Second Order Derivatives
Course Manual
Background to Topic 6
Maximisation and Minimisation
Jacques (4th Edition):
Chapter 4.6 & 4.7
Y
Y=a+bX
a
X
Y= f (X) = a + bX
First Derivative
dY/dX = f  = b
constant slope b
Second Derivative
d2Y/dX2 = f  = 0
constant rate of change -
the change in the slope is
zero
Y=X
>1
Y
X
Y=X
>0 and <1
Y
X
Y= f (X) = X
First Derivative
dY/dX = f  =  X-1 > 0
Positive Slope: change in Y due to change X is
Positive
Second Derivative
d2Y/dX2 = f = (-1) X(-1)-1
or
d2Y/dX2 = (-1)(Y/X2)
d2Y/dX2 = f = 0 if  = 1
constant rate of
change
d2Y/dX2 = f  > 0 if  > 1
increasing rate of change (change Y due to change X is
bigger at higher X – the change in the slope is positive)
d2Y/dX2 = f  < 0 if  < 1
decreasing rate of change (change Y due to change X is
smaller at higher X – the change in the slope is negative)
Maximisation and Minimisation
•Stationary Points
•Second-order derivatives
•Applications
B
Y
Max
A
Min
X*=1
X*=4
X
Stationary points are the turning points or
critical points of a function
Slope of tangent to curve is zero at
stationary points, f (X) = 0
Are these a Max or Min point of the
function?
1) examine slope in region near the
stationary point
Sign of first derivative around a turning point:
Before
At
After
Maximum
plus
zero
minus
Minimum
minus
zero
plus
/dX = f (X) is (–, 0, +)  min
dY
/dX = f (X) is (+, 0, -)  max
dY
2) or calculate the second derivative……
look at the change in the slope beyond the
stationary point
if d2Y/dX2 = f  (X) > 0  a minimum
the change in the slope is positive beyond the
stationary point, so the point is a local minimum
if d2Y/dX2 = f  (X) < 0  a maximum
the change in the slope is negative beyond the
stationary point, so the point is a local maximum
if d2Y/dX2 = f  (X) = 0  indeterminate
the change in the slope is zero beyond the stationary
point - could be a max, or a min, or an inflection
point
e.g. inflection point
Y
f =0 & f =0
X
To find the Max or Min of a function
Y= f(X)
1) First Order Condition (F.O.C.):
set slope dY/dX = f (X) = 0
this identifies the stationary point(s)
2) Second Order Condition (S.O.C.):
check the sign of the second derivative
(gives the change in the slope)
d2Y/dX2 = f  (X) > 0  a minimum
d2Y/dX2 = f  (X) < 0  a maximum
d2Y/dX2 = f  (X) = 0  indeterminate
this identifies whether the slope of the function
is increasing, decreasing, or does not change
after the stationary point(s)
Find the Maxima and Minima of the
following functions:
2
(i) y  x  2 x  1
F.O.C. : slope=0 at stationary point
dy
 2x  2  0
dx
2 x  2
x  1
S.O.C. : check sign of second derivative at x=-1
d2y
 2  0 (slope increases after the stationary point, so must
2
dx
be a minimum at x= -1)
30
25
Y
20
15
10
5
0
-4
-3
-2
-1
0
X
1
2
3
4
Example:
A firm faces the demand curve P=8-0.5Q
3
1
and total cost function TC= /3Q 2
3Q +12Q. Find the level of Q that
maximises total profit and verify
that
this value of Q is where MC=MR
Solution:
P = 8 - 0.5Q
inverse demand function
TR (Q) = P.Q = 8Q - ½Q2
TC (Q) = 1/3Q3 - 3Q2 + 12Q
MAX (Q) = TR (Q) - TC (Q)
 (Q) = -4Q + 2 ½ Q2 – 1/3Q3
MAX
 (Q) = -4Q + 2 ½ Q2 – 1/3Q3
First Order Condition:
d/dQ = f (Q)= - 4 + 5Q – Q2 = 0
(solve quadratic – Q + 5Q – 4 by applying formula: Q 
b
2
b
2
 4ac

