SELF ASSESSMENT SOLUTIONS
TOPIC 6: MAXIMISATION AND MINIMISATION
A
Finding Maxima and Minima
1. Find the stationary points and identify whether these are maximum, minimum or
inflection points.
y   x 2  5x  5 ,
i)
first order condition: slope of function = 0 at stationary point
dy
 2 x  5 =
dx
0
now solve for the value of x at the stationary point:
 x
5
or –2.5.
2
Substitute in this value to the original function to find the value of y. The value of y at
5
2
5
2
this point is y  ( )2  5( )  5  1.25
Thus, stationary point at (-2.5, -1.25)
Second Order Condition: check the sign of second derivative at this point
To find if this is a max or min point, we need to evaluate the change in the slope at x = 2.5
Thus, we check the sign of the second derivative
d2y
dx
2
 2 , which is negative when x  
5
(and in fact, for all values of x). Thus, the
2
change in the slope of the function is negative around the stationary point, which
indicates a maximum.
So (-2.5, -1.25) is a local maximum of y.
1
(ii) y  2 x 2  4
first order condition: slope of function = 0 at stationary point
dy
 4x  0
dx
x0
The value of y at this point is y = 0 +4 = 4
Thus, stationary point at (0, 4)
Second Order Condition: check the sign of second derivative at this point
To find if this is a max or min point, we need to evaluate the change in the slope at x = 0
Thus, we check the sign of the second derivative
d2y
 4  0 at x=0 (and at all values of x). So we have a minimum at (0,4)
dx 2
(iii) y  2 x 3  9 x 2  24 x  10
first order condition: slope of function = 0 at stationary point
dy
 6 x 2  18 x  24  x 2  3x  4  0
dx
This yields a quadratic function. In order to solve for the value of x at the stationary
point, we need to solve the quadratic by applying the formula.
 b  b2  4a c
2a
Where a=1, b= -3, c= -4
3  9  41 4  3  25

2
2
x  4 , x  1
So we have a stationary point at two different values of x. To check whether these are
max or min points, we check change in the slope around these stationary points….
Second Order Condition: check the sign of second derivative at this point
d2y
 12 x  18
dx 2
evaluating the change in the slope around the stationary point where x = 4:
2
d2y
 124   18  20  0 thus, we have a minimum at x = 4 (beyond this stationary
dx 2
point, the slope is increasing)
evaluating the change in the slope around the stationary point where x = -1:
d2y
 12 1  18  20  0 thus we have a maximum at x = -1 (beyond this stationary
dx 2
point, the slope is decreasing )
(iii) y  3x  22
first order condition: slope of function = 0 at stationary point
dy
 23x  2 3  18 x  12  0
dx
18 x  12
2
x
3
Second Order Condition: check the sign of second derivative at this point
d2y
 18  0 when x = 2/3 (and in fact for all values of x)
dx 2
Thus our stationary point is a minimum at x = 2/3
(iv) y  4 x  x
first order condition: slope of function = 0 at stationary point
dy
4

1  0
dx 2 x
42 x 0
2 x 4
x 2
x4
Second Order Condition: check the sign of second derivative at this point
1
d2y
4 1
1
 x 2 
2
4
dx
x x
evaluating this at x = 4:
3
d2y
1
1

   0 maximum at x = 4
2
8
dx
4 4
(v) y  x 3
first order condition: slope of function = 0 at stationary point
dy
 3x 2  0
dx
x0
Second Order Condition: check the sign of second derivative at this point
d2y
 6x
dx 2
evaluating at x=0
d2y
 0 cannot classify therefore must sketch
dx 2
Upward sloping function therefore minimum at x=0
1200
y
1000
800
600
400
200
0
0
1
2
3
4
5
6
7
8
x
10
9
4
3x
x 1
first order condition: slope of function = 0 at stationary point
(vi)
y

2

dy
x 2  1 .3  3x 2 x  3x 2  3  6 x 2  3x 2  3



0
2
2
2
dx
x2  1
x2  1
x2  1






 3x 2  3  0
x2  1
x  1
Second Order Condition: check the sign of second derivative at this point





d2y
x 2  1  6 x    3x 2  3 2 x 2  1 2 x 

4
dx 2
x2  1
2


d 2 y 6 x 5  12 x 3  18 x

4
dx 2
x2  1


evaluate this at x = 1
d 2 y 615  1213  181  24
3


   0 thus maximum at x = 1
2
4
16
2
dx
12  1


evaluate this at x = -1
d 2 y 6 15  12 13  18 1  6  12  18 24 3



  0 minimum at x = -1
4
16
16 2
dx 2
 12  1


5
B
1.
Applications
The average cost curve of a firm is given by AC  15  6Q  Q 2 
1
Q
(i) Derive the total and marginal cost curves
TC  AC.Q

1
TC  15  6Q  Q 2  Q  15Q  6Q 2  Q 3  1
Q

MC 
dTC
dQ
dTC
 15  12Q  3Q 2
dQ
(ii)
If the firm can sell as many units of output as it wishes at a price of 6,
what quantity will it sell if it is to maximise profits?
  TR  TC
TR  P.Q  6Q
d
maximise profits at
0
dQ
  6Q  15Q  6Q 2  Q 3  1
  9Q  6Q 2  Q 3  1
d
 9  12Q  3Q 2  0
dQ
Q 2  4Q  3  0
a=1, b=-4, c=3
4  16  413 4  4

21
2
Q  3, Q  1
Check for max/min using second derivative
d 2
 12  6Q
dQ 2
Q=1
d 2
 12  61  6  0 minimum profit
dQ 2
Q=3
6
d 2
 12  63  6  0 maximum profit
dQ 2
(iii) What profit is made at this level of output? Comment.
Q  3,   9Q  6Q 2  Q 3  1
  93  632  33  1  27  54  27  1  1
This implies the firm is operating at a loss which will not be sustainable in the long run.
2.
Suppose the demand and supply curves for a market are Qd  1200  2 P and
Qs  4 P
(i) Find the equilibrium Price and Quantity
Qd=Qs
1200-2P=4P
1200=6P
P*=200
Qs=Qd=1200-2(200)=1200-400=800=Q*
(ii)
If a purchase tax t is imposed find the new equilibrium and the amount of
tax revenue.
Purchase tax shifts demand curve
Qd=1200-2(P+t)
Qs=4P
Qd=Qs
1200-2(P+t)=4P
1200-2P-2t=4P
6P=1200-2t
P*=200-1/3t
Qd=Qs=4P=4(200-1/3t)=800-4/3t=Q*
Note: The shift in demand as a result of the imposition of a purchase tax t reduces
equilibrium price by 1/3 of the tax and reduces equilibrium quantity by 4/3 of the
tax.
Tax Revenue = T
T=tQ*=t(800-4/3t)=800t-4/3t2
(iii) Find the value of t which maximises tax revenue
dT
0
Maximum T where
dt
7
dT
8
 800  t  0
dt
3
8
800  t
3
2400  8t
300  t
Check it is a maximum using second-order derivatives
d 2T
8
   0 maximum
2
3
dt
The purchase tax which maximises government revenue is t = 300
At this level equilibrium price P*=200-1/3(300)=100 and Q*=800-4/3(300)=400
8