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Search for: Center of gravity, Taylor’s theorem, Local maxima, Local minima,
Saddle points, Lagrange multipliers [If you are using this document on a computer,
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Key Concepts
Center of gravity
The center of gravity is also referred to as the center of mass. This is a point at
which vertical and horizontal moments of a given system balance.
Taylor’s formula
This seeks to extend the Taylor series expansion of a function f (x) of a single
variable at a point x = a to a function f (x, y) of two variables at a point
(a, ).
b
Relative Extrema
Relative extrema is a collective terminology for the relative maximum and mini-
mum values of a function. The singular form of the word is relative extremum,
which may be a relative maximum or a relative minimum value.
Lagrange Multipliers
Lagrange multipliers are the numbers (or parameters) associated with a method
known as the Lagrange multipliers method for solving problems of optimization
(extrema) subject to a given set of constraints.
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Key Theorems and/or Principles
Necessary Condition for Local Extreme Values
If f (x, y) has a relative extreme value (local maximum or local minimum) at
∂f
∂f
a point (a, )
b and if
f =
, f =
exist, then f (a, )
b
x
= 0
x
∂x
y
∂y
and f (a, )
b
.
y
= 0
Second Derivative Test for Local Extreme Values
Let f (x, y) possess continuous first and partial derivatives at a critical point
(a, )
b . At point (a, )
b we define the quantity
f
f
xx
xy
2
D(a, )
b =
= f f − f
.
xx yy
xy
f
f
yx
yy
Then,
• If D > 0 and f
,
then f (a, )
b is a local minimum.
xx > 0
• If D > 0 and f
,
then f (a, )
b is a local maximum.
xx < 0
• If D < 0 then f has neither a local minimum nor a local maximum point
at (a, )
b . The point is known as a saddle point.
Learning Activity
Center of Gravity
In this module we will limit our presentation of the concept of center of gravity
to functions of two variables.
Let us consider a thin, flat plate (a lamina) in the shape of a region D in 2
ℜ ,
whose surface density at any point P ∈ D is a function ρ (x, y) of the coordi-
nates (x, y) of the point. Our interest is to find the coordinates x and y of
the point C ∈ D where, from an engineering point of view, the plate will balance
if a support is placed there.
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The point C (x, y) is called the center of gravity or the center of mass of the
lamina.
Calculation of the coordinates x and y of the center of mass involves calcu-
lating three related quantities, namely:
• The total mass M of the region D ;
• The vertical moment M of the lamina about the x − axis. M is a measure
x
x
of the tendency of the lamina to rotate about the x − axis.
• The horizontal moment M of the lamina about the y − axis. M is a
y
y
measure of the tendency of the lamina to rotate about the y − axis.
After calculating these three quantities, the coordinates of the center of gravity
are then obtained using the formulas:
M
M
x =
y ;
y =
x .
M
M
Calculation of the Total Mass M
The total mass M of a region D in the xy − plane and having surface den-
sity function (mass per unit area) ρ (x, y) is given by the value of the double
integral
M = ∫∫ρ(x, y)dxdy.
D
Example
If the mass of a circular plate of radius r is proportional to its distance from the
center of the circle, find the total mass of the plate.
Solution
The surface density of the plate is indirectly given as
2
2
ρ( ,
x y) = k x + y where
k is a constant of proportionality. The total mass is then calculated from
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r
⎡2π 2
⎤
2
M =
k x2 y dxdy
∫∫
+ 2
= ∫
r dθ dr
⎢ ∫
⎥
=
3
kπr .
3
D
0
⎣ 0
⎦
Calculation of the Vertical Moment M x
The vertical moment of the lamina is given by M =
ρ (x, y) ydxdy
x
∫∫D
Calculation of the Horizontal Moment M y
The vertical moment of the lamina is given by M =
ρ(x, y)xdxdy
y
∫∫D
Example
Determine the coordinates of the center of gravity of the section of the ellipse
2
2
x
y
+
= 1 which lies in the first quadrant, assuming that the surface density
2
2
a
b
ρ( ,
x y) = 1 at all points.
Solution
The total mass
2
2
b a
a
x
⎡
−
⎤
M = ∫∫ dxdy = ⎢ dy⎥dx
a ∫
∫
D
0 ⎢
0
⎥
⎣
⎦
Using the substitution x = a cos θ the above iterated integral gives the value
1
of the total mass as M = abπ
4
Similarly, using the same substitution, the values of the horizontal and vertical
moments
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b a a2 x
⎡ − 2 ⎤
M =
xdxdy =
⎢
dy⎥xdx and
y
∫∫
a ∫
∫
D
0 ⎢
0
⎥
⎣
⎦
2
2
b a
a
x
⎡
−
⎤
M =
ydxdy =
⎢
dy⎥ydx
x
∫∫
a ∫
∫
D
0 ⎢
0
⎥
⎣
⎦
2
a b
2
ab
are found to be M =
and M =
y
3
x
3
4a
4b
These results lead to the coordinates x =
and y =
for the center
π
3
π
3
of gravity of lamina.
