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**Unformatted text preview: **University of Toronto Scarborough
Department of Computer & Mathematical Sciences
MAT A32H Winter 2018
Solutions #5 1. Section 11.3
16. If the cost function is c = c(q) = 0.1q2 + 3q + 2, the marginal cost function is
dc dc
= c ′ (q) = 0.2q + 3. When q = 3,
= c ′ (3) = (0.2)(3) + 3 = 3.6. dq
dq q=3 2000
, the cost function is c = (c)(q) =
q
5 q + 2000 and the marginal cost function is c ′ (q) = 5 for all q > 0. Hence
c ′ (25) = c ′ (250) = 5.
1 2
1
q = 15q −
q , the marginal
24. If the revenue function is r(q) = q 15 −
30
30
q
5
44
dr
= r ′ (q) = 15 − . Now r ′ (5) = 15 −
=
, r ′ (15) =
cost function is
dq
15
15
3
15
150
15 −
= 14 and r ′ (150) = 15 −
= 5.
15
15
32. If the cost function is c = c(q) = 0.4q 2 + 4q + 5, the rate of change of c with
dc
= c ′ (q) = 0.8 q + 4. If q = 2, we have c ′ (2) = (0.8)(2) + 4 = 5.6.
respect to q is
dq
△c
c(3) − c(2)
20.6 − 14.6
Over the interval [2, 3] we have
=
=
= 6.
△q
3−2
1
36. We are given that y = 5 − 3x3 and x = 1.
dy
(a) The rate of change of y with respect to x is
= y ′ = −9 x2 .
dx
y′
1 dy
−9 x2
(b) The relative rate of change of y with respect to x is
=
=
=
y
y dx
5 − 3x3
9
.
3x3 − 5
(c) The rate of change of y with respect to x when x = 1 is y ′ (1) = −9(1)2 =
−9.
′
y
(1) =
(d) The relative rate of change y with respect to x when x = 1 is
y
−9(1)2
9
y ′ (1)
=
= − = −4.5.
3
y(1)
5 − 3(1)
2
(e) The
percentage rate of change of y with respect to x when x = 1 is
y′
(1) × 100% = −450 %.
y
20. If the average cost function is c = 5 + MATA32H page 2 Solutions # 5
dc dc
dq 1
1
dq
which is given to be ; i.e,
=
=⇒
46. The relative rate of change of c is
c
c
q
q
dc
dc
c
and the average cost
=
= c. Hence the marginal cost function
dq
q
dq
function (c) are equal.
2. Section 11.4
2. f (x) = (3x − 1)(7x + 2),
f (x) = (3)(7x + 2) + (3x − 1)(7) = 42x − 1.
′ 4. Q(x) = (x2 + 3x)(7x2 − 5),
Q (x) = (2x + 3)(7x2 − 5) + (x2 + 3x)(14x) = 28x3 + 63x2 − 10x − 15.
′ C(I) = (2I 2 − 3)(3I 2 − 4I + 1),
C ′ (I) = (4I)(3I 2 − 4I + 1) + (2I 2 − 3)(6I − 4) = 24I 3 − 24I 2 − 14I + 12 =
2(12I 3 − 12I 2 − 7I + 6).
√
√
√
1/2
3
x
−
3
x) = (x1/2 + 5x − 2)(x1/3
16. g(x) =
( x + 5x − 2)(
− 3x ),
1 −1/2
1 −2/3 3 −1/2
′
1/3
1/2
1/2
g (x) =
=
x
+ 5 (x − 3x ) + (x + 5x − 2)
x
− x
2
3
2
1
(−135x1/2 + 40x1/3 + 5x−1/6 + 18x−1/2 − 4x−2/3 − 18).
6
2x − 3
,
20. y =
4x + 1
2(4x + 1) − 2x − 3)(4)
14
y′ =
=
.
2
(4x + 1)
(4x + 1)2
x−5
36. y = √ ,
8 x
√
√
(1)(8 x) − (x − 5) √4x
x + √5x
x+5
′
y =
=
=
.
64x
16x
16 x3/2
x−5
1 1/2
Alternate solution: y = √ =
x − 5x−1/2 ,
8
8 x
1
1
x+5
1
5
1
5
−1/2
−3/2
′
=
=
x
+ x
y =
+
.
