Math 1131Q Section 10
Section 4.1
Oct 21, 2010
A tank of water in the shape of a cone is leaking water at a
constant rate of 2 ft3 /hour. The base radius of the tank is 5 ft and
the height of the tank is 14 ft. At what rate is the depth of the
water in the tank changing when the depth of the water is 6 ft?
Section 4.1 (Oct 21, 2010)
Math 1131Q Section 10
2 / 25
A tank of water in the shape of a cone is leaking water at a
constant rate of 2 ft3 /hour. The base radius of the tank is 5 ft and
the height of the tank is 14 ft. At what rate is the depth of the
water in the tank changing when the depth of the water is 6 ft?
The water in the tank forms a smaller cone with the same central
angle as the tank itself. The radius of the water cone at any time
is given by r and the height of the water cone at any time is given
by h.
Section 4.1 (Oct 21, 2010)
Math 1131Q Section 10
2 / 25
The volume of water in the tank at any
1
time t is given by V = πr2 h and we
3
have been given that V 0 = −2.
Section 4.1 (Oct 21, 2010)
Math 1131Q Section 10
3 / 25
The volume of water in the tank at any
1
time t is given by V = πr2 h and we
3
have been given that V 0 = −2.
r
5
=
h
14
Section 4.1 (Oct 21, 2010)
=⇒
r=
Math 1131Q Section 10
5
h
14
3 / 25
The volume of water in the tank at any
1
time t is given by V = πr2 h and we
3
have been given that V 0 = −2.
r
5
=
h
14
=⇒
r=
5
h
14
1
1 5
25 3
V = πr2 h = π( )2 h =
πh
3
3 14
588
Section 4.1 (Oct 21, 2010)
Math 1131Q Section 10
3 / 25
The volume of water in the tank at any
1
time t is given by V = πr2 h and we
3
have been given that V 0 = −2.
r
5
=
h
14
=⇒
r=
5
h
14
1
1 5
25 3
V = πr2 h = π( )2 h =
πh
3
3 14
588
V0 =
Section 4.1 (Oct 21, 2010)
25 2 0
πh h
196
Math 1131Q Section 10
3 / 25
The volume of water in the tank at any
1
time t is given by V = πr2 h and we
3
have been given that V 0 = −2.
r
5
=
h
14
=⇒
r=
5
h
14
1
1 5
25 3
V = πr2 h = π( )2 h =
πh
3
3 14
588
−2 =
25
π(6)2 h0
196
Section 4.1 (Oct 21, 2010)
V0 =
25 2 0
πh h
196
=⇒
h0 =
−98
= −0.1386 ft/hour
225π
Math 1131Q Section 10
3 / 25
Clicker Question
A manufacturer needs to make a cylindrical can that will hold 1.5
liters of liquid. Which can will minimize the amount of material
used in its construction?
(a)
(b)
(c)
Section 4.1 (Oct 21, 2010)
(d)
Math 1131Q Section 10
4 / 25
Use 1 Liter = 1000 cm3 to convert the 1.5 liters into 1500 cm3 .
2π r
r
h
h
V = πhr2 = 1500
h=
1500
πr2
Area = 2πr2 + 2πhr
Section 4.1 (Oct 21, 2010)
Math 1131Q Section 10
5 / 25
Use 1 Liter = 1000 cm3 to convert the 1.5 liters into 1500 cm3 .
2π r
r
h
Goal: Choose r to make area as
small as possible.
h
V = πhr2 = 1500
h=
1500
πr2
Area = 2πr2 + 2πhr
Section 4.1 (Oct 21, 2010)
Math 1131Q Section 10
5 / 25
Use 1 Liter = 1000 cm3 to convert the 1.5 liters into 1500 cm3 .
2π r
r
h
Goal: Choose r to make area as
small as possible.
h
A = 2πr2 + 2πr
V = πhr2 = 1500
1500
πr2
3000
r
How do we find exact value of r
that gives us the most efficient
dimensions?
