PRELIMINARY MATHEMATICS

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PRELIMINARY MATHEMATICS

• LECTURE 5
• Optimization

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Reading

• Chiang, A. C. and K. Wainwright (2005) Chapters 9
  and 11.
• Thomas (1999) Chapters 10-12.
• Jacques (2006) Chapter 5.

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Local maxima and local minima
In this diagram, A and B represent turning points
of the curve y f(x). A is a maximum turning
point and B is a minimum turning point. The value
of y at point A is known as the local maxima
(maximum) and the value of y at point B is called
the local minima (minimum).
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Local maxima and local minima
Note that the slope of the curve y f(x),
measured by is 0 at both turning points.
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Second order condition for local maxima and
minima
If we draw the derivative on a graph we can
confirm that
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Points of inflexion
The condition however is a necessary
condition for local maxima or minima but not a
sufficient condition. In the function below
the sign of at point A is changing from
positive through 0 to positive again, so this is
neither a maximum nor a minimum turning point,
but a point of inflection.
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Second order condition for local maxima and
minima
What distinguishes the two points is the
slope/gradient of the derivative function
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Second order condition for local maxima and
minima
At point A, which is the point of local maximum,
the slope of the curve is negative, while at
point B, which is the point of local minimum, the
slope is positive.
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Second order condition for local maxima and
minima
The slope of is measured by the
derivative of known as the second
derivative, denoted as
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Second order condition for local maxima and
minima
we can see that at the local maximum on the
original curve, and
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Second order condition for local maxima and
minima
we can see that at the local maximum on the
original curve, and
and at the local minimum,
and
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Second order condition for a point of inflection
A point of inflexion can be characterised by
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Second order condition for a point of inflection
A point of inflexion can be characterised by
However it is also possible to have a local
maximum or a local minimum with
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Second order condition for local maxima and
minima
Hence, the obtained second order condition the
local maxima, and
and local minima,
and is a second order sufficient condition but
not necessary because it is also possible to have
a local maximum or a local minimum with
and
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n th order condition for local maxima and minima
n th derivative test for relative extrema or
inflection point. If at a
stationary point, continue differentiating
until a non-zero higher-order derivative is
obtained.
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n th order condition for local maxima and minima
n th derivative test for relative extrema or
inflection point. If at a
stationary point, continue differentiating
until a non-zero higher-order derivative is
obtained. If when evaluated at
the stationary point then

• if n is an odd number , we have a point of
  inflection
• if n is an even number AND ? a
  local maximum
• if n is an even number AND ?
  a local minimum

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nth derivative test an example.

• Let us consider y x 6.
• The first order necessary condition is obtained
  as
• and x 0,
• Next check the second order sufficient condition
• Evaluating at x 0 ,

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nth derivative test an example.
Because we cannot exclude the possibility that
the function is at a local maximum or local
minimum, we carry out the nth derivative test to
check if this is relative extrema or inflection
points. Third order derivative , at x 0,
Forth order derivative , at x 0,
Fifth order derivative , at x 0,
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nth derivative test an example.
Sixth order derivative Because the sixth
order derivative is non-zero, whereby the
non-zero is obtained at an order which is an even
number, and because the non-zero derivative is
positive, we have a local minimum.
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nth derivative test an example.
This is confirmed by drawing a diagram
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Recap Differentiation of bivariate functions

• To recap How to evaluate stationary points
• Step 1. First order condition
• Equate the first derivative to zero and find
  all stationary points.