Self Exercise
1. Assuming a constant surface density ρ (x, y) = k find the coordinates of
the center of mass of the upper half of a circular lamina of radius r .
2. Find the center of gravity of a thin homogeneous plate covering the region
D enclosed between the x − axis and the curve
2
x = 2y − y .
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Taylor’s Formula
On page 354 of our basic reference: The Calculus Bible by Brigham Young we
encountered Taylor’s formula for a function y = f (x) of a single variable about
a point x = a . The relevant theorem (Theorem 8.7.1) states that, if f and all its
derivatives '
'
' '
( n+1
f , f , f ,... f
are continuous for all x in a small neighborhood
of a point x = a , then
n 1
f (x) = ∑ (x − a)k (k)
f
(a) + Rn
k=0 !
k
1
where R =
(x − a)n 1+ (n+ )1
f
(ξ ) , with ξ ∈ (a, x) .
n
(n + )!
1
We now wish to extend this theorem to functions of two variables. Specifically,
we wish to expand a function z = f (x, y) of two variables in a Taylor series
about a point (a, )
b in the domain
2
D ⊂ ℜ of the function.
Let z = f (x, y) be a function of two variables which is continuous, together with
all its partial derivatives up to the
st
(n + )
1 order inclusive, in some neighborhood
of the point (a, ).
b Then, like in the case of a function of a single variable, the
function f (x, y) can be represented in the form of a sum of an n-th degree
polynomial in powers of (x − a) , ( y − )
b , and a remainder term R .
n
Under this assumption the function f (x, y) can be expressed in a Taylor formula
about the point (a, )
b as follows:
k
n 1 ⎡
∂
∂ ⎤
f (x, y) =
(x − a)
+ ( y − b
∑ ⎢
)
⎥ f (a, )
b + R ,
n
k 0
!
= k
x
∂
y
⎣
∂ ⎦
n
1
( + )
1
⎡
∂
∂ ⎤
where R =
(x − a)
+ ( y − )
b
f (ξ,η)
n
(n + )!
1 ⎢
x
∂
y⎥
⎣
∂ ⎦
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for some point (ξ ,η ) lying in the disc centered at (a, )
b and containing the
point (x, y) .
Observation
The learner should pay special attention to the operator
D = (x a ∂
− )
+ ( y b ∂
− )
which appears in the above Taylor expansion
x
∂
y
∂
formula. Powers of this operator are formed in accordance with the binomial
expansion of the term
n
(a + b) . Typically,
2
2
2
2
2
⎡
∂
∂ ⎤
2 ∂
∂
2
D = (x − a)
+ (y − )
b
= (x − a)
+ 2(x − a)(y − )
b
+ (y
)
b ∂
−
⎢
⎥
2
2
x
∂
y
∂
x
⎣
⎦
∂
x
∂ y
∂
y
∂
Do This
Write down all the terms of the differential operator D3.
Maximum, Minimum and Saddle Points Functions of Two Variables
Let z = f (x, y) be a function of two variables defined over a region
2
D ⊂ ℜ
We begin by making three observations:
1. The maximum and minimum values of a function in a given domain are
collectively referred to as extreme values or simply as extrema of the function.
Without specifying the nature of the extreme value, the singular form of the word
extrema is extremum.
2. An extremum of a function z = f (x, y) may be absolute or relative.
3. An extremum of a function z = f (x, y) can occur only at
i) a boundary point of the domain of f ,
ii) at interior points where f = f = 0 or
x
y
iii) at points where f and f fail to exist .
x
y
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Definitions
i) A point P(x ,
is said to be a point of absolute maximum if
0 y )
0 ∈ D
f (P ) ≥ f (Q ) for all points Q ∈ D .
ii) A point P(x ,
is said to be a point of absolute minimum if
0 y )
0 ∈ D
f (P ) ≤ f (Q ) for all points Q ∈ D .
iii) A point P(x ,
is said to be a point of absolute maximum if
0 y )
0 ∈ D
f (P ) ≥ f (Q ) for all points Q ∈ D which are in a small neighborhood
of point
iv) A point P(x ,
is said to be a point of absolute minimum if
0 y )
0 ∈ D
f (P ) ≤ f (Q ) for all points Q ∈ D which are in a small neighborhood
of point
As can be noted from the above definitions, the difference between absolute and
relative maximum or minimum lies in the set of points Q one compares the fixed
point P with. If Q is restricted to a small neighborhood of point P then we
speak of relative extrema (minimum or maximum).