A = 2πr2 +
h=
1500
πr2
Area = 2πr2 + 2πhr
Section 4.1 (Oct 21, 2010)
Math 1131Q Section 10
5 / 25
A = area
A = 2πr2 +
3000
r
r
Section 4.1 (Oct 21, 2010)
Math 1131Q Section 10
6 / 25
A = area
A = 2πr2 +
3000
r
r
The exact value of r that gives us the most efficient dimensions
(minimum area) occurs when the graph of A has a horizontal
tangent. That is, when A0 (r) = 0.
Section 4.1 (Oct 21, 2010)
Math 1131Q Section 10
6 / 25
A = area
A = 2πr2 +
3000
r
r
(a)
(b)
(c)
Section 4.1 (Oct 21, 2010)
(d)
Math 1131Q Section 10
7 / 25
A = area
3000
r
3000
A0 = 4πr − 2 = 0
r
4πr3 − 3000
A0 =
=0
r2
A = 2πr2 +
r
(a)
(b)
(c)
Section 4.1 (Oct 21, 2010)
(d)
Math 1131Q Section 10
7 / 25
A = area
3000
r
3000
A0 = 4πr − 2 = 0
r
4πr3 − 3000
A0 =
=0
r2
3000 1/3
r=(
) ∼ 6.2035
4π
1500
∼ 12.4070
h=
πr2
A = 2πr2 +
r
(a)
(b)
(c)
(d)
Answer to clicker question = (c): h is about double r.
Section 4.1 (Oct 21, 2010)
Math 1131Q Section 10
7 / 25
Theorem (Extreme Value Theorem)
A continuous function always reaches a minimum value and a
maximum value on a closed interval.
Section 4.1 (Oct 21, 2010)
Math 1131Q Section 10
8 / 25
Theorem (Extreme Value Theorem)
A continuous function always reaches a minimum value and a
maximum value on a closed interval.
Extreme values may occur at
end points, points with
horizontal tangents, or where no
tangent exists
Section 4.1 (Oct 21, 2010)
Math 1131Q Section 10
8 / 25
Theorem (Extreme Value Theorem)
A continuous function always reaches a minimum value and a
maximum value on a closed interval.
Extreme values may occur at
end points, points with
horizontal tangents, or where no
tangent exists
This is not always true for
functions that fail to be
continuous (the graph has
breaks)
Section 4.1 (Oct 21, 2010)
Math 1131Q Section 10
8 / 25
Definition
If f is defined on an interval I containing d then we say that f
has an absolute maximum value on I at d if f (d) ≥ f (x) for
every x in I.
Section 4.1 (Oct 21, 2010)
Math 1131Q Section 10
9 / 25
Definition
If f is defined on an interval I containing d then we say that f
has an absolute maximum value on I at d if f (d) ≥ f (x) for
every x in I.
Definition
If f is defined on an interval I containing a then we say that f
has an absolute minimum value on I at a if f (a) ≤ f (x) for
every x in I.
Section 4.1 (Oct 21, 2010)
Math 1131Q Section 10
9 / 25
Identify the absolute minimum and absolute maximum for the
function. f (x) = x2 on [−1, 2].
Section 4.1 (Oct 21, 2010)
Math 1131Q Section 10
10 / 25
Identify the absolute minimum and absolute maximum for the
function. f (x) = x2 on [−1, 2].
Absolute minimum at x = 0
Section 4.1 (Oct 21, 2010)
Absolute maximum at x = 2
Math 1131Q Section 10
10 / 25
Local Maxima
f (x) has a local maximum at x = b if f (b) is the maximum
value of f on some small open interval containing b.
Section 4.1 (Oct 21, 2010)
Math 1131Q Section 10
11 / 25
Local Maxima
f (x) has a local maximum at x = b if f (b) is the maximum
value of f on some small open interval containing b.
Local Minima
f (x) has a local minimum at x = c if f (c) is the minimum value
of f on some small open interval containing c.
Section 4.1 (Oct 21, 2010)
Math 1131Q Section 10
11 / 25
How do we find the points where
f (x) = x4 − 4x2 + 5, −2 ≤ x ≤ 2 has its local minimum and
local maximum values?
