Topic 8
Unconstrained Optimisation: Functions of Several Variables
1 . Find the stationary points of the following functions, and determine whether
each is a maximum, minimum or saddle point.
(a)
z  x 2  y 2  4y  3x  50
First Order Necessary Condition for a stationary point:
set first partial derivatives equal to 0, so Zx = 0 and Zy = 0
Then, taking first derivatives and setting them equal to zero, we have
zx  2x  3  0
(1)
zy  2y  4  0
(2)
Solving these equations we get:
From (1), x   32 ; from (2), y  2 .
So there is a stationary point at (- 3/2 , 2)
Second Order Sufficient Condition for a maximum/minimum/saddle point:
MAX:
Zxx < 0
AND
Zxx Zyy – (Zxy)2 >0
MIN:
Zxx > 0
AND
Zxx Zyy – (Zxy)2 >0
SADDLE:
Zxx Zyy – (Zxy)2 < 0
Checking the second derivatives:
Zxx = – 2
< 0 at all values of x (and hence when x = - 3/2)
Zyy = – 2
Zxy = Zyx = 0
Checking Zxx Zyy – (Zxy)2 >0 when x = - 3/2 and y = 2 we have
( - 2 . - 2) – (0)2 > 0
Confirming that (- 3/2 , 2) is a maximum
(b)
z  2x  5y  5x 2  2y 24xy  1000
1
First Order Necessary Condition for a stationary point:
set first partial derivatives equal to 0, so Zx = 0 and Zy = 0
Then, taking first derivatives and setting them equal to zero, we have
zx  2  10 x  4y  0
(1)
zy  5  4y  4x  0
(2)
Solving these equations we get:
4y =
+ 2 – 10x
= + 5 – 4x
Thus, 6x = - 3 and so x = - ½
Hence from eq (1), y = ½ - 5/2 x = ½ - (5/2 . – ½ ) = ½ + 5/4 = 7/4
Second Order Sufficient Condition for a maximum/minimum/saddle point:
MAX:
Zxx < 0
AND
Zxx Zyy – (Zxy)2 >0
MIN:
Zxx > 0
AND
Zxx Zyy – (Zxy)2 >0
SADDLE:
Zxx Zyy – (Zxy)2 < 0
Checking the second derivatives:
Zxx = 10
> 0 at all values of x (and hence when x = - ½ )
Zyy = 4
Zxy = Zyx = 4
Checking Zxx Zyy – (Zxy)2 >0 when x = - ½ and y = 7/4 we have
( 10 . 4 ) – (4)2 > 0
Confirming that this is a minimum
2
2. A firm charges different prices for its good to its domestic and export
markets. In the domestic market its demand function is given by
PD  20  2QD while in the export market its demand function is given by
PE  15  1.5QE . The total cost function is TC  5QD  QE .
i) What level of output should the firm sell in each market in order to
maximise profit? [hint: total profit = Total Revenue Domestic market + Total
Revenue Export Market - Total Cost ]
ii) What price will the firm charge in each market if it is to maximise its
profits?
The first step is to find an expression for the profit function for the firm. To do this
we need to find an expression for total revenue.
Total revenue in the domestic market will be given by:
TRD  PD QD  20  2QD QD  20QD  2QD2
Total revenue in the export market will be given by:
TRE  PE QE  15  1.5QE QE 2  15QE  1.5QE2
Total revenue from sales in both markets will be given by:
TR  TRD  TRE  20QD  2QD2  15QE  1.5QE2
Total cost is given by TC  5QD  5QE
Profit is therefore given by   TR  TC  20QD  2QD2  15QE  1.5QE2  5QD  5QE
  15QD  2QD2  10QE  1.5QE2
First order conditions:

 15  4Q D  0
Q D
thus, 4QD  15 and hence QD  15 4

 10  3Q E  0
Q E
thus, 3QE  10 and hence QE  10 3
3
Second order conditions:
 2
 4  0
QD2
suggests a max, but need to check second condition
 2
 3  0
QE2
 2
0
Q D Q E
  2

 Q 2
 D
   2
.
  Q 2
 E
   2

  Q Q
  D E
2

2
   4
.  3   0  12  0


therefore we have a Maximum at QD  15 4 and QE  10 3 . These describe the
levels of output the firms should sell into each market in order to maximise
profits
ii) In order to find the price, substitute optimal QD and QE into demand function
to find values for price in each market.
PD  20  2QD  20  215 4  12.5
PE  15  1.5QE  15  1.510 3  10
3. A firms inverse demand curve is described by P  40  2 A  2Q where A
is expenditure on advertising, P is the price of its product and Q is the
quantity
demanded.
Total
costs
of
production
are
given
by
TC  Q 2  AQ  A 2 . What levels of output Q and advertising expenditure
A should the firms choose in order to maximise profit?
i) objective: maximise profit subject to Q and A
  TR  TC
TR  P.Q
TR  40  2 A  2Q .Q  40Q  2 AQ  2Q 2
  40Q  2 AQ  2Q 2  Q 2  AQ  A 2
  40Q  AQ  3Q 2  A 2
4
First order conditions:

 40  A  6Q  0
Q

 Q  2A  0
A
(1)
(2)
From equation (1) we find that Q  2 A
Substitute this into equation (2) and we get:
40  A  62 A  0
40  11A  0
40  11A
A  11 40
now substitute this into Q = 2A and we find:
Q  2 A  211 40  22 40  11 20
One stationary point at A  11 40 , Q  11 20
Second order conditions:
 2
 6  0
suggests a max, but we need to check the second condition
Q 2
 2
 2  0
A 2
 2
1
QA
  2    2


 Q 2 . A 2


2
   2 
2


.  2  1  11  0
  QA    6
 

Thus, a maximum at Q  11 20 and A  11 40
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