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Recap Differentiation of bivariate functions
To recap How to evaluate stationary
points Step 2. Second order sufficient
conditions Evaluate the second order derivative
at each stationary points. If we have a
local maximum If we have a local minimum
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Recap Differentiation of bivariate functions
To recap How to evaluate stationary
points Step 3. If at any stationary point the
second order derivative is zero, carry on
differentiating until a non-zero higher order
derivative is found. If n is an odd number, we
have a point of inflection If n is an even
number AND we have a local maximum if n is
an even number AND we have a local minimum
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Second order partial derivatives
Second order partial derivatives Given a function
z f (x, y), the second order (direct) partial
derivative signifies that the function has been
partially differentiated with respect to one of
the independent variables twice while the other
independent variable has been held
constant The notation with a double
subscript signifies that the primitive function f
is differentiated partially with respect to x
twice. The alternative notation resembles
that of but with the partial derivatives
symbol.
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Second order partial derivatives
Second order partial derivatives The cross
partial derivatives signifies that the function
has been partially differentiated with respect to
one of the independent variables and then in turn
partially differentiated with respect to another
independent variable. The two cross
partial derivatives are identical with each
other, as long as the two cross partial
derivatives are both continuous (Youngs
theorem).
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Unconstrained optimisation
Recall that for a function of a single variable
such as y f (x) any point for which and
was a local minimum. Similarly, any point for
which and was a local maximum.
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Unconstrained optimisation
Now consider a function of two variables z f
(x, y) When does z have a local minimum?
Extrapolating our arguments of local minima of a
single variable function, we may expect z to have
a local minimum when it is not possible to obtain
a smaller value of z by making slight changes in
the values of x and y. There are three ways in
which we can make slight changes in x and
y (a) we can keep x constant and make y
vary (b) we can keep y constant and make x
vary (c) we can make both x and y vary at the
same time. If z has a local minimum at point x
a and y b, then we should be able to change the
values of x and y by any of the methods (a), (b),
or (c), and yet not make the value of z smaller.
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Unconstrained optimisation
Let us first consider case (a). In this case, the
value of x is held constant. At a point where
and we cannot obtain a smaller z by varying
y alone.
Similarly, if we consider case (b), when and
we cannot obtain a smaller z by varying x
alone.
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Unconstrained optimisation
Case (c) is slightly complicated. The condition
that must be satisfied here is or
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Unconstrained optimisation
Hence, the second order conditions are as
follows
A point at which the first partial derivatives
are zero is a local maximum if
a local minimum if
Note that if then we have
neither a maxima nor a minima but what is
called a saddle point
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Determinantal test

• In the two-variable case, the total differential
  is
• We can also derive the second-order total
  differential as

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Determinantal test
In the two-variable case, the total differential
is We can also derive the second-order total
differential as
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Determinantal test
If d 2 z gt 0 (positive definite), local
minima If d 2 z lt 0 (negative definite), local
maxima
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Determinantal test
Since the variables x and y appear only in
squares, this can be seen as a quadratic form. As
discussed in lecture 1, the condition for d 2 z
to be positive/ negative definite may be stated
by use of determinants. The quadratic form can be
expressed in matrix form as
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Determinantal test
The determinant with the second order partial
derivatives as its elements is called a Hessian
determinant.
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Determinantal test
The principal minors in this case are
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Determinantal test
Hence the second order conditions are d 2 z gt
0 (positive definite), iff and
, then local minima. d 2 z lt 0 (negative
definite), iff and , then local maxima.
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Determinantal test
If we consider a function of three choice
variables, The Hessian determinant is
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Determinantal test
And the principal minors are
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Determinantal test
d 2 z gt 0 (positive definite), iff and
, then local minima. d 2 z lt 0 (negative
definite), iff and , then local
maxima.
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Unconstrained optimisation application
Application profit maximisation Let us consider
a firm that is a price taker, produces and sells
two goods, goods 1 and goods 2 in perfectly
competitive markets, and has a profit
function What would be the level of output
for goods 1 and 2 satisfying the first order
condition for maximum profit?
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Unconstrained optimisation application
Application profit maximisation Let us consider
a firm that is a price taker, produces and sells
two goods, goods 1 and goods 2 in perfectly
competitive markets, and has a profit
function What would be the level of output
for goods 1 and 2 satisfying the first order
condition for maximum profit?
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Unconstrained optimisation application
Application profit maximisation Setting both
condition equal to zero,
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Unconstrained optimisation application
Application profit maximisation Setting both
condition equal to zero, Rewrite as
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Unconstrained optimisation application
Application profit maximisation Setting both
condition equal to zero, Rewrite
as Express this in terms of matrices
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Unconstrained optimisation application
Application profit maximisation Use Cramers
rule
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Unconstrained optimisation application
Application profit maximisation Let us consider
a firm that is a price taker, produces and sells
two goods, goods 1 and goods 2 in perfectly
competitive markets, and has a profit
function What would be the level of output
for goods 1 and 2 satisfying the first order
condition for maximum profit? The level of
output/supply for good 1 must be 40 and for good
2 be 24 in order to satisfy the first order
conditions.
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Unconstrained optimisation application
Application profit maximisation Let us consider
a firm that is a price taker, produces and sells
two goods, goods 1 and goods 2 in perfectly
competitive markets, and has a profit
function What would be the level of output
for goods 1 and 2 satisfying the first order
condition for maximum profit? The level of
output/supply for good 1 must be 40 and for good
2 be 24 in order to satisfy the first order
conditions. Next, check the second order
conditions to verify if we have local maxima.
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Unconstrained optimisation application
Application profit maximisation Let us consider
a firm that is a price taker, produces and sells
two goods, goods 1 and goods 2 in perfectly
competitive markets, and has a profit
function Let us consider whether the
stationary point we found at Q1 40 and Q2 24
is a local maxima or a local minima.
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Unconstrained optimisation application
Application profit maximisation The Hessian
matrix can be written as Hence the
profit is at local maximum at Q1 40 and Q2 24