Applied Mathematics II
Chapter-3:
Differential Calculus of Functions of Variables
Nigussie Chanie Admassu
nigussieadm@gmail.com
February 22, 2023
Bahir Dar Institute of Technology
1
Chapter 3: Differential Calculus of Functions of Variables
3.1 Functions of Several Variables
3.2 Limits and Continuity
3.3 Partial Derivatives
3.4 The Chain Rule
3.5 Directional Derivatives and the Gradient Vector
3.6 Maximum and Minimum Values
3.7 Lagrange Multipliers
3.1 Functions of Several
Variables
2
3.1 Functions of Several Variables
Definition-3.1:
• A function f of two variables, x and y, is a rule that assigns a
unique real number
z = f (x, y)
to each point (x, y) in some set D in the xy-plane.
Example-3.1:
f (x, y) = x2 − 2xy + y 2 − 4 is a function of two variables.
3
• A function f of three variables, x, y, and z, is a rule that assigns a
unique real number
w = f (x, y, z)
to each point (x, y, z) in some set D in three-dimensional space.
Example-3.2:
f (x, y, z) = xeyz is a function of three variables.
• A function f of n variables, x1 , · · · , xn is a rule that assigns a
unique real number
w = f (x1 , · · · , xn )
to each "point" (x1 , · · · , xn ) in some set D in n-dimensional space.
• In both cases, D is the domain of f.
4
To find the domain D of f , consider the following:
• expressions in the denominator must be non-zero.
• expressions in the nth root if n is even must be non-negative.
• expressionsin the logarithms must be positive.
Example-3.3:
2
)
Find the domain of f (x, y) = √ln(y−x
2
2
4−x −y
Solution
y − x2 > 0 ∧ 4 − x2 − y 2 > 0
⇒ Domain of f = {(x, y)| y > x2 ∧ x2 + y 2 < 4}.
5
Graph, Level Curve, Surface
Definition-3.2:
i) The set of points (x, y, f (x, y) ) in space, for (x, y) in the domain
of f , is called the graph of f.
• The graph of a function z=(x,y) of two variables is called a surface.
ii) Given a function f(x,y) and a number k in the range of f,a level
curve of a function of two variables for the value k is defined to
be the set of points satisfying the equation f(x,y)=k.
• A level curve can be described as the intersection of the horizontal
plane z = k with the surface defined by f.
• Level curves are also known as contour lines.
iii) For a function w=f(x,y,z), the level surface of value k is the
surface S in R3 on which f(x,y,z)=k.
• Level surface are basically the same as level curves in principle,
except that the domain of f (x, y, z) is in 3D-space.
• Therefore, the set f (x, y, z) = k describes a surface in 3D-space
rather than a curve in 2D-space.
6
Example-3.4:
p
Describe the level curves of the function. f (x) = x2 + y 2 .
Solution
• A level curve is the graph of the equation
x2 + y 2 = k 2 , k ≥ 0, which describes a
circle with radius k.
• Taking different values of k, we obtain
k = 1,
x2 + y 2 = 1
k = 2,
x2 + y 2 = 4
k = 3,
x2 + y 2 = 9
k = 4,
x2 + y 2 = 16
7
Quadratic Surfaces:
A quadratic surface is a surface given by the general second-degree
equation:
Ax2 + By 2 + Cz 2 + Dxy + Exz + F yz + Gx + Hy + Iz + J = 0,
where A, B, · · · , J are all constants.
1. Ellipsoids are closed, bounded
quadratic surfaces defined by
the equation
x2
y2
z2
+
+
= 1.
a2
b2
c2
When a = b = c, the ellipsoid
is a sphere.
8
2. Paraboloid:
Quadratic surfaces
with equation
x2
y2
+ 2 = z, are
2
a
b
elliptic paraboloids and
x2
y2
− 2 = z, are
2
a
b
hyperbolic paraboloid.
3. Cone:
Quadratic surfaces of
x2
y2
z2
the form 2 + 2 = 2
a
b
c
are elliptic cones.
9
4. Cylinders:
−Elliptic Cylinder :
−P arabolic Cylinder :
−Hyperbolic Clinder :
x2
y2
+ 2 =1
2
a
b
y = ax2
x2
y2
−
=1
a2
b2
10
3.2 Limits and Continuity
3.2 Limits and Continuity
• We say that a function f (x, y) approaches the limit L as (x, y)
approaches (a, b) and write lim f (x, y) = L.
(a,b)
11
Properties of Limits of Functions of Two Variables:
Let
lim
f (x, y) = L and
lim
g(x, y) = M. Then
(x,y)→(a,b)
1.
2.
3.
4.