Section 4.1 (Oct 21, 2010)
Math 1131Q Section 10
12 / 25
How do we find the points where
f (x) = x4 − 4x2 + 5, −2 ≤ x ≤ 2 has its local minimum and
local maximum values?
Find points x in (−2, 2) where tangent is horizontal.
NOTE: Local minima and local maxima can only occur at interior
points and not at end points of the interval.
Section 4.1 (Oct 21, 2010)
Math 1131Q Section 10
12 / 25
How do we find the points where
f (x) = x4 − 4x2 + 5, −2 ≤ x ≤ 2 has its local minimum and
local maximum values?
Find points x in (−2, 2) where tangent is horizontal.
NOTE: Local minima and local maxima can only occur at interior
points and not at end points of the interval.
f 0 (x) = 4x3 − 8x
f 0 (x) = 4x(x2 −√2
√
f 0 (x) = 4x(x − 2)(x + 2)
f 0 (x) = 0
x
when
x = 0,
√
− 2,
Section 4.1 (Oct 21, 2010)
√
2
Math 1131Q Section 10
12 / 25
In general, to find points at which the local minimum and
maximum values occur, we need to consider all critical points.
These are the points where either
f 0 (x) = 0 or
f 0 (x) is not defined
Section 4.1 (Oct 21, 2010)
Math 1131Q Section 10
13 / 25
In general, to find points at which the local minimum and
maximum values occur, we need to consider all critical points.
These are the points where either
f 0 (x) = 0 or
f 0 (x) is not defined
Local max and min can occur at
a cusp or other points where the
function does not have a
derivative.
x
They may also occur where the
derivative is zero.
Section 4.1 (Oct 21, 2010)
Math 1131Q Section 10
13 / 25
Clicker Question
Find all the critical points of the function
 3
if x < 0
 x − 3x − 1
f (x) =
 2
x − 4x + 8
if x ≥ 0
(a) −1, 1, 2
(b) −1, 2
(c) −1, 0, 1, 2
(d) −1, 0, 2
(e) None of these
Section 4.1 (Oct 21, 2010)
Math 1131Q Section 10
14 / 25
f (x) =
 3
 x − 3x − 1
if x < 0
x2 − 4x + 8
if x ≥ 0

f 0 (x) = 3x2 − 3,
x<0
3(x2 − 1) = 0
3(x − 1)(x + 1) = 0
x = −1
Section 4.1 (Oct 21, 2010)
Math 1131Q Section 10
15 / 25
f (x) =
 3
 x − 3x − 1
if x < 0
x2 − 4x + 8
if x ≥ 0

f 0 (x) = 3x2 − 3,
x<0
3(x2 − 1) = 0
3(x − 1)(x + 1) = 0
x = −1
Section 4.1 (Oct 21, 2010)
f 0 (x) = 2x − 4,
x>0
2(x − 2) = 0
x=2
Math 1131Q Section 10
15 / 25
f (x) =
 3
 x − 3x − 1
if x < 0
x2 − 4x + 8
if x ≥ 0

f 0 (x) = 3x2 − 3,
x<0
f 0 (x) = 2x − 4,
3(x2 − 1) = 0
x>0
2(x − 2) = 0
3(x − 1)(x + 1) = 0
x=2
x = −1
f 0 (0) does not exist because f is not continuous at x = 0.
lim f (x) = −1
x→0−
Section 4.1 (Oct 21, 2010)
lim f (x) = 8
x→0+
Math 1131Q Section 10
15 / 25
f (x) =
 3
 x − 3x − 1
if x < 0
x2 − 4x + 8
if x ≥ 0

f 0 (x) = 3x2 − 3,
x<0
f 0 (x) = 2x − 4,
3(x2 − 1) = 0
x>0
2(x − 2) = 0
3(x − 1)(x + 1) = 0
x=2
x = −1
f 0 (0) does not exist because f is not continuous at x = 0.
lim f (x) = −1
x→0−
lim f (x) = 8
x→0+
Critical points x = −1, 0, 2
Section 4.1 (Oct 21, 2010)
Math 1131Q Section 10
Answer = d
15 / 25
Identify the absolute extrema and relative extrema for the
function f (x) = x3 on [−2, 2].
