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1 | 2918-2921 | Hence the sum of squares of numbers is
minimum when the numbers are 15
2 and
15
15
15
2
2
−
= Remark Proceeding as in Example 34 one may prove that the two positive numbers,
whose sum is k and the sum of whose squares is minimum, are
2and
2
k
k Example 23 Find the shortest distance of the point (0, c) from the parabola y = x2,
where 1
2 ≤ c ≤ 5 Solution Let (h, k) be any point on the parabola y = x2 |
1 | 2919-2922 | Remark Proceeding as in Example 34 one may prove that the two positive numbers,
whose sum is k and the sum of whose squares is minimum, are
2and
2
k
k Example 23 Find the shortest distance of the point (0, c) from the parabola y = x2,
where 1
2 ≤ c ≤ 5 Solution Let (h, k) be any point on the parabola y = x2 Let D be the required distance
between (h, k) and (0, c) |
1 | 2920-2923 | Example 23 Find the shortest distance of the point (0, c) from the parabola y = x2,
where 1
2 ≤ c ≤ 5 Solution Let (h, k) be any point on the parabola y = x2 Let D be the required distance
between (h, k) and (0, c) Then
2
2
2
2
D
(
0)
(
)
(
)
h
k
c
h
k
c
=
−
+
−
=
+
− |
1 | 2921-2924 | Solution Let (h, k) be any point on the parabola y = x2 Let D be the required distance
between (h, k) and (0, c) Then
2
2
2
2
D
(
0)
(
)
(
)
h
k
c
h
k
c
=
−
+
−
=
+
− (1)
Since (h, k) lies on the parabola y = x2, we have k = h2 |
1 | 2922-2925 | Let D be the required distance
between (h, k) and (0, c) Then
2
2
2
2
D
(
0)
(
)
(
)
h
k
c
h
k
c
=
−
+
−
=
+
− (1)
Since (h, k) lies on the parabola y = x2, we have k = h2 So (1) gives
D ≡ D(k) =
2
(
)
k
k
c
+
−
or
D′(k) =
2
1
2(
)
2
(
)
k
c
k
k
c
+
−
+
−
Now
D′(k) = 0 gives
2
1
c2
k
−
=
Observe that when
2
1
c2
k
−
<
, then 2(
)
1
0
k
−c
+ <
, i |
1 | 2923-2926 | Then
2
2
2
2
D
(
0)
(
)
(
)
h
k
c
h
k
c
=
−
+
−
=
+
− (1)
Since (h, k) lies on the parabola y = x2, we have k = h2 So (1) gives
D ≡ D(k) =
2
(
)
k
k
c
+
−
or
D′(k) =
2
1
2(
)
2
(
)
k
c
k
k
c
+
−
+
−
Now
D′(k) = 0 gives
2
1
c2
k
−
=
Observe that when
2
1
c2
k
−
<
, then 2(
)
1
0
k
−c
+ <
, i e |
1 | 2924-2927 | (1)
Since (h, k) lies on the parabola y = x2, we have k = h2 So (1) gives
D ≡ D(k) =
2
(
)
k
k
c
+
−
or
D′(k) =
2
1
2(
)
2
(
)
k
c
k
k
c
+
−
+
−
Now
D′(k) = 0 gives
2
1
c2
k
−
=
Observe that when
2
1
c2
k
−
<
, then 2(
)
1
0
k
−c
+ <
, i e , D ( )
0
′k
< |
1 | 2925-2928 | So (1) gives
D ≡ D(k) =
2
(
)
k
k
c
+
−
or
D′(k) =
2
1
2(
)
2
(
)
k
c
k
k
c
+
−
+
−
Now
D′(k) = 0 gives
2
1
c2
k
−
=
Observe that when
2
1
c2
k
−
<
, then 2(
)
1
0
k
−c
+ <
, i e , D ( )
0
′k
< Also when
2
1
c2
k
−
>
, then D ( )
0
′k
> |
1 | 2926-2929 | e , D ( )
0
′k
< Also when
2
1
c2
k
−
>
, then D ( )
0
′k
> So, by first derivative test, D (k) is minimum at
2
1
c2
k
−
= |
1 | 2927-2930 | , D ( )
0
′k
< Also when
2
1
c2
k
−
>
, then D ( )
0
′k
> So, by first derivative test, D (k) is minimum at
2
1
c2
k
−
= Rationalised 2023-24
APPLICATION OF DERIVATIVES
169
Hence, the required shortest distance is given by
2
2
1
2
1
2
1
4
1
D
2
2
2
2
c
c
c
c
c
−
−
−
−
=
+
−
=
ANote The reader may note that in Example 35, we have used first derivative
test instead of the second derivative test as the former is easy and short |
1 | 2928-2931 | Also when
2
1
c2
k
−
>
, then D ( )
0
′k
> So, by first derivative test, D (k) is minimum at
2
1
c2
k
−
= Rationalised 2023-24
APPLICATION OF DERIVATIVES
169
Hence, the required shortest distance is given by
2
2
1
2
1
2
1
4
1
D
2
2
2
2
c
c
c
c
c
−
−
−
−
=
+
−
=
ANote The reader may note that in Example 35, we have used first derivative
test instead of the second derivative test as the former is easy and short Example 24 Let AP and BQ be two vertical poles at
points A and B, respectively |
1 | 2929-2932 | So, by first derivative test, D (k) is minimum at
2
1
c2
k
−
= Rationalised 2023-24
APPLICATION OF DERIVATIVES
169
Hence, the required shortest distance is given by
2
2
1
2
1
2
1
4
1
D
2
2
2
2
c
c
c
c
c
−
−
−
−
=
+
−
=
ANote The reader may note that in Example 35, we have used first derivative
test instead of the second derivative test as the former is easy and short Example 24 Let AP and BQ be two vertical poles at
points A and B, respectively If AP = 16 m, BQ = 22 m
and AB = 20 m, then find the distance of a point R on
AB from the point A such that RP2 + RQ2 is minimum |
1 | 2930-2933 | Rationalised 2023-24
APPLICATION OF DERIVATIVES
169
Hence, the required shortest distance is given by
2
2
1
2
1
2
1
4
1
D
2
2
2
2
c
c
c
c
c
−
−
−
−
=
+
−
=
ANote The reader may note that in Example 35, we have used first derivative