5.
lim
(x,y)→(a,b)
(f (x, y) + g(x, y)) = L + M
(x,y)→(a,b)
lim
(f (x, y)g(x, y)) = LM
(x,y)→(a,b)
lim
(x,y)→(a,b)
lim
(x,y)→(a,b)
lim
f (x, y)
L
=
,
M 6= 0
g(x, y)
M
kf (x, y) = kL, k=constant.
(f (x, y))k = Lk , k=constant.
(x,y)→(a,b)
Example-3.5:
Evaluate the limit of:
2
x + y3
a)
lim
cos
x+y+1
(x,y)→(0,0)
x
e sin y
c)
lim
y
(x,y)→(0,0)
cos y + 2
y − sin x
x2 − xy
√
d)
lim
√ .
x− y
(x,y)→(0,0)
b)
lim
(x,y)→(π/2,0)
12
Solution
a)
lim
cos
(x,y)→(0,0)
x2 + y 3
x+y+1
x2 + y 3
= cos
lim
(x,y)→(0,0) x + y + 1
= cos 0 = 1
b)
lim
(x,y)→(π/2,0)
lim(x,y)→(π/2,0) (cos y + 2)
cos y + 2
3
=
=
y − sin x
lim(x,y)→(π/2,0) (y − sin x)
−1
= −3
c)
sin y
ex sin y
=
lim
ex
lim
=1
y
(x,y)→(0,0)
(x,y)→(0,0)
(x,y)→(0,0) y
lim
√
√
x(x − y)( x + y)
x2 − xy
√
√
d)
lim
lim
√ =
√ √
√
x − y (x,y)→(0,0) ( x − y)( x + y)
(x,y)→(0,0)
√
√
x(x − y)( x + y)
=
lim
x−x
(x,y)→(0,0)
√
√
=
lim
x( x + y) = 0
(x,y)→(0,0)
13
Two-Path Test for Nonexistence of a Limit:
• If a function f (x, y) has different limits along two different paths
as (x, y) approaches (x0 , y0 ), then lim f (x, y) does not exist.
(x0 ,y0 )
Example-3.6:
Evaluate the limit:
Solution
Let f (x, y) =
lim
x3 y
+ y2
(x,y)→(0,0) x6
x3 y
x6 +y 2 .
Then
0
= 0, f or x 6= 0.
x6
T heref ore, f (x, y) → 0 as (x, y) → (0, 0) along the x − axis.
If y = 0, then f (x, 0) =
14
N ow, we approach (0, 0) along the curve : y = x3 , f or x 6= 0.
1
x6
= , x 6= 0
2
3
+ (x )
2
1
as (x, y) → (0, 0) along the graph of y = x3 .
T herf ore, f (x, y) →
2
Since we have obtained different limits along different paths, the given
limit does not exist.
f (x, x3 ) =
x6
Evaluating limit using polar coordinates:
• If
lim
(x,y)→(0,0)f (x,y)
exists, then we can evaluate the limit by
converting in to polar coordinates.
• x = r cos θ,
y = r sin θ
and
r=
p
2
2
• If (x, y) → (0, 0) then r = x + y → 0+ .
•
lim
(x,y)→(0,0)
f (x, y) =
lim
(x,y)→(0,0)
p
x2 + y 2 .
= lim+ f (r cos θ, r sin θ)
r→0
15
Example-3.7
Evaluate the limit:
sin(x2 + y 2 )
x2 + y 2
(x,y)→(0,0)
lim
Solution
sin(x2 + y 2 )
sin r2
2r cos r2
= lim+ cos r2 = 1.
= lim+
= lim+
2
2
2
r→0
r→0
r→0
x +y
r
2r
(x,y)→(0,0)
lim
Exercise: Find the limit, if it exists, or show that the limit does not
exist..
a)
x4 − y 4
(x,y)→(0,0) x2 + y 2
b)
c)
x2 sin2 y
(x,y)→(0,0) x2 + 2y 2
d)
lim
lim
x2 + sin2 y
(x,y)→(0,0) 2x2 + y 2
lim
lim
(x,y)→(0,0)
x2 + y 2
p
x2 + y 2 + 1 − 1
16
Continuity:
Definition-3.3:
A function f of two variables is called continuous at (x0 , y0 ) if
lim
(x,y)→(x0 ,y0 )
f (x, y) = f (x0 , y0 ).
We say f is continuous on D
(x0 , y0 ) in D.
(1)
if f is continuous at every point
• all polynomial functions of two variables are continuous on R2 .
• any rational function is continuous on its domain because it is a
quotient of continuous functions.
17
Example-3.8:
Determinethe set of points at which the function
xy

, (x, y) 6= (0, 0)
x2 + xy + y 2
f (x, y) =
is continuous.
