# Partial Differentiation & Input Output Analysis Questionnaire

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Applied Statistics and Mathematics in Economics & Business
Introduction
BS1501
Functions of more than one variable
Till now we have only considered functions dependent on one variable, e.g. y = 3x + 8. Most
functions involve more than one variable, e.g. profit may depend on labour and time.
A function f of two variables, x and y, can be written f (x, y), e.g. f (x, y) = xy + 12x.
To evaluate this function, we specify the numerical values of the variables, x and y, e.g. if x =
3 and y = 4, then f (3, 4) = 3  4 + 12  3 = 48. Note that f (3, 4) and f (4, 3) probably have
different values, in this case f (4, 3) = 4  3 + 12  4 = 60.
For functions of one variable we constructed differentiation, for functions of more than one
f
variable we use partial differentiation, e.g.
(read as ‘partial dee f by partial dee x’ or
x
sometimes just ‘partial f by partial x’). It is found by differentiating with respect to x, the
associated y is temporarily treated as a constant (acts like a constant). For example,
f (x, y) = 12x + xy then
f
= 12 + y
x
More examples;
f (x, y) = x3 + 36xy
f (x, y) = x3 y + 12xy2
then
then
f
= 3×2 + 36y
x
f
= 3×2 y + 12y2
x
and
and
f
= 36x
y
f
= x3 + 24xy
y
See Exercise 13
Example
Write out all first and second order partial derivatives for the function
f(x, y) = 8×3 y2
We may need to investigate the turning points of these functions of more than variable.
Example
Consider the function f (x, y) = 10x – x2 + 10y – y2 , what can we say about the turning points
of this function? The graph of this function is given below.
From the graph we see that this function
has a maximum (top of the hill shape
shown). Can we find for what values of x
and y is the function at its maximum?
Looking at the arrows labelled 1 and 2
(tangents at the top), we see a point of
intersection where the maximum occurs.
Considering just arrow 1, its slope in the x
direction is 0, similarly for the arrow 2 its
slope in the y direction is 0.
That is, the points of x and y such that the function f (x, y) is optimal is in mathematical terms
when;
f
f
= 0 and
= 0.
x
y
f
f
= 10 – 2x = 0 and
= 10 – 2y = 0, hence there is an optimal ‘turning’
x
y
point at x = 5 and y = 5.
In our example,
Is this optimum a maximum, minimum
or something else, possibly a point of
inflection or saddle point? For example,
f (x, y) = x2 – y2 (graphically shown).
To investigate the nature of the turning points, we need to differentiate the first order partial
derivatives, there are a number of outcomes to this (for f (x, y) = 10x – x2 + 10y – y2 );
2f
= –2,
 x2
2f
 x2
and
2f
 y2
2f
= 0,
 x y
are similar to
second order derivatives in one
2f
2f
variable calculus,
and
 x y
 y x
are the cross partial derivatives,
2f
2f
here, in general
=
.
 x y  y x
2f
= –2,
 y2
2f
= 0,
 y x
f
f
= 0 and
= 0, have been satisfied, the following constraints, if met, will be
x
y
sufficient for determining whether an identified turning point is a maximum or a minimum.
Assuming,
2
For a maximum:
 2f 
2f
2f
2f  2 f

 ,
< 0, < 0 and    x2  x2  y 2  y2   x y  For a minimum: 2f 2f 2f  2 f  2f   . > 0,
> 0 and


 x2
 x 2  y 2   x y 
 y2
2
The latter inequality is the associated Discriminant. Hence in our example we have,
2f
2f
=
–2
< 0, = –2 < 0  x2  y2 2 and  2f  2f 2f   .  = 4 >
0
=
 x2
 y2
  x y 
So we can say that at the turning point, x = 5 and y = 5, there is a maximum, which is verified
by the graphical representation of f (x, y).
Example
Find the turning point and its nature for the following function,
f (x, y) = 3×2 + 4xy – 14x – 20y + 4y2 + 1000.
Firstly, partially differentiate with respect to x and y separately, giving (when set to zero),
f
f
= 6x + 4y – 14 = 0 and
= 4x – 20 + 8y = 0.
x
y
Rewriting the above equations, then,
6x + 4y = 14
4x + 8y = 20
Solving these two simultaneous equations we
get the solution, x = 1 and y = 2. That is, there
is a turning point when x = 1 and y = 2.
However, is it a maximum or minimum?
Partially differentiating again we get,
2f
2f
=
6
>
0,
= 8 > 0,
 x2
 y2
and
2f
= 4,
 x y
2
 2f 
2f
2f