test instead of the second derivative test as the former is easy and short Example 24 Let AP and BQ be two vertical poles at
points A and B, respectively If AP = 16 m, BQ = 22 m
and AB = 20 m, then find the distance of a point R on
AB from the point A such that RP2 + RQ2 is minimum Solution Let R be a point on AB such that AR = x m |
1 | 2931-2934 | Example 24 Let AP and BQ be two vertical poles at
points A and B, respectively If AP = 16 m, BQ = 22 m
and AB = 20 m, then find the distance of a point R on
AB from the point A such that RP2 + RQ2 is minimum Solution Let R be a point on AB such that AR = x m Then RB = (20 – x) m (as AB = 20 m) |
1 | 2932-2935 | If AP = 16 m, BQ = 22 m
and AB = 20 m, then find the distance of a point R on
AB from the point A such that RP2 + RQ2 is minimum Solution Let R be a point on AB such that AR = x m Then RB = (20 – x) m (as AB = 20 m) From Fig 6 |
1 | 2933-2936 | Solution Let R be a point on AB such that AR = x m Then RB = (20 – x) m (as AB = 20 m) From Fig 6 16,
we have
RP2 = AR2 + AP2
and
RQ2 = RB2 + BQ2
Therefore
RP2 + RQ2 = AR2 + AP2 + RB2 + BQ2
= x2 + (16)2 + (20 – x)2 + (22)2
= 2x2 – 40x + 1140
Let
S ≡ S(x) = RP2 + RQ2 = 2x2 – 40x + 1140 |
1 | 2934-2937 | Then RB = (20 – x) m (as AB = 20 m) From Fig 6 16,
we have
RP2 = AR2 + AP2
and
RQ2 = RB2 + BQ2
Therefore
RP2 + RQ2 = AR2 + AP2 + RB2 + BQ2
= x2 + (16)2 + (20 – x)2 + (22)2
= 2x2 – 40x + 1140
Let
S ≡ S(x) = RP2 + RQ2 = 2x2 – 40x + 1140 Therefore
S′(x) = 4x – 40 |
1 | 2935-2938 | From Fig 6 16,
we have
RP2 = AR2 + AP2
and
RQ2 = RB2 + BQ2
Therefore
RP2 + RQ2 = AR2 + AP2 + RB2 + BQ2
= x2 + (16)2 + (20 – x)2 + (22)2
= 2x2 – 40x + 1140
Let
S ≡ S(x) = RP2 + RQ2 = 2x2 – 40x + 1140 Therefore
S′(x) = 4x – 40 Now S′(x) = 0 gives x = 10 |
1 | 2936-2939 | 16,
we have
RP2 = AR2 + AP2
and
RQ2 = RB2 + BQ2
Therefore
RP2 + RQ2 = AR2 + AP2 + RB2 + BQ2
= x2 + (16)2 + (20 – x)2 + (22)2
= 2x2 – 40x + 1140
Let
S ≡ S(x) = RP2 + RQ2 = 2x2 – 40x + 1140 Therefore
S′(x) = 4x – 40 Now S′(x) = 0 gives x = 10 Also S″(x) = 4 > 0, for all x and so S″(10) > 0 |
1 | 2937-2940 | Therefore
S′(x) = 4x – 40 Now S′(x) = 0 gives x = 10 Also S″(x) = 4 > 0, for all x and so S″(10) > 0 Therefore, by second derivative test, x = 10 is the point of local minima of S |
1 | 2938-2941 | Now S′(x) = 0 gives x = 10 Also S″(x) = 4 > 0, for all x and so S″(10) > 0 Therefore, by second derivative test, x = 10 is the point of local minima of S Thus, the
distance of R from A on AB is AR = x =10 m |
1 | 2939-2942 | Also S″(x) = 4 > 0, for all x and so S″(10) > 0 Therefore, by second derivative test, x = 10 is the point of local minima of S Thus, the
distance of R from A on AB is AR = x =10 m Example 25 If length of three sides of a trapezium other than base are equal to 10cm,
then find the area of the trapezium when it is maximum |
1 | 2940-2943 | Therefore, by second derivative test, x = 10 is the point of local minima of S Thus, the
distance of R from A on AB is AR = x =10 m Example 25 If length of three sides of a trapezium other than base are equal to 10cm,
then find the area of the trapezium when it is maximum Solution The required trapezium is as given in Fig 6 |
1 | 2941-2944 | Thus, the
distance of R from A on AB is AR = x =10 m Example 25 If length of three sides of a trapezium other than base are equal to 10cm,
then find the area of the trapezium when it is maximum Solution The required trapezium is as given in Fig 6 17 |
1 | 2942-2945 | Example 25 If length of three sides of a trapezium other than base are equal to 10cm,
then find the area of the trapezium when it is maximum Solution The required trapezium is as given in Fig 6 17 Draw perpendiculars DP and
Fig 6 |
1 | 2943-2946 | Solution The required trapezium is as given in Fig 6 17 Draw perpendiculars DP and
Fig 6 16
Fig 6 |
1 | 2944-2947 | 17 Draw perpendiculars DP and
Fig 6 16
Fig 6 17
Rationalised 2023-24
MATHEMATICS
170
CQ on AB |
1 | 2945-2948 | Draw perpendiculars DP and
Fig 6 16
Fig 6 17
Rationalised 2023-24
MATHEMATICS
170
CQ on AB Let AP = x cm |
1 | 2946-2949 | 16
Fig 6 17
Rationalised 2023-24
MATHEMATICS
170
CQ on AB Let AP = x cm Note that ∆APD ~ ∆BQC |
1 | 2947-2950 | 17
Rationalised 2023-24
MATHEMATICS
170
CQ on AB Let AP = x cm Note that ∆APD ~ ∆BQC Therefore, QB = x cm |
1 | 2948-2951 | Let AP = x cm Note that ∆APD ~ ∆BQC Therefore, QB = x cm Also, by
Pythagoras theorem, DP = QC =
2
100
−x |
1 | 2949-2952 | Note that ∆APD ~ ∆BQC Therefore, QB = x cm Also, by
Pythagoras theorem, DP = QC =
2
100
−x Let A be the area of the trapezium |
1 | 2950-2953 | Therefore, QB = x cm Also, by
Pythagoras theorem, DP = QC =
2
100
−x Let A be the area of the trapezium Then