2a
)
Optimal Q solves as: Q*=1 and Q*= 4
Second Order Condition:
d2/dQ2 = f (Q) = 5 – 2Q
Sign ?
f  = 3 > 0
if Q*= 1 (Min)
f = - 3 < 0
if Q*= 4 (Max)
So profit is max at output Q = 4
TR (Q) = 8Q - ½Q2
dTR
MR 
 8Q
dQ
Q = 4 then MR = 4
TC (Q) = 1/3Q3 - 3Q2 + 12Q
MC =
dTC
 Q 2  6Q  12
dQ
Q = 4 then MC = 16 – 24 +12 = +4
At Q = 4, MR = MC
Maximisation and Minimisation
Tax Example
 Jacques (4th Edition):
Chapter 4.6
Supply and Demand Equations of a good
are given, respectively, as
P- t = 8 + QS
P = 80 – 3QD
A tax t per unit, imposed on suppliers, is
being considered.
At what value of t does the government
maximise tax revenue in market
equilibrium?
Set Supply equal to Demand
In equilibrium, QD = QS
Q + 8 + t = 80 – 3Q
Solve for Q
Qe = 18 – ¼ t
Tax Rev. T = t.Qe = t(18 – ¼ t)
MAX
T(t) = 18t – ¼ t2
t*
 First Order Condition for max:
dT/dt = 18 – ½ t = 0
 t* = 36
 Second Order Condition for max:
d2T/dt2 = -½ < 0 (Max)
Results
t* = 36
Qe = 18 – ¼ t* = 9
T = t*.Qe = 18t* – ¼ t*2 = 324
Pe = Qe + 8 + t*= 53
If t = 0, then
Qe = 18
Pe = Qe + 8 = 26
Is the full burden of the tax passed on to
consumers?
Ex-ante (no tax) Pe = 26
Ex-post (t* =36) Pe = 53
Yet the tax is t*= 36 but the price increase
is only 27 (75% paid by consumer)
Another Example…Manual, Section 5, Q 6
Cost Producing Q units output given
C  8K 
2 2
Q
K
capital K:
(a) if K=20 in Short Run, find the level
of Q at which AC is minimised.
Show that MC and AC are equal at
this point.
C = (8*20) + (2/20)Q2 = 160 + 0.1Q2
 AC = C/Q = 160/Q + 0.1Q
and MC = dC/dQ = 0.2Q
First Order Condition:
AC is at min when dAC/dQ = 0
slope AC: dAC/dQ = - 160/Q2 + 0.1 = 0
 Q2 = 1600
 Q = 40
Second Order Condition:
d2AC/dQ2 >0  min.
d2AC/dQ2 = + 320/Q3
at Q = 40, d2AC/dQ2 = 320 / 403 >0
 min AC at Q = 40
 AC at Q=40: 160/40 + (0.1*40) = 8
and MC at Q = 40: 0.2*40 = 8
 MC = AC at min AC when Q=40
b) In Long Run, K changes. What level
of K minimises costs when Q = 1000?
C  8K 
2 2
Q
K
dC/dK = 8 – (2Q2 / K2) = 0
F.O.C.
 8K2 = 2Q2
 K2 = ¼ Q2
 optimal K = ¼Q2 = ½ Q
 if Q = 1000, optimal K = 500
 if Q = Q0, optimal K = Q0/2
min cost producing Q0 =
 Q0
 8 
2






2
  
.Q02 
  Q0

2


= 4Q0 + 4Q0
= 8Q0
Questions Covered:
Section 6: Maximisation and
Minimisation
Q1: Identifying the max and min of
various functions
Q2: Identifying the max and min of
various functions – sketch graphs
Q3: Finding value of t that maximises tax
revenues, given D and S functions (as in
lecture example)
Q4: Similar to Q3 above, but where Qd =
a – bP and Qs = c + dP and find impact of
t on sales and value of t that maximises tax
revenue
Q5: Showing that max of total revenue
curve occurs where elasticity of demand is
unity.
Q6: Identifying AC min output level for a
given K. And min Costs given optimal K.
(lecture example).
Q7: (a) Identifying all local max and min
of various functions. (b) Identifying profit
max output level.
Q8: (a) Differentiate various functions. (b)
more on finding t that maximises tax
revenue