1.
Determine whether solutions exist for each of the following quadratic
equations. Where they do find the solution(s).
Firstly determine whether solutions exist using the following criteria:
Two solutions
b2  4ac  0
2
One solution
b  4ac  0
2
No solution
b  4ac  0
Secondly find the solution where possible using the formula:
x
(i)
 b  b 2  4ac
2a
x2  2x  0
a=1, b=-2, c=0
b2  4ac   22  410  4  0
two solutions exist
 b  b 2  4ac 2  4 2  2


2a
21
2
22
x
2
2
22
x
0
2
x
(ii) 3x  6x  1  0
Multiply out the quadratic
3x 2  3x  6  0
Divide across by 3
x2  x  2  0
a=1, b=-1, c=-2
b2  4ac   12  41 2  9  0
two solutions exist
 b  b 2  4ac 1  9 1  3


2a
21
2
1 3
x
2
2
1 3
x
 1
2
x
(iii) 9 x 2  24 x  16  0
a=9, b=-24, c=16
b2  4ac   242  4916  576  576  0
x
 b  b 2  4ac 24  0 24


 1.33
2a
29
18
one solution
(iv) 3x 2  2 x  3  0
a=3, b=2, c=3
b2  4ac  22  433  4  36  32  0
no solution
(v) 2 x 2  11x  21  0
a=2, b=11, c=-21
b2  4ac  112  42 21  121  168  289  0
two solutions
 b  b 2  4ac  11  289  11  17


2a
22
4
 11  17
 11  17
x
 1.5
x
 7
4
4
x
(vi)  2 x 2  x  10  0
a=-2, b=1, c=10
b2  4ac  12  4 210  81  0
two solutions
 b  b2  4ac  1  81  1  9


2a
2 2
4
1 9
1 9
x
 2
x
 2 .5
4
4
x
2 A firms demand function for a good is given by P = 107-2Q and their total
cost function is given by TC = 200+3Q .
i) Obtain an expression for total revenue profit in terms of Q
Total Revenue = P.Q
TR = (107-2Q)*Q = 107Q-2Q2
Profit = TR-TC
Profit = 107Q-2Q2-200-3Q = -2Q2+104Q-200
ii)
For what values of Q does the firm break even
Firm breaks even where Profit = 0
-2Q2+104Q-200 = 0
a = -2, b=104, c=-200
Q
 104 
Q  2, Q  50
1042  4 2 200  104 

2 2 
10816  1600  104  96

4
4
iii)
Illustrate the answer to (ii) using sketches of the total cost function,
the total revenue function and the profit function
2000
TC / TR / Profit
1500
Proft =
1150
1000
500
TC
TR
0
0
10
30
20
40
Q = 26
50
Profit
Profit
-500
Note: Break even where Profit = 0 or TR=TC.
iv)
From the graph estimate the maximum profit and the level of output
for which profit is maximised
Maximum profit at max point on profit curve.
Max profit = 1150 at Q = 26
3. What is the profit maximising level of output for a firm with the marginal
cost function MC = 1.6Q2-15Q+60 and a marginal revenue function MR =
280-20Q?
Profit is maximised where MR=MC
280-20Q = 1.6Q2-15Q+60
1.6Q2+5Q-220=0
a=1.6, b=5, c=-220
Q
5
52  41.6 220  5 

21.6
25  1408  5  37.85

3.2
3.2
Q  10.27, Q  13.39
Profit maximising level of output is Q = 10.27 (can’t have negative output)
Q
60
4. The demand function for a good is given as Q = 130-10P. Fixed costs
associated with producing that good are €60 and each unit produced costs an
extra €4.
i) Obtain an expression for total revenue and total costs in terms of Q
TR = P.Q
Q = 130-10P
10P = 130-Q
P = 13-Q/10
TR = (13-Q/10)*Q = 13Q-0.1Q2
TC = FC+VC
TC = 60+4Q
ii)
For what values of Q does the firm break even
Firm breaks even where TR = TC
13Q-0.1Q2=60+4Q
-0.1Q2+9Q-60=0
a=-0.1, b=9, c=-60
Q
9
92  4 0.1 60  9  81  24

2 0.1
 0.2

 9  7.55
 0.2
Q  7.25, Q  82.75
iii)
iv)
Obtain an expression for profit in terms of Q and sketch its graph
Use the graph to confirm your answer to (ii) and to estimate maximum
profit and the level of output for which profit is maximised
Profit = TR-TC
Profit = 13Q-0.1Q2-60-4Q=-0.1Q2+9Q-60
200
Profit
150
Profit Max
Profit = 143
100
Profit
50
Q
0
0
-50
-100
10
Break Even
Q = 7.25
20
30
40
50
Profit Max
Q = 45
60
70
80
Break Even
Q = 82.75
90