 0,
(x, y) = (0, 0)
Solution
• Domain of f (x, y) = R2 .
• f (0, 0) = 0.
xy
does not exist.
+ xy + y 2
(x,y)→(0,0)
• Therefore f is not continuous at (0, 0) and
•
lim
f (x, y) =
lim
(x,y)→(0,0) x2
f is continuous on R2 except at the origin (0, 0).
18
3.3 Partial Derivatives
3.3 Partial Derivatives
Definition-3.4:
If f is a function of two variables, its partial derivatives fx and fy are
the functions and defined by
f (x + h, y) − f (x, y)
h→0
h
f (x, y + h) − f (x, y)
fy (x, y) = lim
h→0
h
fx (x, y) = lim
Notations For Partial Derivatives:
If z = f (x, y), we write
∂f
∂
=
f (x, y) =
∂x
∂x
∂
∂f
fy (x, y) = fy =
=
f (x, y) =
∂y
∂y
fx (x, y) = fx =
∂z
= Dx f
∂x
∂z
= Dy f
∂y
19
Rule for Finding Partial Derivatives of z = f (x, y):
1. To find fx , regard y as a constant and differentiate f (x, y) with
respect to x.
2. To find fy , regard x as a constant and differentiate f (x, y) with
respect to y.
Example-3.9:
Find fx (1, 2) and fy (1, 2) if f (x, y) = x2 + x3 y 2 − 3y + 5.
Solution
Holding y constant and differentiating with respect to x, we get
fx = 2x + 3x2 y 2
and so
fx (1, 2) = 2(1) + 3(1)2 (2)2 = 14.
Holding x constant and differentiating with respect to y, we get
fy (x, y) = 2x3 y − 3
fy (1, 2) = 2(1)3 (2) − 3 = 4 − 3 = 1.
20
• Partial derivatives can be interpreted as rates of change. If
z = (x, y), then
∂z
represents the rate of change of z with respect to x when y is
∂x
fixed.
∂z
•
represents the rate of change of z with respect to y when x is
∂y
fixed.
•
Rules of Partial Derivatives:
1. Sum Rule: (f + g)x = fx + gx
and
(f + g)y = fy + gy
2. Product Rule: (f g)x = fx g + f gx and (f g)y = fy g + f gy
fx g − f gx
f
fy g − f gy
f
=
and
=
3. Quotient Rule:
g x
g2
g y
g2
4. Chain Rule:
(h ◦ f )x (x, y) = h0 (f (x, y))fx (x, y) and
(h ◦ f )y (x, y) = h0 (f (x, y))fy (x, y)
21
Example-3.10:
Find fx and fy if f (x, y) = sin
xy
x−y
.
Solution
xy
xy
y(x − y) − (xy)(1)
= cos
x−y x
x−y
(x − y)2
y2
xy
=−
.
cos
(x − y)2
x−y
xy
xy
x(x − y) − (xy)(−1)
xy
= cos
fy (x, y) = cos
x−y
x−y y
x−y
(x − y)2
x2
xy
=
cos
.
2
(x − y)
x−y
fx (x, y) = cos
xy
x−y
22
Higher Derivatives:
• If f is a function of two variables, then its partial derivatives fx
and fy are also functions of two variables, so we can consider their
partial derivatives (fx )x , (fx )y , (fy )x and (fy )y which are called
the second partial derivatives of f.
• If z = f (x, y), we use the following notation:
∂ ∂f
∂2f
∂2z
(fx )x = fxx =
=
=
= zxx ,
∂x ∂x
∂x2
∂x2
∂2f
∂2z
∂ ∂f
=
=
= zxy ,
(fx )y = fxy =
∂y ∂x
∂y∂x
∂y∂x
∂2z
∂ ∂f
∂2f
=
= zyx ,
(fy )x = fyx =
=
∂x ∂y
∂x∂y
∂x∂y
∂ ∂f
∂2f
∂2z
(fy )y = fyy =
=
=
= zyy .
2
∂y ∂y
∂y
∂y 2
23
Theorem
Suppose f is defined on a disk D that contains the point (x0 , y0 ). If the
functions fxy and fyx are both continuous on D, then
fxy (x0 , y0 ) = fyx (x0 , y0 )
Example-3.11
Find fxyy and fxyz if f (x, y, z) = x ln xy 2 z 3 .