A ≡ A(x) = 1
2 (sum of parallel sides) (height)
=
(
2)
1 (2
10 10)
100
2
x
x
+
+
−
=
(
2)
(
10)
100
x
x
+
−
or
A′(x) =
(
)
2
( 2 )2
(
10)
100
2 100
x
x
x
x
−
+
+
−
−
=
2
2
2
10
100
100
x
x
x
−
−
+
−
Now
A′(x) = 0 gives 2x2 + 10x – 100 = 0, i |
1 | 2951-2954 | Also, by
Pythagoras theorem, DP = QC =
2
100
−x Let A be the area of the trapezium Then
A ≡ A(x) = 1
2 (sum of parallel sides) (height)
=
(
2)
1 (2
10 10)
100
2
x
x
+
+
−
=
(
2)
(
10)
100
x
x
+
−
or
A′(x) =
(
)
2
( 2 )2
(
10)
100
2 100
x
x
x
x
−
+
+
−
−
=
2
2
2
10
100
100
x
x
x
−
−
+
−
Now
A′(x) = 0 gives 2x2 + 10x – 100 = 0, i e |
1 | 2952-2955 | Let A be the area of the trapezium Then
A ≡ A(x) = 1
2 (sum of parallel sides) (height)
=
(
2)
1 (2
10 10)
100
2
x
x
+
+
−
=
(
2)
(
10)
100
x
x
+
−
or
A′(x) =
(
)
2
( 2 )2
(
10)
100
2 100
x
x
x
x
−
+
+
−
−
=
2
2
2
10
100
100
x
x
x
−
−
+
−
Now
A′(x) = 0 gives 2x2 + 10x – 100 = 0, i e , x = 5 and x = –10 |
1 | 2953-2956 | Then
A ≡ A(x) = 1
2 (sum of parallel sides) (height)
=
(
2)
1 (2
10 10)
100
2
x
x
+
+
−
=
(
2)
(
10)
100
x
x
+
−
or
A′(x) =
(
)
2
( 2 )2
(
10)
100
2 100
x
x
x
x
−
+
+
−
−
=
2
2
2
10
100
100
x
x
x
−
−
+
−
Now
A′(x) = 0 gives 2x2 + 10x – 100 = 0, i e , x = 5 and x = –10 Since x represents distance, it can not be negative |
1 | 2954-2957 | e , x = 5 and x = –10 Since x represents distance, it can not be negative So,
x = 5 |
1 | 2955-2958 | , x = 5 and x = –10 Since x represents distance, it can not be negative So,
x = 5 Now
A″(x) =
2
2
2
2
( 2 )
100
( 4
10)
( 2
10
100)
2 100
100
x
x
x
x
x
x
x
−
−
−
−
− −
−
+
−
−
=
3
3
2
2
2
300
1000
(100
)
x
x
x
−
−
−
(on simplification)
or
A″(5) =
3
3
2
2
2(5)
300(5)
1000
2250
30
0
75 75
75
(100
(5) )
−
−
−
−
=
=
<
−
Thus, area of trapezium is maximum at x = 5 and the area is given by
A(5) =
2
2
(5
10) 100
(5)
15 75
75 3 cm
+
−
=
=
Example 26 Prove that the radius of the right circular cylinder of greatest curved
surface area which can be inscribed in a given cone is half of that of the cone |
1 | 2956-2959 | Since x represents distance, it can not be negative So,
x = 5 Now
A″(x) =
2
2
2
2
( 2 )
100
( 4
10)
( 2
10
100)
2 100
100
x
x
x
x
x
x
x
−
−
−
−
− −
−
+
−
−
=
3
3
2
2
2
300
1000
(100
)
x
x
x
−
−
−
(on simplification)
or
A″(5) =
3
3
2
2
2(5)
300(5)
1000
2250
30
0
75 75
75
(100
(5) )
−
−
−
−
=
=
<
−
Thus, area of trapezium is maximum at x = 5 and the area is given by
A(5) =
2
2
(5
10) 100
(5)
15 75
75 3 cm
+
−
=
=
Example 26 Prove that the radius of the right circular cylinder of greatest curved
surface area which can be inscribed in a given cone is half of that of the cone Solution Let OC = r be the radius of the cone and OA = h be its height |
1 | 2957-2960 | So,
x = 5 Now
A″(x) =
2
2
2
2
( 2 )
100
( 4
10)
( 2
10
100)
2 100
100
x
x
x
x
x
x
x
−
−
−
−
− −
−
+
−
−
=
3
3
2
2
2
300
1000
(100
)
x
x
x
−
−
−
(on simplification)
or
A″(5) =
3
3
2
2
2(5)
300(5)
1000
2250
30
0
75 75
75
(100
(5) )
−
−
−
−
=
=
<
−
Thus, area of trapezium is maximum at x = 5 and the area is given by
A(5) =
2
2
(5
10) 100
(5)
15 75
75 3 cm
+
−
=
=
Example 26 Prove that the radius of the right circular cylinder of greatest curved
surface area which can be inscribed in a given cone is half of that of the cone Solution Let OC = r be the radius of the cone and OA = h be its height Let a cylinder
with radius OE = x inscribed in the given cone (Fig 6 |
1 | 2958-2961 | Now
A″(x) =
2
2
2
2
( 2 )
100
( 4
10)
( 2
10
100)
2 100
100
x
x
x
x
x
x
x
−
−
−
−
− −
−
+
−
−
=
3
3
2
2
2
300
1000
(100
)
x
x
x
−
−
−
(on simplification)
or
A″(5) =
3
3
2
2
2(5)
300(5)
1000
2250
30
0
75 75
75
(100
(5) )
−
−
−
−
=
=
<
−
Thus, area of trapezium is maximum at x = 5 and the area is given by
A(5) =
2
2
(5
10) 100
(5)
15 75
75 3 cm
+
−
=
=
Example 26 Prove that the radius of the right circular cylinder of greatest curved
surface area which can be inscribed in a given cone is half of that of the cone Solution Let OC = r be the radius of the cone and OA = h be its height Let a cylinder
with radius OE = x inscribed in the given cone (Fig 6 18) |
1 | 2959-2962 | Solution Let OC = r be the radius of the cone and OA = h be its height Let a cylinder
with radius OE = x inscribed in the given cone (Fig 6 18) The height QE of the cylinder
is given by
Rationalised 2023-24
APPLICATION OF DERIVATIVES
171
QE
OA = EC
OC (since ∆QEC ~ ∆AOC)
or
QE
h = r
x
r
−
or
QE =
(
)
h r
x
r
−
Let S be the curved surface area of the given
cylinder |
1 | 2960-2963 | Let a cylinder
with radius OE = x inscribed in the given cone (Fig 6 18) The height QE of the cylinder
is given by
Rationalised 2023-24
APPLICATION OF DERIVATIVES
171
QE
OA = EC