8 2
2
16 x1/2 x3/2
16 x3/2
6. 42. 48. 27x2 + 15x − 2
(9x − 1)(3x + 2)
=
,
4 − 5x
4 − 5x
(54x + 15)(4 − 5x) − (27x2 + 15x − 2)(−5)
−135x2 + 216x + 50
y′ =
=
.
(4 − 5x)2
(4 − 5x)2
y= 1
+
x−1 + a−1
x
=
1
−1
−1
x −a
−
x
(1)(a − x) − (a + x)(−1)
2a
f ′ (x) =
=
.
2
(a − x)
(a − x)2 For a 6= 0 a constant, f (x) = 1
a
1
a = a+x
,
a−x MATA32H Solutions # 5 page 3 √
√
20 I + 0.5 I 3 − 0.4I
√
68. C =
I +5
The marginal propensity to consume is
√
√
√
√
1
10
3
√
√
( I + 0.75 I − 0.4)( I + 5) − (20 I + 0.5 I − 0.4I) 2 I
dC
√
.
=
dI
( I + 5)2 d C When I = 100, we have
≈ 0.393.
d I I=100
dC
so, when I = 100, we have
The marginal propensity to save is 1 −
dI d S ≈ 1 − 0.393 ≈ 0.607.
d I I=100
√
√
√
( I + 2)( I − 4) √
I −2 I −8
√
=
= I −4
70. S = √
I +2
I +2
The marginal propensity to save is
dS
1
1
= I −1/2 = √ .
dI
2
2 I 1
d S = √
≈ 0.04082.
When I = 150, we have d I I=150 2 150
1
The marginal propensity to consume is 1 − √ so, when I = 150, we have
2 I d C ≈ 1 − 0.04082 ≈ 0.9592.
d I I=150
72. 3. dc
d
(c) = 0. Hence 0 =
=
Given a cost function c = f (q) we assume
dq
dq
q dd qc − (c)(1)
dc
dc
c
dc
d c
=
− c = 0 =⇒ q
= c =⇒
= = c.
=⇒ q
2
dq q
q
dq
dq
dq
q
Therefore, the marginal cost function is the same as the average cost, as required. Section 11.5
y = 2u3 − 8u and u = 7x − x3 .
d y Chain d y d u
·
= (6u2 − 8)(7 − 3x2 ) = (6(7x − x3 )2 − 8)(7 − 3x2 ) = 2(3x6 −
=
d x Rule d u d x
42x4 + 147x2 − 4)(7 − 3x2 ).
√
4. y = 4 z and z = x5 − x4 + 3.
1
5x4 − 4x3
5x4 − 4x3
d y Chain d y d z
p
·
= z −3/4 (5x4 −4x3 ) =
=
.
=
d x Rule d z d x
4
4 (x5 − x4 + 3)3/4
4 4 (x5 − x4 + 3)3
√
2
6. z = u2 + u + 9 and
u = 2s −1.
1
d z Chain d z d u
·
= 2u + √
(4s).
=
d s Rule d u d s
2 u
1
d z 2
= 2+
(−4) = −10.
When s = −1, u = 2(−1) − 1 = 1 and
d s s=−1
2 2. MATA32H page 4 Solutions # 5 √
y = 3x2 − 7 = (3x2 − 7)1/2 ,
1
3x
y ′ = (3x2 − 7)−1/2 (6x) = √
.
2
3x2 − 7
√
3
22. y = 8x2 − 1 = (8x2 − 1)1/3 ,
16 x
1
.
y ′ = (8x2 − 1)−2/3 (16x) =
3
3 (8x2 − 1)2/3
−2/3
3
2
30. y =
=
3
3x
−
x
,
2/3
2
(3x
−x)
−5/3
−2 (6x − 1)
2
.
3x2 − x
(6x − 1) =
y′ = 3 −
3
(3x2 − x)5/3
√
36. y = 4 x3 1 − x2 = 4 x3 (1 − x2 )1/2 ,
20. y ′ = 12 x2 (1 − x2 )1/2 + 2x3 (1 − x2 )−1/2 (−2x) = 12 x2 √ 4 x2 (3 − 4x2 )
√
.