Solution
fx = ln xy 2 z 3 + 1,
fxyy = −
2
,
y2
fxy =
2
y
fxyz = 0
24
3.4 The Chain Rule
3.4 The Chain Rule
1) If y = f (x) and x = g(t), where f and g are differentiable functions,
then y is indirectly a differentiable function of t and
dy
dy dx
=
.
dt
dx dt
2) The Chain Rule Case - 1: Suppose that z = f (x, y) is a
differentiable function of x and y, where x = g(t) and y = h(t) are
both differentiable functions of t. Then z is a differentiable function
of t and
dz
∂z dx
dz dy
=
+
dt
∂x dt
∂y dt
(2)
25
Example-3.12:
The radius of a right circular cone is increasing at a rate of 3 m/s while
its height is decreasing at a rate of 1.5 m/s. At what rate is the volume
of the cone changing when the radius is 1m and the height is 2m.
Solution
V =
1 2
πr h.
3
dr
dh
= 3m/s,
= −1.5m/s,
r = 1m,
h = 2m.
dt
dt
dV
∂V dr dV dh
=
+
.
dt
∂r dt
∂h dt
2π dr π 2 dh
π
=
rh + r
=
2(1)(2)(3m/s) + (1)2 (−1.5m/s)
3
dt
3 dt
3
= 3.5 π m/s.
26
3) The Chain Rule Case - 2: Suppose that z = f (x, y) is a
differentiable function of x and y, where x = g(u, v) and y = h(u, v)
are both differentiable functions of u and u. Then
∂z ∂x ∂z ∂y
∂z
=
+
∂u
∂x ∂u ∂y ∂u
∂z
∂z ∂x ∂z ∂y
=
+
∂v
∂x ∂v
∂y ∂v
(3)
(4)
27
Example-3.13:
Suppose f is a differentiable function of x and y and
g(u, v) = f (eu + sin v, eu + cos v). If fx (1, 2) = 2 and fy (1, 2) = 5, then
calculate gu (0, 0) and gv (0, 0).
Solution
x = eu + sin v,
xu = eu ,
xu (0, 0) = 1,
y = eu + cos v.
xv = cos v,
xv (0, 0) = 1,
yu = eu ,
yu (0, 0) = 1,
yv = − sin v.
yv (0, 0) = 0.
gu (0, 0) = fx (1, 2)xu (0, 0) + fy (1, 2)yu (0, 0)
= (2)(1) + (8)(1) = 10.
gv (0, 0) = fx (1, 2)xv (0, 0) + fy (1, 2)yv (0, 0)
= (2)(1) + (8)(0) = 2.
28
4) Implicit Differentiation:
• The Chain Rule can be used to give a more complete description of
the process of implicit differentiation.
• We suppose that an equation of the form F (x, y) = 0 defines y
implicitly as a differentiable function of x, that is,y = f (x), where
F (x, f (x)) = 0 for all x in the domain of f.
• If F is differentiable, we can apply Case 1 of the Chain Rule to
differentiate both sides of the equation F (x, f (x)) = 0 with respect
to x. Since both x and y are functions of x, we obtain
∂F dy
∂F dx
+
= 0.
∂x dx
∂y dx
• But
dx
∂F
dy
= 1, so if
6= 0 we solve for
and obtain
dx
∂y
dx
Fx
dy
=− .
dx
Fy
(5)
29
• Now we suppose that z is given implicitly as a function z = f (x, y)
by an equation of the form F (x, y, z) = 0. This means that
F (x, y, f (x, y)) = 0 for all (x, y) in the domain of . If F and f are
differentiable, then we can use the Chain Rule to differentiate the
equation F (x, y, z) = 0 as follows:
∂F ∂x ∂F ∂y
∂F ∂z
+
+
=0
∂x ∂x
∂y ∂x
∂z ∂x
• But,
• If
∂x
=1
∂x
and
∂y
= 0, so this equation becomes
∂x
∂F
∂F ∂z
+
=0
∂x
∂z ∂x
∂F
∂z
6= 0, we solve for
and obtain
∂z
∂x
∂z
Fx
=−
∂x
Fz
(6)
30
• If
∂F
6= 0, then
∂z
Fy
∂z
=−
∂y
Fz
Example-3.14:
∂z
∂z
Find
and
∂x (1,0,2)
∂y
(7)
2
if x2 y 3 + z 2 y 5 = yex + z 3 − 2
(0,−1,2)
Solution
Let
2
F (x, y, z) =x2 y 3 + z 2 y 5 − yex − z 3 + 2. T hen
2
2
Fx = 2xy 3 − 2xyex , Fy = 3x2 y 2 + 5z 2 y 4 − ex , Fz = 2zy 5 − 3z 2
Fx (0, −1, 2) = 0,
∂z
∂x
=−
(1,0,2)
Fy (0, −1, 2) = 19,
∂z
Fx (0, −1, 2)
= 0, and
Fz (0, −1, 2)
∂y
Fz (0, −1, 2) = −16.