OC (since ∆QEC ~ ∆AOC)
or
QE
h = r
x
r
−
or
QE =
(
)
h r
x
r
−
Let S be the curved surface area of the given
cylinder Then
S ≡ S(x) = 2
(
)
xh r
x
r
π
−
=
2
2
(
)
h rx
x
πr
−
or
2
S ( )
(
2 )
4
S ( )
h
x
r
x
r
h
x
r
π
′
=
−
− π
′′
=
Now S′(x) = 0 gives
x =r2 |
1 | 2961-2964 | 18) The height QE of the cylinder
is given by
Rationalised 2023-24
APPLICATION OF DERIVATIVES
171
QE
OA = EC
OC (since ∆QEC ~ ∆AOC)
or
QE
h = r
x
r
−
or
QE =
(
)
h r
x
r
−
Let S be the curved surface area of the given
cylinder Then
S ≡ S(x) = 2
(
)
xh r
x
r
π
−
=
2
2
(
)
h rx
x
πr
−
or
2
S ( )
(
2 )
4
S ( )
h
x
r
x
r
h
x
r
π
′
=
−
− π
′′
=
Now S′(x) = 0 gives
x =r2 Since S″(x) < 0 for all x, S
0
r2
′′
<
|
1 | 2962-2965 | The height QE of the cylinder
is given by
Rationalised 2023-24
APPLICATION OF DERIVATIVES
171
QE
OA = EC
OC (since ∆QEC ~ ∆AOC)
or
QE
h = r
x
r
−
or
QE =
(
)
h r
x
r
−
Let S be the curved surface area of the given
cylinder Then
S ≡ S(x) = 2
(
)
xh r
x
r
π
−
=
2
2
(
)
h rx
x
πr
−
or
2
S ( )
(
2 )
4
S ( )
h
x
r
x
r
h
x
r
π
′
=
−
− π
′′
=
Now S′(x) = 0 gives
x =r2 Since S″(x) < 0 for all x, S
0
r2
′′
<
So
x =r2
is a
point of maxima of S |
1 | 2963-2966 | Then
S ≡ S(x) = 2
(
)
xh r
x
r
π
−
=
2
2
(
)
h rx
x
πr
−
or
2
S ( )
(
2 )
4
S ( )
h
x
r
x
r
h
x
r
π
′
=
−
− π
′′
=
Now S′(x) = 0 gives
x =r2 Since S″(x) < 0 for all x, S
0
r2
′′
<
So
x =r2
is a
point of maxima of S Hence, the radius of the cylinder of greatest curved surface area
which can be inscribed in a given cone is half of that of the cone |
1 | 2964-2967 | Since S″(x) < 0 for all x, S
0
r2
′′
<
So
x =r2
is a
point of maxima of S Hence, the radius of the cylinder of greatest curved surface area
which can be inscribed in a given cone is half of that of the cone 6 |
1 | 2965-2968 | So
x =r2
is a
point of maxima of S Hence, the radius of the cylinder of greatest curved surface area
which can be inscribed in a given cone is half of that of the cone 6 4 |
1 | 2966-2969 | Hence, the radius of the cylinder of greatest curved surface area
which can be inscribed in a given cone is half of that of the cone 6 4 1 Maximum and Minimum Values of a Function in a Closed Interval
Let us consider a function f given by
f (x) = x + 2, x ∈ (0, 1)
Observe that the function is continuous on (0, 1) and neither has a maximum value
nor has a minimum value |
1 | 2967-2970 | 6 4 1 Maximum and Minimum Values of a Function in a Closed Interval
Let us consider a function f given by
f (x) = x + 2, x ∈ (0, 1)
Observe that the function is continuous on (0, 1) and neither has a maximum value
nor has a minimum value Further, we may note that the function even has neither a
local maximum value nor a local minimum value |
1 | 2968-2971 | 4 1 Maximum and Minimum Values of a Function in a Closed Interval
Let us consider a function f given by
f (x) = x + 2, x ∈ (0, 1)
Observe that the function is continuous on (0, 1) and neither has a maximum value
nor has a minimum value Further, we may note that the function even has neither a
local maximum value nor a local minimum value However, if we extend the domain of f to the closed interval [0, 1], then f still may
not have a local maximum (minimum) values but it certainly does have maximum value
3 = f (1) and minimum value 2 = f (0) |
1 | 2969-2972 | 1 Maximum and Minimum Values of a Function in a Closed Interval
Let us consider a function f given by
f (x) = x + 2, x ∈ (0, 1)
Observe that the function is continuous on (0, 1) and neither has a maximum value
nor has a minimum value Further, we may note that the function even has neither a
local maximum value nor a local minimum value However, if we extend the domain of f to the closed interval [0, 1], then f still may
not have a local maximum (minimum) values but it certainly does have maximum value
3 = f (1) and minimum value 2 = f (0) The maximum value 3 of f at x = 1 is called
absolute maximum value (global maximum or greatest value) of f on the interval
[0, 1] |
1 | 2970-2973 | Further, we may note that the function even has neither a
local maximum value nor a local minimum value However, if we extend the domain of f to the closed interval [0, 1], then f still may
not have a local maximum (minimum) values but it certainly does have maximum value
3 = f (1) and minimum value 2 = f (0) The maximum value 3 of f at x = 1 is called
absolute maximum value (global maximum or greatest value) of f on the interval
[0, 1] Similarly, the minimum value 2 of f at x = 0 is called the absolute minimum
value (global minimum or least value) of f on [0, 1] |
1 | 2971-2974 | However, if we extend the domain of f to the closed interval [0, 1], then f still may