 is
and so

 x2
 y2
  x y 
satisfied. In this case 6  8  42 so is
also satisfied. It follows, the above
function has a minimum at the point, x
= 1 and y = 2, the value of f (x, y) at this
point is;
f (x, y) = 3  12 + 4  1  2 – 14  1
– 20  2 + 4  22 + 1000
= 3 + 8 – 14 – 40 + 16 + 1000
= 973
See Exercise 14
Applied Statistics and Mathematics in Economics & Business
Lagrange Multipliers
BS1501
Optimisation
Many types of problems or decisions have certain constraints associated with them, in the
form limited capitol, labour, machinery, time or other resources.
The problem of optimising our use of these resources to achieve maximum profit or minimum
labour and time expenditures etc., is of importance to us.
Here we will illustrate one way in which we can maximise (minimise) the values of functions
depending on certain constraints.
Introduction
The method of Lagrange Multipliers aims to optimise an objective function:
f (x, y)
(e.g. f (x, y) = xy)
(throughout this section we will consider objective functions dependent on two variables
only), subject to a constraint:
h (x, y) = M
(e.g. x + 4y = 16)
The graph shows the objective function f (x, y) = xy and constraint x + 4y = 16. In the left
graph, along a curve its value stays the same for different values of x and y (e.g. f (4, 1) = 4
and f (6, 2) = 12).
This family of curves moves in a north east direction as the value of f (x, y) increases. The
straight line represents x + 4y = 16. Our goal is to find the highest valued curve of f (x, y)
which touch’s the straight line (but must otherwise lie on one side of the curve).
In the above example we see that the values x = 8 and y = 2 achieves this, giving the
maximum value of f (x, y) to be f (8, 2) = 16.
How can we do this mathematically, points to note are:
At the constrained optimum point c = (xc, yc), the highest level curve of f (x, y) is tangent to
the constraint h (x, y) = M, i.e. they have the same slope at this point, this can be written as:
f
h
( xc , yc )
(x , y )
x
x c c
=
f
h
( xc , yc )
(x , y )
y
y c c
Rewriting the equation as
where  is the common value of the two
quotients. Hence we can rewrite the above
as two equations, i.e.
f
f
( xc , yc )
(x , y )
x
y c c
=
=
h
h
(x , y )
(x , y )
x c c
y c c
f
h
( xc , yc ) − 
(x , y ) = 0
x
x c c
f
h
( xc , yc ) − 
(x , y ) = 0
y
y c c
there are three unknowns in these two equations i.e. xc, yc and , so we need a third equation.
So we use the equation h (x, y) − M = 0.
Hence we are now in a position to describe the method, which is as follows
i)
Define the new function (called the Lagrangian Function)
g (x, y, ) = f (x, y) + (M − h (x, y))
note: (M − h (x, y)) will not change the value of g (x, y, ), since (M − h (x, y)) = 0.
ii) Solve the simultaneous equations formed by calculating
g
f
h
( x, y ,  ) =
( x, y ) − 
( x, y ) = 0
x
x
x
g
f
h
( x, y ,  ) =
( x, y ) − 
( x, y ) = 0
y
y
y
g
( x, y,  ) = M – h (x, y) = 0

to determine the x, y and . The scalar  is called the Lagrange Multiplier.
Example
Use Lagrange multipliers to find the optimal value of the objective function
xy
subject to the constraint
x + 4y = 16
Using our theory then we have
f (x, y) = xy,
h (x, y) = x + 4y,
M = 16
the Lagrangian function is given by
g (x, y, ) = xy + (16 − x − 4y)
= xy + 16 − x − 4y
Working out the three partial derivatives of g gives
g
g
g
=y −=0
= x − 4 = 0
= 16 − x − 4y = 0
x
y

hence we need to solve the 3 equations with 3 unknowns
y −  =0
x
− 4 = 0
x + 4y
= 16
[1]
[2]
[3]
To find  put y =  from [1] and put x = 4 from [2], in [3]
4 + 4 = 16
so
8 = 16
giving
 = 2.
It follows that y = 2 and x = 8.
Therefore the optimal solution is when x = 8 and y = 2, the corresponding value for the
objective function xy subject to the constraint x + 4y = 16 is f (8, 2) = 8  2 = 16
Up till now we have no idea whether the optimal solution is maximum or minimum, from our
discussion, we have a condition which if satisfied tells us that the function at this point is a
maximum or minimum, this being:
For a maximum:
g
= 0,
x
g
= 0,
y
g
= 0,