=−
(0,−1,2)
19
Fy (0, −1, 2)
=
Fz (0, −1, 2)
16
31
3.5 Directional Derivatives and
the Gradient Vector
3.5 Directional Derivatives and the Gradient Vector
The Gradient Vector:
1. If f is a function of two variables x and y, then the gradient of f is
the vector function ∇f defined by
∇f (x, y) = hfx (x, y), fy (x, y)i =
∂f
∂f
i+
j
∂x
∂y
(8)
2. For a function of three variables, the gradient vector, denoted by ∇f
or grad f , is
∇f (x, y, z) = hfx (x, y, z), fy (x, y, z), fz (x, y, z)i
∂f
∂f
∂f
=
i+
j+
k
∂x
∂y
∂z
(9)
(10)
32
Directional Derivatives:
The directional derivative of f at (x0 , y0 ) in the direction of a unit vector
~u = hu1 , u2 i is
f (x0 + hu1 , y0 + hu2 ) − f (x0 , y0 )
h→0
h
D~u f (x0 , y0 ) = lim
if this limit exists.
• We see that if ~u = ~i = h1, 0i , then
D~i f = lim
h→0
f (x + h · 1, y + h · 0) − f (x, y)
f (x + h, y) − f (x, y)
= lim
h→0
h
h
= fx .
• If ~u = ~j = h0, 1i , then Dj f = fy .
33
Theorem
1. If f is a differentiable function of x and y, then f has a directional
derivative in the direction of any unit vector ~u = hu1 , u2 i and
Du f (x, y) = fx (x, y)u1 + fy (x, y)u2
= hfx , fy i · hu1 , u2 i = ∇f · u.
= ||∇f || cos θ,
where θ is the angle between the vectors ∇f and u.
2. If f (x, y, z) is a differentiable function then the directional derivative
of f in the direction of any unit vector ~u = hu1 , u2 , u3 i is:
Du f (x, y, z) = fx (x, y, z)u1 + fy (x, y, z)u2 + fz (x, y, z)u3 .
= hfx , fy , fz i · hu1 , u2 , u3 i = ∇f · u.
= ||∇f || cos θ,
where θ is the angle between the vectors ∇f and u.
34
Example-3.15:
find the gradient of f (x, y, z) = (x + 2y + 3z)3/2 at the point (1, 1, 2)
and find the rate of change of f at (1, 1, 2) in the direction of the vector
v = 2j − k
Solution
3
9
fx = (x + 2y + 3z)1/2 , fy = 3(x + 2y + 3z)1/2 , fz = (x + 2y + 3z)1/2 .
2
2
27
fx (1, 1, 2) = fy (1, 1, 2) = 9, fz (1, 1, 2) =
.
2
9
27
∇f (1, 1, 2) =
, 9,
.
2
2
2
v
1
= 0, √ , − √
.
u=
||v||
5
5
9
Dv f (1, 1, 2) = ∇f (1, 1, 2) · u = − .
2
35
Maximizing the Directional Derivatives:
Suppose we have a function f of two or three variables and we consider
all possible directional derivatives of f at a given point. These give the
rates of change of f in all possible directions. We can then ask the
questions: In which of these directions does change fastest and what is
the maximum rate of change?
Theorem
• Suppose f is a differentiable function of two or three variables. The
maximum value of the directional derivative Du f (P ) is k∇f (P )k
and it occurs when u has the same direction as the gradient vector
∇f (P ).
• A differentiable function f decreases most rapidly at a point P in
the direction opposite to the gradient vector, that is, in the direction
of −∇f (P ).
36
Example-3.17:
1. Find the maximum rate of change of f (x, y, z) = tan(x + 2y + 3z)
at the point (−5, 1, 1) and the direction in which it occurs.
2. Find the directional derivative of the function
f (x, y, z) = x3 y 2 z 5 − 2xz + yz + 3x
at P (−1, −2, 1) in the direction of the negative z-axis.
Solution
• fx = sec2 (x + 2y + 3z),
fx (−5, 1, 1) = 1
1. • fy = 2 sec2 (x + 2y + 3z), fy (−5, 1, 1) = 2
• fz = 3 sec2 (x + 2y + 3z), fz (−5, 1, 1) = 3
√
• ∇f (−5, 1, 1) = h1, 2, 3i and k∇f (−5, 1, 1)k = 14.
√
• ∴, the maximum rate of change is k∇f (−5, 1, 1)k = 14.
• It occurs in the direction of h1, 2, 3i .
2. Let v = h0, 0, −1i be the vector from P (−1, −2, 1) in the direction
of the negative z-axis.
