DSC1520/203/1/2017
Tutorial letter 203/1/2017 Quantitative Modelling 1 DSC1520 Semesters 1 Department of Decision Sciences
Solutions to Assignment 3
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Dear Student Here are the solutions to the third compulsory assignment. Please contact me if you have any questions. My contact details are as follows: Office: Room 4-37, Hazelwood Campus, Unisa Tel: +27 12 433 4602 E-mail:
[email protected] Kind regards Dr Victoria Mabe-Madisa
2
DSC1520/203/1
Question 1 f (x) = x−2 − 4x−1 , therefore f ′ (x) = − x23 +
4 . x2
[Option 3]
Question 2 f ′ (x) = 12e6x + 12xe6x .6 = 12e6x (1 + 6x).
[Option 3]
Question 3 12x2 +16x 4x3 +8x2
f ′ (x) =
=
3x2 +4x x3 +2x2
[Option 4]
Question 4 f ′ (x) = 3e4x
2 −3x
2 (8x − 3) + 8x − 3 = (8x − 3) 3e4x −3x + 1 .
[Option 2]
Question 5 R
x+4 x2
=
R
1 x
+ 4x−2 = ln x −
4 x
[Option 3]
+ c.
Question 6 R
4 (3x+1)3
=
R
4(3x + 1)−3 = 4 (3x+1) −2
−2
1 3
2 = − 3(3x+1) 2 + c
[Option 2]
Question 7 Z
1
3x+1
18e 0
e3x+1 1 dx = 18 = 6(e4 − e1 ) = 311. 3 0
[Option 1]
Question 8 MR =
d dx
2x3 −
x2 2
+ 10x + 15 = 6x2 − x + 10. When x = 5, M R = 155.
[Option 2]
Question 9 d dL
6L2 − 0,2L3 = 12L − 0,6L2 = L(12 − 0,6L) = 0. Therefore L = 0 or L = 20.
[Option 2]
When L = 0, Q = 0 ∴ minimum output. Hence maximum output is achieved when L = 20.
Question 10 f ′ (x) = 6x2 −6x−12 = 0 gives (x−2)(x+1) = 0 or x = 2 and x = −1. f ′′ (x) = 12x−6 giving f ′′ (2) = 18 > 0 and f ′′ (−1) = −18 < 0. At x = 2, f is a minimum and at x = −1 it is a maximum. [Option 4] 3
Question 11 If Q = 10, then P = 92. PS = PQ −
Z
0
10
Q2 − 2Q + 12 dQ
10 Q3 − Q2 + 12Q) 0 = 10 × 92 − ( 33 10 2 = 920 − − 10 + 120 3 1 000 = 920 − − 100 + 120 2 = 920 − 353,33 = 566,67.
[Option 2]
Question 12 √
At equilibrium 6Q + 10 = 74 − Q2 which gives Q = −6± 236+256 = 5,544, Q = 5,544 ≈ 6 and P = 43,26 R6 (work with rounded Q). Now, CS = 0 74 − Q2 dQ − (6 × 43,26) = 372 − 259,56 = 112,44. [Option 2]
Question 13 4 T R = P Q = 120Q − 5Q2 and profit π = T R − T C = − 30 Q3 − 3Q2 + 102Q − 60.
[Option 3]
Question 14 dπ dQ
= −0,4Q2 − 6Q + 102 = 0 or 0,2Q2 + 3Q − 51 = 0 √ −3± 32 −4(0,2)(−51) giving Q = = 10,14 or − 25. 2(0,2)
[Option 2]
Question 15 P Given P = 44e−0,7Q from which we get ln 44 = −0,7Q or Q =
εd =
dQ P dP Q
1 = − 0,7P
P 2
1 1 1 = − 0,7 2 = − 1,4 = −0,714.
Since |εd | = 0,714 < 1, demand is inelastic.
4
lnP −0,7
−
ln 44 −0,7
=
ln P −0,7
+ 5,41 giving
dQ dP
1 = − 0,7P .
[Option 2]