Basic Question:
How can one locate the points P at which the function
f (x, y) has extreme values?
Assuming that f (x, y) has continuous first partial derivatives, the answer to the
above question is simple:
All maximum and minimum points which lie in the interior of the domain of
definition of f (x, y) must be among the critical points of the function.
The critical points of f (x, y) are those points which simultaneously satisfy the
two equations
∂f
f
∂
= 0 ;
= 0
∂x
y
∂
Once we have obtained all the critical points we shall then have to classify which
among them are maximum points and among them are minimum points. We shall
also have to say something about any critical points which represent neither a
maximum nor a minimum point of the function. An easy method of classifying
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critical points involves the second partial derivatives of f (x, y). The method
is given in Section 2.4 above and for ease of reference, we have reproduce it
here.
Second Derivative Test for Local Extreme Values
Let f (x, y) possess continuous first and second partial derivatives at a critical
point
P(x , y ) . At P we define the quantity
0
0
f
f
xx
xy
2
D(a, )
b =
= f f − f
.
xx yy
xy
f
f
yx
yy
• If D > 0 and f
, then f (a, )
b is a local minimum.
xx > 0
• If D > 0 and f
, then f (a, )
b is a local maximum.
xx < 0
• If D < 0 then f has neither a local minimum nor a local maximum point
at (a, )
b . The point is known as a saddle point.
Worked Example
Classify the critical points of the function f (x, y) = x3 + y3 − 3xy .
Solution
The first and second partial derivatives are
∂f
f
∂
= 3x 2 − 3 y ;
= 3 y 2 − 3x
∂x
y
∂
2
∂ f
∂ 2 f
∂ 2 f
2
∂ f
= 6x ;
=
= − 3 ;
= 6 y .
2
x
∂
∂x∂y
∂y∂x
2
y
∂
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∂f
f
∂
Solving the simultaneous nonlinear equations
= 0 ;
= 0 one gets
∂x
y
∂
the following critical points: (0,0) and
)
1
,
1
(
. Applying the second derivative
test on each point we find:
0
− 3
For point P (0,0) : D( ,
0 0) =
= 9
− < 0
− 3
0
We therefore conclude that the point P (0,0) is neither a local minimum nor a
local maximum, but is a saddle point.
6
− 3
For point P
)
1
,
1
(
: D
)
1
,
1
(
=
= 25 > 0
− 3
6
Since f
)
1
,
1
(
we conclude that P
)
1
,
1
(
is a point of relative minimum.
xx
> 0
Self Exercise
1. Test the surface
4
4
z = f ( ,
x y) = 4xy − x − y for maxima, minima and saddle
points. Calculate the values of the function at the critical points.
2. D e t e r m i n e t h e r e l a t i v e e x t r e m a f o r t h e f u n c t i o n
f ( ,
x y)
3
3
= x + y + 3 2
y − 3x − 9y + 2
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Lagrange Multipliers
Constrained Extrema For Functions Of Two Variables
In Section 2.5.3 we discussed the problem of determining the extrema (maximum
and minimum values) of a function f (x, y) . We did so without imposing any
condition on the set D of possible values of x and y .
In this Section we revisit the problem of finding an extremum (singular form of
extrema) of a function f (x, y) but with a difference. Here we impose a condi-
tion (or a set of conditions) on the variables x and y . Typically we may wish
to find the extrema of f (x, y) only for points lying on a particular curve in the
domain D of the function, such as
g(x, y) = 0
The problem of finding extreme values of a function f (x, y) subject to a
constraint of the general form g(x, y) = 0 is a typical example of a constrained
optimization problem.
Example
Find the maximum volume of a rectangular box having a square base and open
at the top, if the side length of the base is x cm, the height is y cm and the total
area of the material to be used is 108
2
cm
Reduction Method
A possible method of solving the general constrained minimization problem
posed here is the following:
(i) Solve the equation g(x, y) = 0 for one of the variables, whichever is most
convenient to solve for.
(ii) Substitute the result into the function f (x, y) being optimized. The substitu-
tion will eliminate one of the variables from the expression for f (x, y) reducing
the problem to a single variable extremum problem.
To illustrate this we apply the method on the example given above.