
∴ x ≈ 5.7 km
Answer:
The pipeline meets the shore at a point C, 5.7 km from point A, directly
across from P.
25. The printed area of a page in a book will be 81 cm2. The margins at the
top and bottom of the page will each be 3 cm deep. The margins at the
sides of the page will each be 2 cm wide. What page dimensions will
minimize the amount of paper?
Solution
Let x cm be the length of the printed area, then the
81
height of the printed area would be cm, and the
𝑥
area of the page would be
324
A = (x + 4) (81/x + 6) = 6x +
+ 105
324
𝑥
∴ A (x) = 6x +
+ 105
𝑥
Determine the value of x that minimizes the amount
of paper.
324
A' (x) = 6 – 2
𝑥
324
A' (x) = 0 ⇒ 6 – 2 = 0 ⇒ x2 = 54 ⇒ x ≈ 7.35 cm
𝑥
x
t > 7.35
0 ≤ x < 7.35 7.35
–
0
+
A' (x)
A (x)
min
↘
↗
∴ x ≈ 7.35 cm
The page dimensions that minimize the amount of paper are x + 4 ≈11.35
81
cm and + 6 ≈ 17.02 cm
𝑥
Answer: 11.35 cm by 17.02 cm
26. A rectangular rose garden will be surrounded by a brick wall on three
sides and by a fence on the fourth side. The area of the garden will be
1000 m2. The cost of the brick wall is $ 192/m. The cost of the fencing is $
48/m. Find the dimensions of the garden so that the cost of the materials
will be as low as possible.
Solution
Let x m be the length of the garden, y m be its width. We get:
1000
A = xy ⇒ y =
𝑥
The cost of the materials equals
384000
C = 192(x + 2y) + 48x = 240x + 384y = 240x +
.
𝑥
Determine the value of x which minimizes C(x) = 240x +
C' (x) = 240 –
384000
𝑥2
C' (x) = 0 ⇒ 240 –
384000
x
0 ≤ x < 40
C' (x)
–
C (x)
↘
1000
𝑥2
= 0 ⇒ x2 = 1600 ⇒ x = 40 m
40
0
min
∴ x = 40 m, y =
= 25 m
40
Answer: 40m by 25 m
x > 40
+
↗
384000
𝑥
.
27. Two towns, Ancaster and Dundas, are 4 km and 6 km, respectively,
from an old railroad line that has been made into a bike trail. Points C and
D on the trail are the closest points to the two towns, respectively. These
points are 8 km apart. Where a rest stop should be built to minimize the
length of new trail that must be built from both towns to the rest stop?
Solution
Let BC = x km, then BD = 8 – x km.
The length of the new trail that must be built from both towns to the rest
stop equals:
L (x) = √𝑥 2 + 16 + √(8 − 𝑥 )2 + 36
= √𝑥 2 + 16 + √𝑥 2 − 16𝑥 + 100, 0 ≤ x ≤ 8.
Determine the value of x ∈ [0, 8] that minimizes L (x).
𝑥
𝑥−8
L'(x) = 2
+ 2
√𝑥 + 16
L'(x) = 0 ⇒
𝑥
√𝑥 − 16𝑥+100
𝑥−8
√𝑥 2 + 16
+
√𝑥 2 − 16𝑥+100
= (8 – x) √𝑥 2
=0
x √𝑥 2 − 16𝑥 + 100
+ 16
Square both sides:
x2 (x2 –16x + 100) = (8 – x)2 (x2 + 16)
x4 – 16x3 + 100x2 = (64 – 16x + x2) (x2 + 16)
x4 – 16x3 + 100x2 = x4 – 16x3 + 80x2 – 256x + 1024
20x2 + 256x – 1024 = 0
5x2 + 64x – 256 = 0
x = 3.2 ∈ [0, 8] or x = –16 ∉ [0, 8]
L (0) = 4 + 10 = 14 km,
L (8) = √80 + 10 ≈ 18.9 km
L (3.2) = √(3.2)2 + 16 + √(3.2)2 − 16(3.2) + 100
≈ 5.12 + 7.68 = 12.8 km
∴ the value of x ∈ [0, 8] that minimizes L (x) is x = 3.2.
Answer: 3.2 km from point C