Topic 10
Integration Solutions
1. A profit maximising firm has MR  34  3Q and MC  Q  10Q  26 .
2
i) How much will the profit maximising firm produce (i.e. where MR = MC)?
Step 1: set MR=MC and find output that maximises profit, q*
Q 2  10Q  26  34  3Q
Q 2  7Q  8  0
Solve the quadratic for value of Q using formula
Q
 b  b 2  4a c 
2a 
:
a=1, b=-7, c=-8
Q
7  49  41 8 7  9

21
2
so
Q  1 (inadmissible) or Q  8
Thus 8 units produced by profit max firm
ii)
Find expressions for total revenue TR and total Cost TC, and hence profits.
Integrate MR and MC to find TR & TC, and thus profits
  TR  TC
3
TR   MR.dQ   34  3Q dQ  34Q  Q 2  c
2
In this case, the constant of integration c  0 , since the firm makes no revenue when Q=0
So
TR  34Q 
3 2
Q
2


1
TC   MC .dQ   Q 2  10Q  26 dQ  Q 3  5Q 2  26Q  F
3
F = the constant of integration = Fixed Costs
1
 = TR - TC
3
2
1
3
  34Q  Q 2  Q 3  5Q 2  26Q  F
1
3
7
2
   Q 3  Q 2  8Q  F
iii) What level of profits will the firm make when producing the profit maximising level
of output (found in part (i))
substitute in q* to TR and TC to get profit max values when producing q*
Substituting in Q  8 for profit max.
 
1 3 7 2
8  8  88  170 2  224  64  117 1  F
3
2
3
3
iv) Find the level of Fixed Costs at which the firm will make zero profits.
Set profit =0 (thus TR – TC = 0), & solve for F
Setting
  0 , gives
Thus, value of F at  = 0 is
1
0  117  F
3
F  117
1
3
2. A firm which has NO fixed costs has MC and MR given as follows:
MC=2Q2 – 6Q + 6;
MR = 22 – 2Q;
Find total profit for profit maximising firm when MR=MC?
Solution:
1) Find profit max output Q where MR = MC
2
22 – 2Q = 2Q2 – 6Q + 6
gives Q2 – 2Q – 8 = 0
Solve quadratic for Q, by using formula, or
(Q - 4)(Q + 8) = 0 so Q = +4 or Q =-2
so Q = +4
(since Q=-2 inadmissable)
2) Find TR and TC
TR( Q )   22  2QdQ
TR( Q )  22 dQ  2 QdQ
TR( Q )  22Q  Q2  c
TR = c when Q=0; but TR = 0 when Q = 0; so therefore c = 0
so TR = 22Q – Q2
MC = f (Q) = 2Q2 – 6Q + 6


TC ( Q )   2Q 2  6Q  6 dQ
TC ( Q )  2 Q 2dQ  6 QdQ 6 dQ
2
TC ( Q )  Q3  3Q 2  6Q  F
3
2 3
2
(from question) so…. TC( Q )  Q  3Q  6Q
3
F = Fixed Cost = 0
3. Find profit = TR-TC, by substituting in value of q* when MR = MC
Profit = TR – TC
TR if q*=4:
TC if q* =4:
22(4) - 42 = 88-16 = 72
2
/3 (4)3 – 3(4)2 + 6(4)
= 2/3(64) – 48 + 24
= 182/3
so total profit when producing at MR=MC at q*=4 is
TR – TC = 72 - 182/3 = 53 1/3
3.
The inverse demand and supply functions for a good are, respectively:
p D  100  1 q
2
and
p S  20  q
3
(a)
Find the market equilibrium values of p and q.
For market equilibrium, pD  100  21 q  pS  20  q
So 120 = 3/2 q
Hence 3q = 240 and so q = 80
Since q = 80 and p = 100 – ½ (80) = 60
Equilibrium q = 80 and p = 60
(b) Sketch the diagram and highlight Consumer and Producer Surplus at equilibrium
Demand curve: p D  100  1 q so intercepts: at q = 0, p = 100 so (0,100) and at p =
2
0 , q = 50 so (50, 0)
Supply curve: p S  20  q , so intercept: at q = 0, p = - 20
CS S
P
100
p*=60
PS
D
0
-20
q* = 80
50
Q
(c) Find the consumers’ surplus when the market is in equilibrium.
Difference between value to consumers and to the market…. Area above price line and under
Demand curve
4
CS   Dq dq  p * q *
q*
0
80
Consumer surplus (CS) 
 (100  21 q )dq  pq
0
 100q  41 q 2 
 100q  41 q 2 
 80  60  8000  1600  4800  1600

q 80 
 q 0
d) Find the producers’ surplus when the market is in equilibrium
Producer Surplus
Difference between market value and total cost to producers… area below price line and
above Supply curve
PS  p * q *  S q .dq
q*
0
PS  6080  
80
0

 20  qdQ
PS  4800   20q  12 q 2

80
0
PS  4800   20(80)  12 (6400)   200  12 0
PS  4800  (1600  3200)  4800  1600  3200  3200
(e) What is the measure of Total Surplus?
Total Surplus = CS + PS = 1600 + 3200 = 4800
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