not have a local maximum (minimum) values but it certainly does have maximum value
3 = f (1) and minimum value 2 = f (0) The maximum value 3 of f at x = 1 is called
absolute maximum value (global maximum or greatest value) of f on the interval
[0, 1] Similarly, the minimum value 2 of f at x = 0 is called the absolute minimum
value (global minimum or least value) of f on [0, 1] Consider the graph given in Fig 6 |
1 | 2972-2975 | The maximum value 3 of f at x = 1 is called
absolute maximum value (global maximum or greatest value) of f on the interval
[0, 1] Similarly, the minimum value 2 of f at x = 0 is called the absolute minimum
value (global minimum or least value) of f on [0, 1] Consider the graph given in Fig 6 19 of a continuous function defined on a closed
interval [a, d] |
1 | 2973-2976 | Similarly, the minimum value 2 of f at x = 0 is called the absolute minimum
value (global minimum or least value) of f on [0, 1] Consider the graph given in Fig 6 19 of a continuous function defined on a closed
interval [a, d] Observe that the function f has a local minima at x = b and local
Fig 6 |
1 | 2974-2977 | Consider the graph given in Fig 6 19 of a continuous function defined on a closed
interval [a, d] Observe that the function f has a local minima at x = b and local
Fig 6 18
Rationalised 2023-24
MATHEMATICS
172
minimum value is f(b) |
1 | 2975-2978 | 19 of a continuous function defined on a closed
interval [a, d] Observe that the function f has a local minima at x = b and local
Fig 6 18
Rationalised 2023-24
MATHEMATICS
172
minimum value is f(b) The function also has a local maxima at x = c and local maximum
value is f (c) |
1 | 2976-2979 | Observe that the function f has a local minima at x = b and local
Fig 6 18
Rationalised 2023-24
MATHEMATICS
172
minimum value is f(b) The function also has a local maxima at x = c and local maximum
value is f (c) Also from the graph, it is evident that f has absolute maximum value f (a) and
absolute minimum value f (d) |
1 | 2977-2980 | 18
Rationalised 2023-24
MATHEMATICS
172
minimum value is f(b) The function also has a local maxima at x = c and local maximum
value is f (c) Also from the graph, it is evident that f has absolute maximum value f (a) and
absolute minimum value f (d) Further note that the absolute maximum (minimum)
value of f is different from local maximum (minimum) value of f |
1 | 2978-2981 | The function also has a local maxima at x = c and local maximum
value is f (c) Also from the graph, it is evident that f has absolute maximum value f (a) and
absolute minimum value f (d) Further note that the absolute maximum (minimum)
value of f is different from local maximum (minimum) value of f We will now state two results (without proof) regarding absolute maximum and
absolute minimum values of a function on a closed interval I |
1 | 2979-2982 | Also from the graph, it is evident that f has absolute maximum value f (a) and
absolute minimum value f (d) Further note that the absolute maximum (minimum)
value of f is different from local maximum (minimum) value of f We will now state two results (without proof) regarding absolute maximum and
absolute minimum values of a function on a closed interval I Theorem 5 Let f be a continuous function on an interval I = [a, b] |
1 | 2980-2983 | Further note that the absolute maximum (minimum)
value of f is different from local maximum (minimum) value of f We will now state two results (without proof) regarding absolute maximum and
absolute minimum values of a function on a closed interval I Theorem 5 Let f be a continuous function on an interval I = [a, b] Then f has the
absolute maximum value and f attains it at least once in I |
1 | 2981-2984 | We will now state two results (without proof) regarding absolute maximum and
absolute minimum values of a function on a closed interval I Theorem 5 Let f be a continuous function on an interval I = [a, b] Then f has the
absolute maximum value and f attains it at least once in I Also, f has the absolute
minimum value and attains it at least once in I |
1 | 2982-2985 | Theorem 5 Let f be a continuous function on an interval I = [a, b] Then f has the
absolute maximum value and f attains it at least once in I Also, f has the absolute
minimum value and attains it at least once in I Theorem 6 Let f be a differentiable function on a closed interval I and let c be any
interior point of I |
1 | 2983-2986 | Then f has the
absolute maximum value and f attains it at least once in I Also, f has the absolute
