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For all real values of x, the minimum value of 
is
(A) 0 (B) 1
(C) 3 (D) 
Let
∴By second derivative test, f is the minimum at x = 1 and the minimum value is given by 
.
The correct answer is D.
At what points in the interval [0, 2Ï€) does the function sin 2x attain its maximum value?
Let f(x) = sin 2x.
Then, we evaluate the values of f at critical points
and at the end points of the interval [0, 2π].
Hence, we can conclude that the absolute maximum value of f on [0, 2π] is occurring at
and
.
Find two positive numbers x and y such that their sum is 35 and the product x2y5 is a maximum
Let one number be x. Then, the other number is y = (35 − x).
Let P(x) = x2y5. Then, we have:
x = 0, x = 35, x = 10
When x = 35,
and y = 35 − 35 = 0. This will make the product x2y5 equal to 0.
When x = 0, y = 35 − 0 = 35 and the product x2y2 will be 0.
∴ x = 0 and x = 35 cannot be the possible values of x.
When x = 10, we have:
∴ By second derivative test, P(x) will be the maximum when x = 10 and y = 35 − 10 = 25.
Hence, the required numbers are 10 and 25.
Find the maximum and minimum values, if any, of the following functions given by
(i) f(x) = (2x − 1)2 + 3 (ii) f(x) = 9x2 + 12x + 2
(iii) f(x) = −(x − 1)2 + 10 (iv) g(x) = x3 + 1
(i) The given function is f(x) = (2x − 1)2 + 3.
It can be observed that (2x − 1)2 ≥ 0 for every x ∈ R.
Therefore, f(x) = (2x − 1)2 + 3 ≥ 3 for every x ∈ R.
The minimum value of f is attained when 2x − 1 = 0.
2x − 1 = 0 ⇒ 
∴Minimum value of f =
= 3
Hence, function f does not have a maximum value.
(ii) The given function is f(x) = 9x2 + 12x + 2 = (3x + 2)2 − 2.
It can be observed that (3x + 2)2 ≥ 0 for every x ∈ R.
Therefore, f(x) = (3x + 2)2 − 2 ≥ −2 for every x ∈ R.
The minimum value of f is attained when 3x + 2 = 0.
3x + 2 = 0 ⇒ 
∴Minimum value of f = 
Hence, function f does not have a maximum value.
(iii) The given function is f(x) = − (x − 1)2 + 10.
It can be observed that (x − 1)2 ≥ 0 for every x ∈ R.
Therefore, f(x) = − (x − 1)2 + 10 ≤ 10 for every x ∈ R.
The maximum value of f is attained when (x − 1) = 0.
(x − 1) = 0 ⇒ x = 1
∴Maximum value of f = f(1) = − (1 − 1)2 + 10 = 10
Hence, function f does not have a minimum value.
(iv) The given function is g(x) = x3 + 1.
Hence, function g neither has a maximum value nor a minimum value.
Find the maximum and minimum values, if any, of the following functions given by
(i) f(x) = |x + 2| − 1 (ii) g(x) = − |x + 1| + 3
(iii) h(x) = sin(2x) + 5 (iv) f(x) = |sin 4x + 3|
(v) h(x) = x + 1, x
(−1, 1)
(i) f(x) = 
We know that 
for every x ∈ R.
Therefore, f(x) = 
for every x ∈ R.
The minimum value of f is attained when
.
∴Minimum value of f = f(−2) = 
Hence, function f does not have a maximum value.
(ii) g(x) =
We know that 
for every x ∈ R.
Therefore, g(x) = 
for every x ∈ R.
The maximum value of g is attained when
.
∴Maximum value of g = g(−1) = 
Hence, function g does not have a minimum value.
(iii) h(x) = sin2x + 5
We know that − 1 ≤ sin 2x ≤ 1.
⇒ − 1 + 5 ≤ sin 2x + 5 ≤ 1 + 5
⇒ 4 ≤ sin 2x + 5 ≤ 6
Hence, the maximum and minimum values of h are 6 and 4 respectively.
(iv) f(x) =
We know that −1 ≤ sin 4x ≤ 1.
⇒ 2 ≤ sin 4x + 3 ≤ 4
⇒ 2 ≤≤ 4
Hence, the maximum and minimum values of f are 4 and 2 respectively.
(v) h(x) = x + 1, x ∈ (−1, 1)
Here, if a point x0 is closest to −1, then we find for all x0∈ (−1, 1).
Also, if x1 is closest to 1, then 
for all x1∈ (−1, 1).
Hence, function h(x) has neither maximum nor minimum value in (−1, 1).
Find the local maxima and local minima, if any, of the following functions. Find also the local maximum and the local minimum values, as the case may be:
(i). f(x) = x2
(ii). g(x) = x3 − 3x
(iii). h(x) = sinx + cosx, 0 <
(iv). f(x) = sinx − cos x, 0 < x < 2π
(v). f(x) = x3 − 6x2+ 9x + 15
(vi). 