 2g
< 0,  x2  2g < 0,  y2  2g  2g   2g      x2  y 2   x y  2 g = 0,   2g > 0,
 x2
 2g
> 0,
 y2
 2g  2g   2g 



 x2  y 2
  x y 
2
For a minimum:
g
= 0,
x
g
= 0,
y
If neither of these sets of conditions hold, then more investigation is needed, this is not
discussed in this course. For this course this is the only check we make to gain information on
whether the optimal solution is a maximum or minimum.
g
g
g
= 0,
= 0,
=0

y
x
since we have an optimal solution when x = 8 and y = 2, so now look at the second
derivatives;
Referring back to our previous example then we already know
 2g
 2g
=
0,
= 0,
 y2
 x2
 2g
= 1,
 x y
so we are unable to determine the nature of the optimal solution using this method.
See Exercise 15
2
 2g  2g   2g 
hence

8
2  36 +
not scarce
7
7
So the Anthracite and Best coal needs were only just met, calculating their shadow prices;
If we were able to produce one less unit of Anthracite
what would be the increase/decrease in price? First solve
Remove U from equation [2], so: [2]  5  [1]
Giving U =
14L = 5,
, putting these values into the
equation for the cost, C = 100U + 150L, we get:
U + 4L = 7
5U + 6L = 30
hence L =
C = 100 
£610.71
[1]
[2]
5
14 hours
+ 150  5
14 =
So decrease in cost would be £621.43  £610.71 = £10.72, hence the Shadow price for
Anthracite is £10.72.
If we were able to produce one less unit of Best coal what
would be the increase/decrease in cost? First solve
Remove U from equation [2], so: [2]  5  [1]
Giving U =
68
equation for the cost, C = 100U + 150L, we get:
[1]
[2]
11 hours
14
11
C = 100 68
14 + 150 14 =
14L = 11,
14 , putting these values into the
U + 4L = 8
5U + 6L = 29
hence L =
£603.57
So decrease in cost would be £621.43  £603.57 = £17.86, hence the Shadow price for Best
coal is £17.86.
See Exercise 31
Applied Statistics and Mathematics in Economics & Business
BS1501
Resource allocation
Linear
Programming
Before we start
Sketch the feasibility area (shading it in, which satisfies) the following inequalities:
y  2,
x + y ≤ 10,
y ≤ 4,
y–x≤1
—————————————-
2. Sketch the feasibility area (shading it in, which satisfies) the following inequalities:
y < 5, y > 3,
x < 5, y – 2x > –4
Introduction
Linear programming is a method for finding the optimal solutions to problems of constrained
optimisation.
Many real life problems will mean many variables to consider, here the examples only use 2
or 3 variables. To describe linear programming we consider a simple example.
Example
A farmer has 100 acres of land, and £5,000 capital available, she has the choice of two crops
she can grow on her land, these being wheat and barley. She estimates that wheat costs £40
per acre and barley £60 per acre to seed and the respective profits per acre are £30 (wheat)
and £40 (barley), what mixture of wheat and barley must she grow for maximum profit?
 Let W be the number of acres of wheat.
 Let B be the number of acres of barley.
 Let P be the profit in total.
looking at the land constraint, we see the maximum amount of land
available to us is 100 acres, so our first constraint is that the amount of
land growing wheat (W) or barley (B) cannot exceed 100 acres, i.e.
The amount of money available is £5,000, this amount will be split up
between the number of acres of wheat (£40 per acre) or barley (£60
per acre) which is grown, so our second constraint is:
The amount of profit (P) we achieve is also dependent on the
number of acres of wheat (£30 profit per acre) and barley (£40
profit per acre) grown and is given by:
W + B  100.
40W + 60B 
5000.
P = 30W + 40B,
our aim is to maximise
this value P.
We note also that we have two non-negative constraints, these being W  0 and B  0, since
we are unable to grow a negative number of acres of wheat and barley.
Summary table
Land
Cash
Profit
Non-negative
Wheat (W acres)
l (acre per acre)
40 (£ per acre)
30 (£30 per acre)
Barley (B acres)
1 (acre per acre)
60 (£60 per acre)
40 (£40 per acre)
Constraint
W + B  100
40W + 60B  5000
maximise P = 30W + 40B
W  0, B  0
We now turn our attention to constructing a
graphical interpretation to the above constraints:
For the land constraint, when W = 0 then B  100,
similarly when B = 0 then W  100, as shown.
The shaded area is that area of the graph which
satisfies W + B  100 (also satisfies the constraints
W  0 and B  0). In a similar way we could also
draw a graph (and subsequent shaded area) for the
constraint 40W + 60B  5000, see graph.
Combining the two graphs we get a graph of the
form shown.