Dv (−1, −2, 1) = 20
37
Tangent Planes to Surfaces
• If F (x, y, z) is differentiable a function at P0 (x0 , y0 , z0 ) and if
∇F (P0 ) 6= 0, then
n = ∇F (P0 ) = Fx (P0 )i + Fy (P0 )j + Fz (P0 )k
(11)
is normal to the level surface F (x, y, z) = 0 at P0 (x0 , y0 , z0 ).
• The equation of the tangent plane to the level surface S of a
function of three variables F (x, y, z) = 0 at the point P0 (x0 , y0 , z0 )
is
Fx (P0 )(x − x0 ) + Fy (P0 )(y − y0 ) + Fz (P0 )(z − z0 ) = 0
(12)
• The normal line to S at P0 is the line passing through P0 and
perpendicular to the tangent plane. The direction of the normal line
is therefore given by the gradient vector ∇F (P0 ) and so its
symmetric equations are
x − x0
y − y0
z − z0
=
=
(13)
Fx (P0 )
Fy (P0 )
Fz (P0 )
38
• Let P0 (x0 , y0 , z0 ) be any point on the surface z = f (x, y). If
z = f (x, y) is differentiable at (x0 , y0 ), then the surface has a
tangent plane at P0 , and this plane has the equation
fx (x0 , y0 )(x − x0 ) + fy (x0 , y0 )(y − y0 ) − (z − f (x0 , y0 )) = 0.
(14)
• n = fx (x0 , y0 )i + fy (x0 , y0 )j − k is a normal vector to the surface
z = f (x, y) at the point P0 (x0 , y0 , z0 ).
• The symmetric equations of the normal line to the surface
z = f (x, y), are
x − x0
y − y0
z − f (x0 , y0 )
=
=
fx (x0 , y0 )
fy (x0 , y0 )
−1
(15)
39
Example-3.18:
Find equations of the tangent plane and the normal line to the surface
yz = ln(x + z) at the point (0, 0, 1.
Solution
Let F (x, y, z) = yz − ln(x + z) = 0. Then
1
,
x+z
Fx (0, 0, 1) = 1,
Fx =
Fy = z,
1
x+z
Fz (0, 0, 1) = 1
Fz = y +
Fy (0, 0, 1) = 1,
The equation of the tangent plane to the surface is
x+y+z =1
and symmetric equations of the normal line to the surface are
y
z−1
x
= =
1
1
1
40
3.6 Maximum and Minimum
Values
3.6 Maximum and Minimum Values.
Definition-3.6:
• A function of two variables has a local maximum at (x0 , y0 ) if
f (x, y) ≤ f (x0 , y0 ) when (x, y) is near (x0 , y0 ). [This means that
f (x, y) ≤ f (x0 , y0 ) for all points (x, y) in some disk with center
(x0 , y0 ).] The number f (x0 , y0 ) is called a local maximum value.
• If f (x, y) ≥ f (x0 , y0 ) when (x, y) is near (x0 , y0 ), then f has a local
minimum at (x0 , y0 ) and f (x0 , y0 ) is a local minimum value.
• If the inequalities hold for all
points in the domain of f, then
f has an absolute maximum (or
absolute minimum) at (x0 , y0 ).
41
• The point (x0 , y0 ) is called a critical point (or stationary point) of
z = f (x, y) if one of the following conditions holds
1. fx (x0 , y0 ) = fy (x0 , y0 ) = 0, or
2. either fx (x0 , y0 ) or fy (x0 , y0 ) does not exist.
• If f has a local maximum or minimum at (x0 , y0 ), then (x0 , y0 ) is a
critical point of f. At a critical point, a function could have a local
maximum or a local minimum or neither.
• The point (x0 , y0 , f (x0 , y0 )) is a saddle point of z = f (x, y) if
fx (x0 , y0 ) = 0 and fy (x0 , y0 ) = 0, but f does not have any local
extremum values (maxima or minima) at (x0 , y0 ).
42
Second Derivatives Test:
Suppose that f (x, y) has continuous second-order partial derivatives in
some open disk containing the point (a, b) and that
fx (a, b) = fy (a, b) = 0. Define the discriminant D for the point (a, b) by
2
D(a, b) = fxx (a, b)fyy (a, b) − [fxy (a, b)]
(16)
(i) If D(a, b) > 0 and fxx (a, b) > 0, then f has a local minimum at (a, b).
(ii) If D(a, b) > 0 and fxx (a, b) < 0, then f has a local maximum at (a, b).
(iii) If D(a, b) < 0, then f has a a saddle point at(a, b).
(iv) If D(a, b) = 0, then no conclusion can be drawn.
Example-3.19:
Find all critical points of f (x, y) = x4 − 6x2 + 4y 3 + 2y 2 − 1 and classify
them.
