BT 1120, Homework I BT 1120, Homework I

Question 1

x+1
x
 - x+1
x+2
=
(x+1)(x+2)-(x+1)(x)
x(x+2)
=
x2 +3x +2 -x2 -x
x(x+2)
=
2x +2
x(x+2)

x
x+3
x2+2x
(x+3)(x+2)
=
x
x+3
 ¸ x2+2x
(x+3)(x+2)
=
x
x+3
 × (x+3)(x+2)
x2+2x
=
x(x+2)
x+2
=
1

Question 2
(i) y = -9x+27
(ii) y = -[17/ 8] x + [37/ 8]

Question 3
(a)Firstly

y
=
ax +b
y
=
-7x +b
Then
8
=
-7(3) +b
8
=
-21 +b
298
=
b
So
y = -7x+29

(b) Firstly

y
=
ax +b
17
=
2a +b
20 = 3a +b
So
20-17
=
a
3
=
a
So the slope is 3.

Question 4
(a)

x-y
=
-1
-x +2y
=
5
y
=
4
x
=
y-1
=
4-1
=
3
So (3,4) is the unique solution.

(b)

2x+2y = 4
Þ
x+y = 2
-12x-12y = -24
Þ
x+y = 2
The two lines are the same, so there are an infinite number of solutions.

(c)

3x-7y
=
2
x+4y
=
11
Multiplying the second equation by (-3) gives:
3x-7y
=
2
-3x-12y
=
-33
So
-19y
=
-31
y
=
31
19
Then find x:
x
=
11-4y
x
=
11-4 31
19
x
=
85
19
x
=
4 9
19
So (4[9/ 19],1 [12/ 19]) is the unique solution.

(d)

y
=
x+4
x-y
=
-4
-2x+2y
=
10
x-y
=
-5
No solution, so the two lines are parallel.

Question 5

P+2Q
=
60
P-Q
=
30
Multiplying the second equation by (-1) gives:
P+2Q
=
60
-P+Q
=
-30
So we get
3Q
=
30
Q
=
10
Hence
P = 30+Q = 30+10 = 40
This means that the equilibrium price is 40 and the equilibrium quantity is 10.

Figure

Figure 1: Plot of the two lines showing the equilibrium point

Question 6
Find the roots of the following quadratic functions:
(i) y = x2-x-6

x
=
-(-1) ±   ___________
Ö(-1)2 -4(1)(-6)
 
2(1)
=
1 ±   ____
Ö1 +24
 
2
=
1 ±   __
Ö25
 
2
=
1 ±5
2
So the roots are x = 3 and x = -2.

(ii) y = 6x2 +24x-24 (10 marks)

x
=
-(24) ±   ____________
Ö(24)2 -4(6)(-24)
 
2(6)
=
-(24) ±   _______
Ö576 +576
 
12
=
-(24) ±   ____
Ö1152
 
12
So the roots are x = 0.828 and x = -4.828.

Question 7
(i) The coefficient of x2 is +1, so the graph is ``U'' shaped.
x = 0 when y = -15 so (0, -15) is on the graph.
y = 0 when x2+2x-15 = 0 solving this gives roots at (-5,0) and (3,0).

Figure
Figure 2: Plot of y = x2 +2x -15

(ii) The coefficient of x2 is -3, so the graph has an inverted ``U'' shape.
x = 0 when y = 36 so (0, 36) is on the graph.
y = 0 when -3x2-3x+36 = 0 solving this gives roots at (-4,0) and (3,0).

Figure
Figure 3: Plot of y = -3x2 -3x +36

Question 8
(a)

TR
=
P·Q
=
(75-3Q)Q
=
75Q -3Q2
(b) If TR=0 then
75Q -3Q2
=
0
3Q2 -75Q
=
0
Q(3Q-75)
=
0
So the roots are Q = 0 and Q = 25. (c) The maximum value occurs at Q = 121/2 when
TR
=
75 (12 1
2
 ) - 3 (12 1
2
 )2
=
468.75

File translated from TEX by TTH, version 2.10.
On 24 Mar 1999, 14:45.