The Domain, Sydney and Total Profit Function Essay

Submitted By pinkcookiebutter
Words: 388
Pages: 2

Part 1

Question 1

a) 3x2 + 8 < 35
3x2 < 27 x2 7 x > 9

OR

x-2 < -7 x < -5

ANS. x< -5 OR x>9

c) (ex)2e-x/ e-3x
(e2x)(e-x)/e-3x
(ex)(e3x)
ANS. e4x

d) log 4x – log 2 = 5 log 4x/2 = 5 log 2x = log 105
2x = 100000
ANS. x = 50000

e) Initially,
F(t) = ekt + A
5000 = e0k+ A
5000 = 1 + A
A = 4999

After two months,
5403 = e2k + 4999 e2k = 404 ln e2k = ln 404
2k = 6.0014 k= 3.0007

After three months,
F(3) = e3(3.0007) + 4999
F(3) = 8120.118 + 4999
F(3) = 13119.118
ANS. 13119.118 ≈ 13119 households

Part 2 Question 2

a) Weekly C(X) = 80x + 200000
b) Domain= 0 ≤ x ≤ 3000
Range = 200000≤ C(X) ≤ 440000
ANS. The domain of x is [0, 3000].
The range of C(X) is [200000, 440000].
c) Monthly C(X) = 4(80x + 200000) [one month = 4 weeks]
= 320x + 800000
d) Weekly R(X) = 240x
e) P(X) = R(X) – C(X)
= 240x – (80x + 200000)
= 160X – 200000
f) Domain = 0 ≤ X ≤ 3000
Range = −200000 ≤ P(X) ≤ 280000
ANS. The domain of x is [0, 3000].
The range of P(X) is [-200000, 280000].

Question 3

a) – 0.5x2 + x + 350 = 0.3x2 + 100
-0.8x2 + x + 250 = 0
8x2 – 10x – 2500 = 0 x = (-b ± √(b2-4ac))/2a x= (10 ±√(100+80000))/16 x= (10 + 283.019)/16 = 18.3137 ≈ 18 [since X cannot be negative]
Substitute x=18.3 to equation p1 = -0.5x2 + x + 350 and p2 = 0.3x2 + 100,
P1 = -0.5 × (18)2 + 18 + 350 = 206
P2 = 0.3 × (18)2 + 100 = 197.2
Price = (206 + 197.2) = 201.6
ANS. The equilibrium price is $201.60 and the quantity is 18.

b) Quantity Demanded
300 = -0.5x2 + x + 350
5x2 – 10x – 500 = 0 x2 – 2x – 100 = 0 x = (2±√(4+400))/2 x = 11.049 ≈ 11

Quantity Supplied
300 = 0.3x2 + 100
3x2 = 2000 x = √(2000/3) = 25.819 ≈ 26

ANS. The quantity demanded is 11 and the quantity supplied is 26.

Part 3

Question 4

a) Due to the increase in cost of production, the total profit function is also