43
Solution:
√
• fx = 4x3 − 12x = 4x(x2 − 3) = 0 ⇒ x = 0, ± 3
and
• fy = 12y + 12y = 12y(y + 1) = 0 ⇒ y = 0, −1.
√
√
• Therefore, (0, 0), (0, −1), (± 3, 0), (± 3, −1) are the critical
points of f .
2
• fxx = 12x2 − 12,
fxy = 0,
fyy = 24y + 12.
2
• The discriminant, D(x, y) = fxx fyy − [fxy ] = 144(x2 − 1)(2y + 1).
Critical point:
(a,b)
fxx (a, b)
D(a, b)
Classification
(0,0)
(0,-1)
√
(± 3, 0)
√
(± 3, −1)
-12
-144
saddle
Point
-12
144
Relative
Maximum
24
288
Relative
Minimum
24
-288
Saddle
Point
44
Definition-3.7:
• We call f (a, b) the absolute maximum of f on the region R if
f (a, b) ≥ f (x, y) for all (x, y) ∈ R.
• Similarly, f (a, b) is called the absolute minimum of f on R if
f (a, b) ≤ f (x, y) for all (x, y) ∈ R.
• In either case, f (a, b) is called an absolute extremum of f.
Extreme Value Theorem for Functions of Two Variables:
If f is continuous on a closed, bounded set R in R2 , then f attains an
absolute maximum value f (x1 , y1 ) and an absolute minimum value
f (x2 , y2 ) at some points (x1 , y1 ) and x1 , y1 ) in R.
To find the absolute maximum and minimum values of a continuous
function on a closed, bounded set R :
1. Find the values of f at the critical points of f in R.
2. Find the extreme values of f on the boundary of R.
3. The largest of the values from steps 1 and 2 is the absolute
maximum value; the smallest of these values is the absolute
minimum value.
45
Example-3.20:
Find the absolute maximum and minimum values of
f (x, y) = 3 + xy − x − 2y on the closed triangular region with vertices
(1, 0), (5, 0), and (1, 4).
Solution:
• Values of f at the critical points of f in R.:
fx = y − 1,
fx = 0 ⇒ y = 0,
fy = x − 2
fy = 0 x = 2.
Therefore, the critical point of f in R is (2, 1) and f (2, 1) = 1.
• Extreme values of f on the boundary of R :
Boundary of R:
S1 : y = 0,
1 ≤ x ≤ 5;
S2 : x = 1,
0 ≤ y ≤ 4;
S3 : y = −x + 5,
1 ≤ x ≤ 5.
46
• The value of f on S1
g1 (x) = f (x, 0) = 3 − x,
⇒
g10 (x)
= −1 6= 0
1 ≤ x ≤ 5.
has no critical point.
∴ , g1 (1) = f (1, 0) = 2, and g1 (5) = f (5, 0) = −2.
• The value of f on S2
g2 (y) = f (1, y) = 2 − y,
0 ≤ x ≤ 4.
⇒ g20 (y) = −1 6= 0 has no critical point.
∴ , g2 (0) = f (1, 0) = 2, and g2 (4) = f (1, 4) = −2.
• The value of f on S3
g3 (x) = f (x, 5 − x) = −x2 + 6x − 7,
⇒
g30 (x)
1 ≤ x ≤ 5.
= −2x + 6 ⇒ x = 3 is a critical point.
∴ , g3 (1) = f (1, 4) = −2, g3 (3) = f (3, 2) = 2 and g3 (5) = f (5, 0) = −2.
47
The absolute maximum value of f on R is f (1, 0) = f (3, 2) = 2 and
the absolute minimum value is f (5, 0) = f (1, 4) = f (5, 0) = −2.
Example-3.21:
Find the absolute maximum and minimum values of f (x, y) = xy 2 on the
region D = {(x, y)|x ≥ x, y ≥ 0, x2 + y 2 ≤ 3}.
Solution
• Values of f at the critical points of f in R.:
fx = y 2 ,
fy = 2xy
⇒ y = 0,
x=0
or
y = 0.
√
⇒ (x, 0), f or 0 ≤ x ≤ 3, are critical points, and
f (x, 0) = 0.
• Extreme values of f on the boundary of R :
Boundary of R:
S1 : y = 0,
0≤x≤
S2 : x = 0,
p
S3 : y = 3 − x2 ,
0≤y≤
0≤x≤
√
√
√
3;
3;
3.
48
• The value of f on S1
h1 (x) = f (x, 0) = 0,
0≤x≤
√
3.
• The value of f on S2
h2 (y) = f (0, y) = 0,
0≤x≤
√
3.
