Topic 9
SELF-ASSESSMENT
Constrained Optimisation - Solutions
1.
Using the substitution method, optimise x  2 xy subject to 2 x  4 y  100
Objective function: x  2 xy constraint: 2 x  4 y  100
The substitution method:
Step 1:
from the constraint…..
2 x  100  4 y
x  50  2 y
Step 2:
substitute in this value of x into the objective function
f  x  2 xy
f  50  2 y  250  2 y y
f  50  2 y  100 y  4 y 2
f  50  98 y  4 y 2
Step 3:
Now we have re-written the objective function as a function of one
variable, while constraining the value of x to being equal to x = 50 – 2y . So optimise
this function with respect to y to find the value of y at the stationary point
df
First Order Condition:
 98  8 y  0
dy
8 y  98
y  12.25
Second Order Condition:
Step 4:
d2 f
 8  0
dy 2
so maximum at y  12.25
Substitute in this value of y into the constraint function to find the value
of x
x  50  2 y  50  212.25  25.5
Solution: maximum where x  25.5 , y  12.25
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2.Using the Lagrange multiplier method, solve the following:
(i) optimise objective function y  x 2 subject to constraint x  y  1
Step 1:
The Lagrangian:
L  y  x 2   1  x  y 
Step 2:
L
 2x    0
x
L
 1   0
y
L
 1 x  y  0

Step 3:
eq.1
eq.2
eq.3
Solve the 3 simultaneous equations:
EQ2:    1
EQ1: 2x  1  0 so 2x  1 and x = ½
EQ3: 1  1 2  y  0 so y = ½
The solution is: x = ½ and y = ½
(ii) Optimise the objective function 3x 2  y 2  2 xy subject to the constraint
6 x y
Step 1:
Step 2:
Step 3:
The Lagrangian:
L  3x 2  y 2  2 xy   6  x  y 
L
 6x  2 y   = 0
x
L
 2 y  2x   = 0
y
L
 6 x y= 0

eq.1
eq.2
eq.3
Solve the 3 simultaneous equations:
Solving EQ1: 2 y  6 x   and EQ2: 2 y  2 x  
So
6x    2x  
2
4x  2
  2x
EQ3: y  6  x
Substituting values of y and  into eq1:
EQ1: 6 x  2 y    0
6x  26  x  2x  0
6x  12  2x  2x  0
6x  12
x2
thus, y  6  x  6  2  4
The solution is:
x  2, y  4
(iii) Optimise the objective function 30 xy subject to the constraint 500  x  25 y
The Lagrange multiplier method:
Step 1:
The Lagrangian:
L  30xy   500  x  25 y 
Step 2:
Step 3:
L
 30 y   = 0
x
L
 30 x  25 =0
y
L
 500  x  25 y = 0

eq.1
eq.2
eq.3
Solve the 3 simultaneous equations:
EQ2: 30x  25
so
6 5x  
EQ3: y  20  x 25
EQ1: 30 y    0
3020  x 25  6 5 x  0
600  6 5 x  6 5 x  0
3
600  12 5 x  0
600  12 5 x  0
600  12 5 x
x  250
y  20  x 25
y  20  250 25  10
Optimum points at:
x  250 , y  10
(iv) Optimise the objective function x 0.25 y 0.25 subject to the constraint
24  x 10  y
The Lagrange multiplier method:
Step 1:
Step 2:
Step 3:
The Lagrangian:
L  x 0.25 y 0.25   24  x 10  y 
L
 0.25 x 0.75 y 0.25   10 = 0
x
L
 0.25 x 0.25 y 0.75   = 0
y
L
 24  x 10  y = 0

eq.1
eq.2
eq.3
Solve the 3 simultaneous equations:
EQ2: 0.25 x 0.25 y 0.75    0
0.25 x 0.25 y 0.75  
EQ1:


0.25 x 0.75 y 0.25  0.25 x 0.25 y 0.75 10  0
2.5 x 0.75 y 0.25  0.25 x 0.25 y 0.75
2.5 y 0.25 y 0.75  0.25 x 0.25 x 0.75
2.5 y  0.25 x
10 y  x
EQ3: 24  x 10  y  0
24  10 y 10  y  0
4
24  2 y  0
y  12
10 y  x
1012  x
120  x
x  120 , y  12
Optimum point at:
(v) optimise y  x 2 subject to y  6 x  9
(Non-linear therefore use Lagrange Multiplier Method)
The Lagrange multiplier method:
Step 1:
The Lagrangian:
L  y  x 2   9  6 x  y 
L
 2 x  6
Step 2:
x
L
 1 
y
L
 9  6x  y

Step 3:
Solve the simultaneous equations:
EQ1:  2x  6  0
EQ2: 1    0
EQ3: 9  6 x  y  0
EQ2:
  1
EQ1:
 2 x  6 1  0
x3
Sub in to EQ3:
9  63  y  0
y9
The solution is x  3 , y  9
(vi) optimise y  2 x subject to y  x 2
(Non-linear therefore use Lagrange Multiplier Method)
The Lagrange multiplier method:
Step 1:
The Lagrangian:
L  y  2x   y  x 2


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Step 2:
Step 3:
L
 2  2 x
x
L
 1 
y
L
 y  x2

Solve the simultaneous equations:
EQ1:  2  2x  0
EQ2: 1    0
EQ3: y  x 2  0
EQ2:
  1
EQ1:
 2  2 x 1  0
x 1
Sub in to EQ3:
y  12  0
y 1
The solution is x  1, y  1
3. A consumer’s utility function is given by U  x1 x2 where x1 is the quantity of
good 1 that is bought and x 2 is the quantity of good 2 that is bought. The
price of good 1 is €10 while the price of good 2 is €2. If the consumer’s
income is €100 what will the consumer’s optimal utility level be?
Budget constraint: p1x1 + p2x2 = M where M is income
Thus 10 x1  2 x2  100
So, Maximise U  x1 x2 subject to 10 x1  2 x2  100
The Lagrange multiplier method:
Step 1:
The Lagrangian:
L  x1 x2   100  10x1  2x2 
Step 2:
L
 x 2  10 = 0
x1
eq.1
6
L
 x1  2 = 0
x 2
L
 100  10 x1  2 x 2 = 0

Step 3:
eq.2
eq.3
Solve the 3 simultaneous equations:
EQ2:
x1  2
x1 2  
EQ1:
x2  10x1 2  0
x2  5x1
EQ3: 100  10x1  25x1   0
20 x1  100
x1  5
x2  5x1  55  25
The optimal value of U is where x1  5 and x2  25
U  x1 x2  525  125
4. A firm’s production function is given by Q  L0.25 K 0.25 where L is the quantity of
labour employed and K is the quantity of capital employed. The price of
labour is €20 and the price of capital is €5. If the producer’s costs are
constrained to €320 find the maximum level of production of the firm.
Maximise Q  L0.25 K 0.25 subject to 20L  5K  320
The Lagrange multiplier method:
Step 1:
The Lagrangian:
L  L0.25 K 0.25   320  20 L  5K 
Step 2:
L
 0.25 L0.75 K 0.25  20
L
L
 0.25 L0.25 K 0.75  5
K
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L
 320  20 L  5 K

Step 3:
Solve the simultaneous equations:
EQ1: 0.25L0.75 K 0.25  20  0
EQ2: 0.25L0.25 K 0.75  5  0
EQ3: 320  20L  5K  0
EQ1:
0.25 L0.75 K 0.25 20  
EQ2:
0.25L0.25 K 0.75 5  
Equate both expressions for  :
0.25L0.75 K 0.25 20  0.25L0.25 K 0.75 5
1.25L0.75 K 0.25  5L0.25 K 0.75
K 0.25 K 0.75  4 L0.25 L0.75
K  4L
EQ3: 320  20L  54L  0
320  40L  0
L8
K  4L  48  32
The optimal value of Qis where L  8 and K  32
Q  L0.25 K 0.25  80.25 320.25  4
5. See lecture overheads
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