Section 4.1 (Oct 21, 2010)
Math 1131Q Section 10
16 / 25
Identify the absolute extrema and relative extrema for the
function f (x) = x3 on [−2, 2].
No local maximum
No local minimum
(Note: Local maxima and
minima occur only at interior
points of the interval.)
Section 4.1 (Oct 21, 2010)
Math 1131Q Section 10
16 / 25
Identify the absolute extrema and relative extrema for the
function f (x) = x3 on [−2, 2].
No local maximum
Absolute maximum at x = 2
No local minimum
(Note: Local maxima and
minima occur only at interior
points of the interval.)
Section 4.1 (Oct 21, 2010)
Absolute minimum at x = −2
Math 1131Q Section 10
16 / 25
Identify the absolute extrema and relative extrema for the
1
function. f (x) = 2 on [−1, 1]
x
Section 4.1 (Oct 21, 2010)
Math 1131Q Section 10
17 / 25
Identify the absolute extrema and relative extrema for the
1
function. f (x) = 2 on [−1, 1]
x
No local maximum
No local minimum
Section 4.1 (Oct 21, 2010)
Math 1131Q Section 10
17 / 25
Identify the absolute extrema and relative extrema for the
1
function. f (x) = 2 on [−1, 1]
x
No absolute maximum
No local maximum
No local minimum
Section 4.1 (Oct 21, 2010)
Absolute minimum at x = −1
and x = 1
Math 1131Q Section 10
17 / 25
Theorem (Extreme Value Theorem)
A function f (x) that is continuous on a closed interval [a, b]
always reaches an absolute maximum value and also an absolute
minimum value.
Section 4.1 (Oct 21, 2010)
Math 1131Q Section 10
18 / 25
Theorem (Extreme Value Theorem)
A function f (x) that is continuous on a closed interval [a, b]
always reaches an absolute maximum value and also an absolute
minimum value.
Absolute max value: the greatest among f (a), f (b) and
the values of f (x) at the critical points
Absolute min value: the least among f (a), f (b) and the
values of f (x) at the critical points
Section 4.1 (Oct 21, 2010)
Math 1131Q Section 10
18 / 25
Theorem (Extreme Value Theorem)
A function f (x) that is continuous on a closed interval [a, b]
always reaches an absolute maximum value and also an absolute
minimum value.
Absolute max value: the greatest among f (a), f (b) and
the values of f (x) at the critical points
Absolute min value: the least among f (a), f (b) and the
values of f (x) at the critical points
To find absolute max & min:
Find the critical points
Make a list of the values of f (x) at the critical pts & end
points
From this list you can pick out the smallest and largest values
Section 4.1 (Oct 21, 2010)
Math 1131Q Section 10
18 / 25
Example:
Find the absolute maximum and mimumum values of the function
f (x) = 2x3 − 3x2 + 4 on [−1, 2].
Section 4.1 (Oct 21, 2010)
Math 1131Q Section 10
19 / 25
Example:
Find the absolute maximum and mimumum values of the function
f (x) = 2x3 − 3x2 + 4 on [−1, 2].
Notice that f is continuous on the given interval. [The EVT
tells us that f has both an absolute min and an absolute max
on [−1.2].]
Section 4.1 (Oct 21, 2010)
Math 1131Q Section 10
19 / 25
Example:
Find the absolute maximum and mimumum values of the function
f (x) = 2x3 − 3x2 + 4 on [−1, 2].
Notice that f is continuous on the given interval. [The EVT
tells us that f has both an absolute min and an absolute max
on [−1.2].]
Find the derivative and use it to find the critical points.
Section 4.1 (Oct 21, 2010)
Math 1131Q Section 10
19 / 25
Example:
Find the absolute maximum and mimumum values of the function
f (x) = 2x3 − 3x2 + 4 on [−1, 2].