• The value of f on S3
h3 (x) = f (x,
⇒
p
3 − x2 ) = 3x − x2 ,
0≤x≤
√
3.
h03 (x)
= 3 − 2x ⇒ x = 3/2 is a critical point.
√
√
∴ , h3 (0) = f (0, 3) = 0, h3 (3/2) = f (3/2, 3/2) = 9/4,
√
√
h3 ( 3) = f ( 3, 0) = 0.
• Therefore, the absolute maximum value of f on R is
√
f (3/2, 3/2) = 9/4 and the absolute minimum value is
√
√
f (x, 0) = f (0, y) = 0 for 0 ≤ x ≤ 3 and 0 ≤ y ≤ 3.
49
Exercise:
1. Find the volume of the largest rectangular box in the first octant
with three faces in the coordinate planes and one vertex in the plane
x + y + z = 6.
2. Find the dimensions of a rectangular box of maximum volume such
that the sum of the lengths of its 12 edges is a constant c.
3. Find the dimensions of the rectangular box with largest volume if
the total surface area is given as 64 cm2 .
4. Find the points on the cone z 2 = x2 + y 2 that are closest to the
point (4, 2, 0).
50
3.7 Lagrange Multipliers
3.7 Lagrange Multipliers
Method of Lagrange Multipliers
To find the maximum and minimum values of f (x, y, z) subject to the
constraint g(x, y, z) = k[assuming that these extreme values exist and
∇g(x, y, z) 6= 0 on the surface g(x, y, z) = k ]:
(a) Find all values of x, y, z and λ such that
∇f (x, y, z) = λ∇g(x, y, z)
and
g(x, y, z) = k
(17)
(18)
• The number λ in Equation 21 is called a Lagrange multiplier.
(b) Evaluate f at all the points (x, y, z) that result from step (a). The
largest of these values is the maximum value of ; the smallest is
the minimum value of f.
To find the extreme values of f (x, y) subject to the constraint
g(x, y) = k, we look for values of x, y, and λ such that
∇f (x, y) = λ∇g(x, y)
and
g(x, y) = k
(19)
51
Example-3.22:
Use Lagrange multipliers to find the extreme values of the function
f (x, y, z) = xyz subject to the constraint x2 + 2y 2 + 3z 2 = 6.
Solution
According to the method of Lagrange multipliers, we solve
∇f = λ∇g, g = 6, where g(x, y, z) = x2 + 2y 2 + 3z 2 . This gives
yz = 2λx
(20)
xz = 4λy
(21)
xy = 6λz
(22)
x2 + 2y 2 + 3z 2 = 6
(23)
The simplest way to solve these equations is multiply equations (24),
(25) and (26) by x, y and z respectively.
52
xyz = 2λx2
(24)
xyz = 4λy
2
(25)
xyz = 6λz
2
(26)
From equations (28), (29) and (30), we have
(27)
x2 = 2y 2 = 3z 2
Using equations (27) and (31), we get
2
√
3x = 6 ⇒ x = ± 2,
√
y = ±1,
z=±
6
3
Therefore f has possible extreme values at the points
√ √
6
± 2, ±1, ±
3
53
2
Evaluating f at these eight points, the maximum value of f is √
√ √ 3
√
√
6
6
occurs at the four points − 2, 1, −
, − 2, −1,
,
3
3
√ √ √
√
6
6
, and − 2, −1,
and the minimum value of
2, −1, −
3
3
√ √
2
6
f is − √ occurs at the four points − 2, −1, −
,
3
3 √ √ √ √
√
√
6
6
6
2, −1,
,
2, −1,
and
2, 1, −
.
3
3
3
Example-3.23:
Find the extreme values of f (x, y) = x2 + 2y 2 on the disk x2 + 2y 2 ≤ 1.
Solution
We compare the values of f at the critical points with values at the
points on the boundary. Since fx = 2x and fy = 4y the only critical
point is (0, 0) .
54
Using Lagrange multipliers, we solve the equations
∇f = λ∇g
g(x, y) = x2 + y 2 = 1
or
2
2x = 2λx
(28)
4y = 2λy
(29)
2
x +y =1
(30)
From (32) we have x = 0 or λ = 1. If x = 0, then (34) gives y = ±1. If
λ = 1, then y = 0 from (33), so then (11) gives x = ±1. Therefore f has
possible extreme values at the points (0, 1), (0, −1), (1, 0) and (−1, 0).
Evaluating f at the critical point and at these four points, we find that
f (0, 0) = 0 f (0, 1) = 2, f (0, −1) = 2, f (1, 0) = 1 f (−1, 0) = 1.
Therefore the maximum value of f on the circle x2 + y 2 = 1 is
f (0, ±1) = 2 and the minimum value is f (0, 0) = 0
55