Where the shaded area in this graph is that area
within which all 4 constraints (inequalities W + B 
100, 40W + 60B  5000, W  0 and B  0) are
satisfied (sometimes called the feasible region)
Turning our attention to the profit constraint i.e.
maximise P = 30W + 40B, we can draw a series of
lines for various values of P by writing the previous
equation as B = 
3
1
W+
P.
4
40
Any two points on one of these lines has the same
profit value, we see that as the lines move away
from the origin the value of P increases.
The above diagram combines the two previous
graphs, we can find the most profitable point within
the feasible area for P given the by the point A.
This point can be found mathematically by solving
the simultaneous equations
W + B = 100
40W + 60B = 5000
[1]
[2]
From [1], let B = W + 100 into equation [2], giving, 40W + 60(W + 100) = 5000
40W  60W + 6000 = 5000
20W = 5000  6000
20W = 1000
W = 50
which gives B = 50, putting these values into the equation for the profit P = 30W + 40B we
get,
P = 30  50 + 40  50 = £3,500.
What happens if the profit on wheat becomes £40
per acre and the profit on barley becomes £35 per
acre?
The profit function changes to be
P = 40W + 35B,
hence we get a slightly different set of lines denoting
the change in the value of profit
Using this above set of lines and the original feasible
region we have
In this case the optimum point is when
W = 100 and B = 0,
i.e. grow only Wheat. Let us consider another
example.
The change in the solution came about from a change in the gradient of the objective
function, and is an issue on the sensitivity of the solution.
Consider the gradients of the lines which define the
feasible region.
The land constraint B = W + 100, gradient is 1.
For the capital constraint gradient is 
2
3
While gradient of objective function line’s gradient is between 1 and 
as found. Objective function B = 
W+
, solution will stay
, between these values.
In the change of this example the objective function
became;
P = 40W + 35B,
i.e.
B=
40
1
W+
P,
35
35
40
, i.e. 1.1428.
35
Hence we would have expected the optimum point
to have changed in this case.
How about at the optimum points, in these cases we have only one bound on the gradient of
2
the optimum function, i.e. at P gradient must be greater than 
and at Q gradient must be
3
less than 1.
Example
Jarvis Ltd. makes two kinds of doors ‘Sturdy’ and ‘Deluxe’, but are limited because of
limitations on the amount of, glass, wood and labour, available. The firm can obtain 600m 2 of
wood per week at a cost of £4 per m 2 of wood and 300m2 of glass at a cost of £16 per m 2,
they have 8 operators, each working a 40 hour week, at an hourly rate of £6.
‘Sturdy’ uses 2m2 of wood and 0.5m2 of glass, while ‘Deluxe’ uses 1.5m2 of wood and 1m2 of
glass. It takes 1 hour of labour time to cut and assemble each door. ‘Sturdy’ sells for £30 and
‘Deluxe’ sells for £40.
What is the optimum number of ‘Sturdy’ and ‘Deluxe’ doors to sell that maximises profit?
Summarising data
Wood
Glass
labour
Non-negative
Sturdy (S doors)
2 (m2)
0.5 (m2)
1 (hour)
Deluxe (D doors)
1.5 (m2)
1 (m2)
1 (hour)
Constraint
2S + 1.5D  600
0.5S + 1D  300
S + D  320
S  0, D  0
The wood, glass and labour constraints give us the
following feasibility region
To calculate the profit function we need to remove all the resource costs from the selling
price of each door
Sturdy:
£30  2  £4  0.5  £16  £6 = £8
profit
Deluxe:
£40  1.5  £4  1  £16  £6 = £12 profit
Hence profit (P) is given by;
P = 8S + 12D
In the same way as the previous example we can
create a series of lines for various values of P, any
two points on a single line create the same amount
of profit;
Combining these graphs
We can yet again solve this numerically by first solving where the optimum profit touches
two of lines making up the feasibility region
0.5S + 1D = 300
[1]
1S + 1D = 320
[2]
To remove D,
[2]  [1]
0.5S = 20 hence S = 40, so D = 280
Putting these values into the equation for the profit i.e. P = 8S + 12D we get
P = 8  40 + 12  280 = £3,680
Q8.
Q7.
Q9.
zan-hust thε-26 = (hik) f
hчу) = 142
М – ер 50
8 (1, 2) (#2 – зазначу-ч-)
г 8
t» ( 50-х-2)

оз – kazg- = 1
ne
.
9.
a
бо
– се
о: 4– a5 =
— 64 + 84
+ 8x -x=0

82-84 = 2 о
8 – 4 = 2
e=h-x-es
Now, eq” 0 teq (2)
-)
x=la E r2 =h2
POCO
SMOT ON DOC MODO
94, 95-) – 8
1 – 8 м, 4 )
8y-on-
do
8y-6y=
2lambda
2y=2lmabda
y=lambda
8n-by-d-o

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partial differentiation

input output analysis

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