Notice that f is continuous on the given interval. [The EVT
tells us that f has both an absolute min and an absolute max
on [−1.2].]
Find the derivative and use it to find the critical points.
Compute the values of f at the critical points within [−1.2]
and at the end points.
Section 4.1 (Oct 21, 2010)
Math 1131Q Section 10
19 / 25
f (x) = 2x3 − 3x2 + 4 on [−1, 2]
Step 1. Find the critical points
f 0 (x) = 6x2 − 6x
6x(x − 1) = 0
Section 4.1 (Oct 21, 2010)
Math 1131Q Section 10
20 / 25
f (x) = 2x3 − 3x2 + 4 on [−1, 2]
Step 1. Find the critical points
f 0 (x) = 6x2 − 6x
6x(x − 1) = 0
Critical Points
x = 0, 1
Section 4.1 (Oct 21, 2010)
Math 1131Q Section 10
20 / 25
f (x) = 2x3 − 3x2 + 4 on [−1, 2]
Step 1. Find the critical points
f 0 (x) = 6x2 − 6x
6x(x − 1) = 0
Critical Points
x = 0, 1
Section 4.1 (Oct 21, 2010)
Step 2. Find the values of f (x)
at the critical points and end
points
f (0) = 4 and f (1) = 3
f (−1) = −1 and f (2) = 8
Math 1131Q Section 10
20 / 25
f (x) = 2x3 − 3x2 + 4 on [−1, 2]
Step 1. Find the critical points
f 0 (x) = 6x2 − 6x
6x(x − 1) = 0
Critical Points
x = 0, 1
Step 2. Find the values of f (x)
at the critical points and end
points
f (0) = 4 and f (1) = 3
f (−1) = −1 and f (2) = 8
On the interval [−1, 2] the function f attains its maximum value
of 8 at the right end point x = 2. It attains its minimum value of
-1 at the left end point x = −1.
Section 4.1 (Oct 21, 2010)
Math 1131Q Section 10
20 / 25
Determine absolute minimum and maximum for
Q(y) = 3y(y + 4)2/3 on [−5, 1]
We first find the
derivative so we
can use it to find
the critical points.
Section 4.1 (Oct 21, 2010)
Math 1131Q Section 10
21 / 25
Determine absolute minimum and maximum for
Q(y) = 3y(y + 4)2/3 on [−5, 1]
We first find the
derivative so we
can use it to find
the critical points.
2
Q0 (y) = 3(y + 4)2/3 + 3y( )(y + 4)−1/3
3
Q0 (y) = 3(y + 4)2/3 +
Q0 (y) =
Section 4.1 (Oct 21, 2010)
2y
(y + 4)1/3
3(y + 4) + 2y
5y + 12
=
1/3
(y + 4)
(y + 4)1/3
Math 1131Q Section 10
21 / 25
Clicker Question
Determine absolute minimum and maximum for
Q(y) = 3y(y + 4)2/3 on [−5, −1]
Section 4.1 (Oct 21, 2010)
Math 1131Q Section 10
22 / 25
Clicker Question
Determine absolute minimum and maximum for
Q(y) = 3y(y + 4)2/3 on [−5, −1]
Q0 (y) =
5y + 12
(y + 4)1/3
(a) Absolute max occurs at y = −1
(b) Absolute min occurs at y = −12/5
(c) Absolute max occurs at y = −5
(d) Absolute min occurs at y = −5
(e) None of these
Section 4.1 (Oct 21, 2010)
Math 1131Q Section 10
22 / 25
Q(y) = 3y(y + 4)2/3 on [−5, −1]
and
Q0 (y) =
5y + 12
(y + 4)1/3
Critical points
y = −4 because Q0 (−4) does not exist
y = −12/5 because Q0 (−12/5) = 0
Section 4.1 (Oct 21, 2010)
Math 1131Q Section 10
23 / 25
Q(y) = 3y(y + 4)2/3 on [−5, −1]
and
Q0 (y) =
5y + 12
(y + 4)1/3
Critical points
y = −4 because Q0 (−4) does not exist
y = −12/5 because Q0 (−12/5) = 0
Test at end points and critical points
Q(−4) = 0 Absolute Max
Q(−5) = −15 Absolute Min
Section 4.1 (Oct 21, 2010)
Math 1131Q Section 10
23 / 25
Q(y) = 3y(y + 4)2/3 on [−5, −1]
and
Q0 (y) =
5y + 12
(y + 4)1/3
Critical points
y = −4 because Q0 (−4) does not exist
y = −12/5 because Q0 (−12/5) = 0
Test at end points and critical points
Q(−4) = 0 Absolute Max
Q(−12/5) = −9.849
Q(−5) = −15 Absolute Min
Q(−1) = −6.241
Section 4.1 (Oct 21, 2010)
Math 1131Q Section 10
23 / 25
A 15 foot ladder is resting against the wall. The bottom is
initially 10 feet away from the wall and is being pushed towards
1
the wall at a rate of ft/sec. How fast is the top of the ladder
4
moving up the wall 12 seconds after we start pushing?