minimum value and attains it at least once in I Theorem 6 Let f be a differentiable function on a closed interval I and let c be any
interior point of I Then
(i) f ′(c) = 0 if f attains its absolute maximum value at c |
1 | 2984-2987 | Also, f has the absolute
minimum value and attains it at least once in I Theorem 6 Let f be a differentiable function on a closed interval I and let c be any
interior point of I Then
(i) f ′(c) = 0 if f attains its absolute maximum value at c (ii) f ′(c) = 0 if f attains its absolute minimum value at c |
1 | 2985-2988 | Theorem 6 Let f be a differentiable function on a closed interval I and let c be any
interior point of I Then
(i) f ′(c) = 0 if f attains its absolute maximum value at c (ii) f ′(c) = 0 if f attains its absolute minimum value at c In view of the above results, we have the following working rule for finding absolute
maximum and/or absolute minimum values of a function in a given closed interval
[a, b] |
1 | 2986-2989 | Then
(i) f ′(c) = 0 if f attains its absolute maximum value at c (ii) f ′(c) = 0 if f attains its absolute minimum value at c In view of the above results, we have the following working rule for finding absolute
maximum and/or absolute minimum values of a function in a given closed interval
[a, b] Working Rule
Step 1: Find all critical points of f in the interval, i |
1 | 2987-2990 | (ii) f ′(c) = 0 if f attains its absolute minimum value at c In view of the above results, we have the following working rule for finding absolute
maximum and/or absolute minimum values of a function in a given closed interval
[a, b] Working Rule
Step 1: Find all critical points of f in the interval, i e |
1 | 2988-2991 | In view of the above results, we have the following working rule for finding absolute
maximum and/or absolute minimum values of a function in a given closed interval
[a, b] Working Rule
Step 1: Find all critical points of f in the interval, i e , find points x where either
( )
0
f
′x =
or f is not differentiable |
1 | 2989-2992 | Working Rule
Step 1: Find all critical points of f in the interval, i e , find points x where either
( )
0
f
′x =
or f is not differentiable Step 2: Take the end points of the interval |
1 | 2990-2993 | e , find points x where either
( )
0
f
′x =
or f is not differentiable Step 2: Take the end points of the interval Step 3: At all these points (listed in Step 1 and 2), calculate the values of f |
1 | 2991-2994 | , find points x where either
( )
0
f
′x =
or f is not differentiable Step 2: Take the end points of the interval Step 3: At all these points (listed in Step 1 and 2), calculate the values of f Step 4: Identify the maximum and minimum values of f out of the values calculated in
Step 3 |
1 | 2992-2995 | Step 2: Take the end points of the interval Step 3: At all these points (listed in Step 1 and 2), calculate the values of f Step 4: Identify the maximum and minimum values of f out of the values calculated in
Step 3 This maximum value will be the absolute maximum (greatest) value of
f and the minimum value will be the absolute minimum (least) value of f |
1 | 2993-2996 | Step 3: At all these points (listed in Step 1 and 2), calculate the values of f Step 4: Identify the maximum and minimum values of f out of the values calculated in
Step 3 This maximum value will be the absolute maximum (greatest) value of
f and the minimum value will be the absolute minimum (least) value of f Fig 6 |
1 | 2994-2997 | Step 4: Identify the maximum and minimum values of f out of the values calculated in
Step 3 This maximum value will be the absolute maximum (greatest) value of
f and the minimum value will be the absolute minimum (least) value of f Fig 6 19
Rationalised 2023-24
APPLICATION OF DERIVATIVES
173
Example 27 Find the absolute maximum and minimum values of a function f given by
f (x) = 2x3 – 15x2 + 36x +1 on the interval [1, 5] |
1 | 2995-2998 | This maximum value will be the absolute maximum (greatest) value of
f and the minimum value will be the absolute minimum (least) value of f Fig 6 19
Rationalised 2023-24
APPLICATION OF DERIVATIVES
173
Example 27 Find the absolute maximum and minimum values of a function f given by
f (x) = 2x3 – 15x2 + 36x +1 on the interval [1, 5] Solution We have
f (x) = 2x3 – 15x2 + 36x + 1
or
f ′(x) = 6x2 – 30x + 36 = 6 (x – 3) (x – 2)
Note that f ′(x) = 0 gives x = 2 and x = 3 |
1 | 2996-2999 | Fig 6 19
Rationalised 2023-24
APPLICATION OF DERIVATIVES
173
Example 27 Find the absolute maximum and minimum values of a function f given by
f (x) = 2x3 – 15x2 + 36x +1 on the interval [1, 5] Solution We have
f (x) = 2x3 – 15x2 + 36x + 1
or
f ′(x) = 6x2 – 30x + 36 = 6 (x – 3) (x – 2)
