DSC1520/203/2/2018
Tutorial letter 203/2/2018 Quantitative Modelling 1 DSC1520 Semesters 2 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. Kind regards Dr Victoria Mabe-Madisa E-mail:
[email protected]
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DSC1520/203/2
Question 1 1 1 3 1 3 d x − x2 d 1 1 3√ √ = x 2 − x 2 = x− 2 − x 2 = √ − x. dx d 2 2 x 2 x 2 [Option 2]
Question 2 d 40(20 + 20 ln Q) 800 (20 + 20 ln Q)2 = = (1 + ln Q) . dx Q Q [Option 2]
Question 3 1 4x3 + 2√1 x 4x3 + 12 x− 2 1 d √ =− √ 2 = − 4 √ 2. dx x4 + x (x4 + x) (x + x)
[Option 5]
Question 4 2x d 3 2x 3 3 ln x2 + 1 + e−2x = 2 + e−2x −2 × 3x2 = 2 − 6x2 e−2x . dx (x + 1) x +1 [Option 1]
Question 5 Z
2 = (2x + 1)4
Z
2 (2x + 1)
−4
=
Z
2
(2x + 1)−3 1 1 1 +c=− + c. −3 2 3(2x + 1)3 [Option 3]
Question 6 1
R t10 − t 2 dt = t3
Z
3
7
t −t
− 52
t8 t− 2 t8 2 dt = − 3 +c= + √ + c. 8 8 −2 3 t3 [Option 3]
Question 7 R5 1
2x + x3 dx = x2 +
x4 4
|51 == (5)5 +
(5)4 4
− (1)2 +
(1)4 4
= 181,25 − 1,25 = 180. [Option 3] 3
Question 8 Z
√ Z √ 2 1 x+ 3x dx = 1 + x− 3 dx = x + 3x 3 + c = x + 3 3 x + c. x [Option 1]
Question 9 y′ = 0 y =
−x3
+
9x2
y ′ = −3 x2 − 6x + 8
− 24x + 26
y ′ = −3x2 + 18x − 24 = 0 x =
−18 ±
p
= x2 − 6x + 8 = 0
OR
(x − 2)(x − 4) = 0
182 − 4(−3)(−24) 2(−3)
x = 2 and x = 4
= x and 4 [Option 2]
Question 10 The marginal cost function is the differentiated total cost function. Thus by differentiating the total cost function we can determine the marginal cost function. Now if the total cost function is T C = Q4 − 30Q2 + 300Q + 500? then the marginal cost function is MC =
dT C = 4Q3 − 30 × 2Q + 300. dQ
MC =
dT C = 4Q3 − 30 × 2Q + 300. dQ
then the marginal cost function is
Now the marginal cost function’s value when Q is equal 10 is MC = 4(10)3 − 60(10) + 300 = 4 000 − 600 + 300 = 3 700. [Option 1]
Question 11 Profit function: π = T R − T C = 40Q − 8Q2 − (8 + 16Q − Q2 ) = −7Q2 + 24Q − 8. dπ dQ
Maximum where
= −14Q + 24 = 0, that is Q − 1,714 = 0
Q = 1,714 = 1 714 units. [Option 3]
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Question 12 The given demand function is 1
Qd = 100 − 2P 2 . The price elasticity of demand is given by εd =
dQ P , dP Q
with P and Q the price and quantity respectively. The price of elasticity of demand if the price is P = 400 and Q assumed to be 68 and substituting results in 1 dQ = −P − 2 dP
εd =
1
dQ P dP Q
P = −P − 2 Q
1
= − PQ2 1 2
= − (400) = −0,29. 68 The price elasticity of demand is −0,29. Demand is inelastic since |εd | < 1, a 1%, price increase will result in 0,29% less in units demanded. [Option 1]
Question 13 M LC = 3 + 4L TC = =
R
M LCdL
R7 0
3 + 4LdL
7 3L + 2L2 0
= 3 × 7 + 2 × 72 − (0) = 119 = 119. [Option 4]
Question 14 =
25 Q+2
at P = 5
5 =
25 Q+2
⇒Q=3
P
Area under the curve – area of the rectangle R 3 25 = 0 Q+2 dQ − P0 Q0 or area of rect.
= 25 [ln 5 − ln 2] − 15 = 25 × 0,9163 − 15 = 7,9075. [Option 3]
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Question 15 When Q = 6,P = 562. Consumer surplus is therefore R6 CS = 0 (850 − 8Q2 )dQ − 6 × 562 3 6 − 3 372 = 850Q − 8Q 0 3 = 4 524 − 3 372 = 1 152.
Consumers are willing to pay R1 152 more for 6 bicycles than they actually pay. [Option 4]
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