Section 4.1 (Oct 21, 2010)
Math 1131Q Section 10
24 / 25
A 15 foot ladder is resting against the wall. The bottom is
initially 10 feet away from the wall and is being pushed towards
1
the wall at a rate of ft/sec. How fast is the top of the ladder
4
moving up the wall 12 seconds after we start pushing?
After 12 seconds, the bottom of the ladder has moved a
1
distance = constant rate × time = ( )(12) = 3 ft and so
4
x(12) = 10 − 3 = 7 feet.
Section 4.1 (Oct 21, 2010)
Math 1131Q Section 10
24 / 25
A 15 foot ladder is resting against the wall. The bottom is
initially 10 feet away from the wall and is being pushed towards
1
the wall at a rate of ft/sec. How fast is the top of the ladder
4
moving up the wall 12 seconds after we start pushing?
After 12 seconds, the bottom of the ladder has moved a
1
distance = constant rate × time = ( )(12) = 3 ft and so
4
x(12) = 10 − 3 = 7 feet.
dx
1
= − ft/sec
dt
4
x2 + y 2 = (15)2
Section 4.1 (Oct 21, 2010)
Math 1131Q Section 10
24 / 25
A 15 foot ladder is resting against the wall. The bottom is
initially 10 feet away from the wall and is being pushed towards
1
the wall at a rate of ft/sec. How fast is the top of the ladder
4
moving up the wall 12 seconds after we start pushing?
After 12 seconds, the bottom of the ladder has moved a
1
distance = constant rate × time = ( )(12) = 3 ft and so
4
x(12) = 10 − 3 = 7 feet.
dx
1
= − ft/sec
dt
4
x2 + y 2 = (15)2
The height after 12 seconds,√y(12) can be√
computed from
2
2
2
x + y = (15) . Hence y = 225 − 49 = 176.
Section 4.1 (Oct 21, 2010)
Math 1131Q Section 10
24 / 25
dx
1
= − ft/sec
dt
4
x2 + y 2 = (15)2
Section 4.1 (Oct 21, 2010)
Math 1131Q Section 10
25 / 25
dx
1
= − ft/sec
dt
4
x2 + y 2 = (15)2
d 2
d
(x + y 2 ) = ((15)2 )
dt
dt
2x
dx
dy
+ 2y
=0
dt
dt
dy
x dx
=−
dt
y dt
Section 4.1 (Oct 21, 2010)
Math 1131Q Section 10
25 / 25
dx
1
= − ft/sec
dt
4
x2 + y 2 = (15)2
When t = 12 seconds,
d 2
d
(x + y 2 ) = ((15)2 )
dt
dt
2x
x = 7 and y =
dx
dy
+ 2y
=0
dt
dt
Hence
dy
x dx
=−
dt
y dt
176
1
dy
7
= −√
(− )
dt
176 4
dy
= 0.1319 ft/sec
dt
The ladder is moving up since
Section 4.1 (Oct 21, 2010)
√
dy
> 0.
dt
Math 1131Q Section 10
25 / 25