Note that f ′(x) = 0 gives x = 2 and x = 3 We shall now evaluate the value of f at these points and at the end points of the
interval [1, 5], i |
1 | 2997-3000 | 19
Rationalised 2023-24
APPLICATION OF DERIVATIVES
173
Example 27 Find the absolute maximum and minimum values of a function f given by
f (x) = 2x3 – 15x2 + 36x +1 on the interval [1, 5] Solution We have
f (x) = 2x3 – 15x2 + 36x + 1
or
f ′(x) = 6x2 – 30x + 36 = 6 (x – 3) (x – 2)
Note that f ′(x) = 0 gives x = 2 and x = 3 We shall now evaluate the value of f at these points and at the end points of the
interval [1, 5], i e |
1 | 2998-3001 | Solution We have
f (x) = 2x3 – 15x2 + 36x + 1
or
f ′(x) = 6x2 – 30x + 36 = 6 (x – 3) (x – 2)
Note that f ′(x) = 0 gives x = 2 and x = 3 We shall now evaluate the value of f at these points and at the end points of the
interval [1, 5], i e , at x = 1, x = 2, x = 3 and at x = 5 |
1 | 2999-3002 | We shall now evaluate the value of f at these points and at the end points of the
interval [1, 5], i e , at x = 1, x = 2, x = 3 and at x = 5 So
f (1) = 2(13) – 15(12) + 36 (1) + 1 = 24
f (2) = 2(23) – 15(22) + 36 (2) + 1 = 29
f (3) = 2(33) – 15(32) + 36 (3) + 1 = 28
f (5) = 2(53) – 15(52) + 36 (5) + 1 = 56
Thus, we conclude that absolute maximum value of f on [1, 5] is 56, occurring at
x =5, and absolute minimum value of f on [1, 5] is 24 which occurs at x = 1 |
1 | 3000-3003 | e , at x = 1, x = 2, x = 3 and at x = 5 So
f (1) = 2(13) – 15(12) + 36 (1) + 1 = 24
f (2) = 2(23) – 15(22) + 36 (2) + 1 = 29
f (3) = 2(33) – 15(32) + 36 (3) + 1 = 28
f (5) = 2(53) – 15(52) + 36 (5) + 1 = 56
Thus, we conclude that absolute maximum value of f on [1, 5] is 56, occurring at
x =5, and absolute minimum value of f on [1, 5] is 24 which occurs at x = 1 Example 28 Find absolute maximum and minimum values of a function f given by
4
1
3
3
( )
12
6
,
[ 1, 1]
f x
x
x
x
=
−
∈ −
Solution We have
f (x) =
4
1
3
3
12
6
x
x
−
or
f ′(x) =
31
2
2
3
3
2
2(8
1)
16
x
x
x
x
−
−
=
Thus, f ′(x) = 0 gives
x =81 |
1 | 3001-3004 | , at x = 1, x = 2, x = 3 and at x = 5 So
f (1) = 2(13) – 15(12) + 36 (1) + 1 = 24
f (2) = 2(23) – 15(22) + 36 (2) + 1 = 29
f (3) = 2(33) – 15(32) + 36 (3) + 1 = 28
f (5) = 2(53) – 15(52) + 36 (5) + 1 = 56
Thus, we conclude that absolute maximum value of f on [1, 5] is 56, occurring at
x =5, and absolute minimum value of f on [1, 5] is 24 which occurs at x = 1 Example 28 Find absolute maximum and minimum values of a function f given by
4
1
3
3
( )
12
6
,
[ 1, 1]
f x
x
x
x
=
−
∈ −
Solution We have
f (x) =
4
1
3
3
12
6
x
x
−
or
f ′(x) =
31
2
2
3
3
2
2(8
1)
16
x
x
x
x
−
−
=
Thus, f ′(x) = 0 gives
x =81 Further note that f ′(x) is not defined at x = 0 |
1 | 3002-3005 | So
f (1) = 2(13) – 15(12) + 36 (1) + 1 = 24
f (2) = 2(23) – 15(22) + 36 (2) + 1 = 29
f (3) = 2(33) – 15(32) + 36 (3) + 1 = 28
f (5) = 2(53) – 15(52) + 36 (5) + 1 = 56
Thus, we conclude that absolute maximum value of f on [1, 5] is 56, occurring at
x =5, and absolute minimum value of f on [1, 5] is 24 which occurs at x = 1 Example 28 Find absolute maximum and minimum values of a function f given by
4
1
3
3
( )
12
6
,
[ 1, 1]
f x
x
x
x
=
−
∈ −
Solution We have
f (x) =
4
1
3
3
12
6
x
x
−
or
f ′(x) =
31
2
2
3
3
2
2(8
1)
16
x
x
x
x
−
−
=
Thus, f ′(x) = 0 gives
x =81 Further note that f ′(x) is not defined at x = 0 So the
critical points are x = 0 and
x =81 |
1 | 3003-3006 | Example 28 Find absolute maximum and minimum values of a function f given by
4
1
3
3
( )
12
6
,
[ 1, 1]
f x
x
x
x
=
−
∈ −
Solution We have
f (x) =
4
1
3
3
12
6
x
x
−
or
f ′(x) =
31
2
2
3
3
2
2(8
1)
16
x
x
x
x
−
−
=
Thus, f ′(x) = 0 gives
x =81 Further note that f ′(x) is not defined at x = 0 So the
critical points are x = 0 and
x =81 Now evaluating the value of f at critical points
x = 0, 1
8 and at end points of the interval x = –1 and x = 1, we have
f (–1) =
4
1
3
3
12( 1)
6( 1)
18
−
−
−
=
f (0) = 12 (0) – 6(0) = 0
Rationalised 2023-24
MATHEMATICS
174
f 81
=
4
1
3
3
1
1
9
12
6
8
8
−4
−
=
f (1) =
4
1
3
3
12(1)
6(1)
6
−
=
Hence, we conclude that absolute maximum value of f is 18 that occurs at x = –1
and absolute minimum value of f is
49
− that occurs at
x =81 |
1 | 3004-3007 | Further note that f ′(x) is not defined at x = 0 So the
critical points are x = 0 and
x =81 Now evaluating the value of f at critical points
x = 0, 1
8 and at end points of the interval x = –1 and x = 1, we have
f (–1) =
4
1
3
3
12( 1)
6( 1)
18
−
−
−
=
f (0) = 12 (0) – 6(0) = 0
Rationalised 2023-24
MATHEMATICS
174
f 81
=
4
1
3
3
1
1
9
12
6
8
8
−4
−
=
f (1) =
4
1
3
3
12(1)
6(1)
6
−
=