1 − x2
4
2x
42. y =
,
x+2
3
128 x3
(2)(x + 2) − (2x)(1)
2x
′
=
.
y =4
x+2
(x + 2)2
(x + 2)5
r
1/3
2
2
8x − 3
3 8x − 3
44. y =
,
=
x2 + 2
x2 + 2
−2/3
(16 x)(x2 + 2) − (8 x2 − 3)(2x)
1 8 x2 − 3
′
y =
3
x2 + 2
(x2 + 2)2
38 x
=
.
3 (8 x2 − 3)2/3 (x2 + 2)4/3
56. 1 − x2 − √ z = 2y 2 − 4y + 5, y = 6x − 5 and x = 2t.
d z Chain d z d y d x
· ·
= (4y − 4)(6)(2).
=
d t Rule d y d x d t d z When t = 1, x = 2 and y = 7 so
= (24)(6)(2) = 288.
dt 4 x4
=
1 − x2 t=1 68. 100
100 q
and r = p q =
.
q + 10
q + 10
d r Chain d r d q
·
.
The marginal revenue product is
=
d m Rule d q d m
dr
100(q + 10) − 100 q (1)
1000
250
Now
=
=
=
. When m = 5, q = √
2
2
dq
(q + 10)
(q + 10)
36 125
d r 360
=⇒
.
= 3
d q m=5 √ 961
50 m2 + 11 − (50 m)9m2 + 11)(2m)
550
dq
=
=
. When
Also
2
2
dm
m + 11
(m + 11)3/2
q=√ 50 m
,
m2 + 11 p= MATA32H page 5 Solutions # 5 d q 275
m = 5,
.
= dm m=5 108
d r d q d r =
Hence
d m dq d m m=5 m=5 m=5 = 2750
360 275
·
=
.
961 108
2883 k
, k is a constant, and q = f (m). Since r = p q, the marginal revenue
q
d r Chain d r d q
product is
·
.
=
d m Rule d q d m
dr
dr
dq
Since r = p q = k,
= 0. Hence
= 0·
= 0.
dq
dm
dm
74. We are given S = 340 E 2 − 4360 E + 42800, where E = years of education,
E ≥ 7, and S = salary.
70. p= (a) dS
= 680 E − 4360 so, when E = 16,
dE d S = 6520.
d E E=16 dS
= 680 E −
If salary is changing at the rate of $ 5000 per year, we have
dE
9360
4360 = 5000 =⇒ 680 E = 9360 =⇒ E =
≈ 13.8. Hence it takes
680
about 13.8 years of education to achieve this rate of change in salary.
50 q
.
80. q = 2 m (2 m + 1)3/2 , and r = √
1000 + 3 q
(a) When m = 12, q = 3000 and r = 1500. Hence the price per unit is
r
1500
1
p= =
= = 0.50. The price per unit is 50 cents.
q
3000
2
√
(50)( 1000 + 3q) − 50 q 21 (1000 + 3q)−1/2 (3)
dr
=
.
(b) Marginal revenue is
dq
1000 + 3q 2750
d r 11
=
When m = 12, q = 3000, so
=
.
d q q=3000 10, 000
40
dr
dr dq
dq
(c) The marginal revenue product is
=
·
. Now
= (2)(2m +
dm dq dm
dm
d q 3
1)3/2 + (2m)
(2m + 1)1/2 (2), so
= 610. Therefore, using part
2
d m m=12
!
671
d r d q d r 11
(610) =
=
=
=
(b), we have d m m=12
d q q=3000
d m m=12
40
4
167.75.
324
5 19
+ + .
82. c = c(q) = p
q 2 + 35 q 18
!
5 19
19
324
+ +
= .
(a) The limiting value of the average cost is lim p
q→∞
18
q 2 + 35 q 18
324 q
19 q
.
(b) The cost function is c(q) = c · q = p
+5+
2
18
q + 35
(b) MATA32H Solutions # 5 page 6 The marginalp
cost is
(324) q 2 + 35 − (324 q) 21 (q 2 + 35)−1/2 (2 q) 19
′
+ .
c (q) =
q 2 + 35
18
′
When q = 17, c (17) = 3. When q = 17, the marginal cost is $ 300 per
unit.
(c) If production is increased from 17 to 18, the addition cost is about $ 300
(from part (b)), but revenue will only increase by $ 275, resulting in a loss of
about $ 25. Hence, production should not be increased from 17 to 18. ...

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