Hence, we conclude that absolute maximum value of f is 18 that occurs at x = –1
and absolute minimum value of f is
49
− that occurs at
x =81 Example 29 An Apache helicopter of enemy is flying along the curve given by
y = x2 + 7 |
1 | 3005-3008 | So the
critical points are x = 0 and
x =81 Now evaluating the value of f at critical points
x = 0, 1
8 and at end points of the interval x = –1 and x = 1, we have
f (–1) =
4
1
3
3
12( 1)
6( 1)
18
−
−
−
=
f (0) = 12 (0) – 6(0) = 0
Rationalised 2023-24
MATHEMATICS
174
f 81
=
4
1
3
3
1
1
9
12
6
8
8
−4
−
=
f (1) =
4
1
3
3
12(1)
6(1)
6
−
=
Hence, we conclude that absolute maximum value of f is 18 that occurs at x = –1
and absolute minimum value of f is
49
− that occurs at
x =81 Example 29 An Apache helicopter of enemy is flying along the curve given by
y = x2 + 7 A soldier, placed at (3, 7), wants to shoot down the helicopter when it is
nearest to him |
1 | 3006-3009 | Now evaluating the value of f at critical points
x = 0, 1
8 and at end points of the interval x = –1 and x = 1, we have
f (–1) =
4
1
3
3
12( 1)
6( 1)
18
−
−
−
=
f (0) = 12 (0) – 6(0) = 0
Rationalised 2023-24
MATHEMATICS
174
f 81
=
4
1
3
3
1
1
9
12
6
8
8
−4
−
=
f (1) =
4
1
3
3
12(1)
6(1)
6
−
=
Hence, we conclude that absolute maximum value of f is 18 that occurs at x = –1
and absolute minimum value of f is
49
− that occurs at
x =81 Example 29 An Apache helicopter of enemy is flying along the curve given by
y = x2 + 7 A soldier, placed at (3, 7), wants to shoot down the helicopter when it is
nearest to him Find the nearest distance |
1 | 3007-3010 | Example 29 An Apache helicopter of enemy is flying along the curve given by
y = x2 + 7 A soldier, placed at (3, 7), wants to shoot down the helicopter when it is
nearest to him Find the nearest distance Solution For each value of x, the helicopter’s position is at point (x, x2 + 7) |
1 | 3008-3011 | A soldier, placed at (3, 7), wants to shoot down the helicopter when it is
nearest to him Find the nearest distance Solution For each value of x, the helicopter’s position is at point (x, x2 + 7) Therefore, the distance between the helicopter and the soldier placed at (3,7) is
2
2
2
(
3)
(
7
7)
x
x
−
+
+
−
, i |
1 | 3009-3012 | Find the nearest distance Solution For each value of x, the helicopter’s position is at point (x, x2 + 7) Therefore, the distance between the helicopter and the soldier placed at (3,7) is
2
2
2
(
3)
(
7
7)
x
x
−
+
+
−
, i e |
1 | 3010-3013 | Solution For each value of x, the helicopter’s position is at point (x, x2 + 7) Therefore, the distance between the helicopter and the soldier placed at (3,7) is
2
2
2
(
3)
(
7
7)
x
x
−
+
+
−
, i e ,
2
4
(
3)
x
x
−
+ |
1 | 3011-3014 | Therefore, the distance between the helicopter and the soldier placed at (3,7) is
2
2
2
(
3)
(
7
7)
x
x
−
+
+
−
, i e ,
2
4
(
3)
x
x
−
+ Let
f (x) = (x – 3)2 + x4
or
f ′(x) = 2(x – 3) + 4x3 = 2(x – 1) (2x2 + 2x + 3)
Thus, f ′(x) = 0 gives x = 1 or 2x2 + 2x + 3 = 0 for which there are no real roots |
1 | 3012-3015 | e ,
2
4
(
3)
x
x
−
+ Let
f (x) = (x – 3)2 + x4
or
f ′(x) = 2(x – 3) + 4x3 = 2(x – 1) (2x2 + 2x + 3)
Thus, f ′(x) = 0 gives x = 1 or 2x2 + 2x + 3 = 0 for which there are no real roots Also, there are no end points of the interval to be added to the set for which f ′ is zero,
i |
1 | 3013-3016 | ,
2
4
(
3)
x
x
−
+ Let
f (x) = (x – 3)2 + x4
or
f ′(x) = 2(x – 3) + 4x3 = 2(x – 1) (2x2 + 2x + 3)
Thus, f ′(x) = 0 gives x = 1 or 2x2 + 2x + 3 = 0 for which there are no real roots Also, there are no end points of the interval to be added to the set for which f ′ is zero,
i e |
1 | 3014-3017 | Let
f (x) = (x – 3)2 + x4
or
f ′(x) = 2(x – 3) + 4x3 = 2(x – 1) (2x2 + 2x + 3)
Thus, f ′(x) = 0 gives x = 1 or 2x2 + 2x + 3 = 0 for which there are no real roots Also, there are no end points of the interval to be added to the set for which f ′ is zero,
i e , there is only one point, namely, x = 1 |
1 | 3015-3018 | Also, there are no end points of the interval to be added to the set for which f ′ is zero,
i e , there is only one point, namely, x = 1 The value of f at this point is given by
f (1) = (1 – 3)2 + (1)4 = 5 |
1 | 3016-3019 | e , there is only one point, namely, x = 1 The value of f at this point is given by
f (1) = (1 – 3)2 + (1)4 = 5 Thus, the distance between the solider and the helicopter is
(1)
5
f
= |
1 | 3017-3020 | , there is only one point, namely, x = 1 The value of f at this point is given by
f (1) = (1 – 3)2 + (1)4 = 5 Thus, the distance between the solider and the helicopter is
(1)
5
f
= Note that 5 is either a maximum value or a minimum value |
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