(vii). 
(viii). 
(i) f(x) = x2
Thus, x = 0 is the only critical point which could possibly be the point of local maxima or local minima of f.
We have
, which is positive.
Therefore, by second derivative test, x = 0 is a point of local minima and local minimum value of f at x = 0 is f(0) = 0.
(ii) g(x) = x3 − 3x
∴ g'(x) = 3x2 − 3
Now,
g'(x) = 0 ⇒ 3x2 = 3 ⇒ x = ±±1
g''(x) = 6x
g''(1) = 6 > 0
g''(−1) = −6 < 0
By second derivative test, x = 1 is a point of local minima and local minimum value of g at x = 1 is g(1) = 13 − 3 = 1 − 3 = −2. However,
x = −1 is a point of local maxima and local maximum value of g at
x = −1 is g(−1) = (−1)3 − 3 (− 1) = − 1 + 3 = 2.
(iii) h(x) = sinx + cosx, 0 < x <
Therefore, by second derivative test,
is a point of local maxima and the local maximum value of h at is 
(iv) f(x) = sin x − cos x, 0 < x < 2π
∴ f'(x) = cosx+sinx
f"(x)=0⇒cosx=−sinx⇒tanx=−1⇒
f''(x)=−sinx+cosx
f''
f''
Therefore, by second derivative test,
is a point of local maxima and the local maximum value of f at is 
However,
is a point of local minima and the local minimum value of f at is 
.
(v) f() =
3 − 6
2 + 9
+ 15
∴f'()=3
-12
+9
f'()=0=3(
-4
+3)=0
=3(-1)(
-3)=0
=1,3
Now f"()=6
-12=6(
-2)
f"(1)=6(1-2)=-6<0
f"(3)=6(3-2)=6>0
Therefore, by second derivative test, x = 1 is a point of local maxima and the local maximum value of f at x = 1 is f(1) = 1 − 6 + 9 + 15 = 19. However, x = 3 is a point of local minima and the local minimum value of f at x = 3 is f(3) = 27 − 54 + 27 + 15 = 15.
(vi) 
Since x > 0, we take x = 2.
Therefore, by second derivative test, x = 2 is a point of local minima and the local minimum value of g at x = 2 is g(2) =
(vii) 
Now, for values close to x = 0 and to the left of 0,
Also, for values close to x = 0 and to the right of 0,
.
Therefore, by first derivative test, x = 0 is a point of local maxima and the local maximum value of
.
(viii) 
Therefore, by second derivative test,
is a point of local maxima and the local maximum value of f at is
Prove that the following functions do not have maxima or minima:
(i) f(x) = ex (ii) g(x) = logx
(iii) h(x) = x3 + x2 + x + 1
We have,
(i) f(x) = ex
Now, if
. But, the exponential function can never assume 0 for any value of x.
Therefore, there does not exist c∈ R such that
Hence, function f does not have maxima or minima.
We have,
(ii) g(x) = log x
Therefore, there does not exist c∈ R such that
.
Hence, function g does not have maxima or minima.
We have,
(iii) h(x) = x3 + x2 + x + 1
Now,
h(x) = 0 ⇒ 3x2 + 2x + 1 = 0 ⇒ 
Therefore, there does not exist c∈ R such that
.
Hence, function h does not have maxima or minima.
Find the absolute maximum value and the absolute minimum value of the following functions in the given intervals:
(i) 
(ii) 
(iii) 
(iv) 
(i) The given function is f(x) = x3.
Then, we evaluate the value of f at critical point x = 0 and at end points of the interval [−2, 2].
f(0) = 0
f(−2) = (−2)3 = −8
f(2) = (2)3 = 8
Hence, we can conclude that the absolute maximum value of f on [−2, 2] is 8 occurring at x = 2. Also, the absolute minimum value of f on [−2, 2] is −8 occurring at x = −2.
(ii) The given function is f(x) = sin x + cos x.
Then, we evaluate the value of f at critical point
and at the end points of the interval [0, π].
Hence, we can conclude that the absolute maximum value of f on [0, π] is
occurring atand the absolute minimum value of f on [0, π] is −1 occurring at x = π.
(iii) The given function is
Then, we evaluate the value of f at critical point x = 4 and at the end points of the interval
.
Hence, we can conclude that the absolute maximum value of f onis 8 occurring at x = 4 and the absolute minimum value of f on is −10 occurring at x = −2.
(iv) The given function is
Now,
2(x − 1) = 0 ⇒ x = 1
Then, we evaluate the value of f at critical point x = 1 and at the end points of the interval [−3, 1].
Hence, we can conclude that the absolute maximum value of f on [−3, 1] is 19 occurring at x = −3 and the minimum value of f on [−3, 1] is 3 occurring at x = 1.
Find the maximum profit that a company can make, if the profit function is given by
p(x) = 41 − 72x − 18x2
The profit function is given as p(x) = 41 − 72x − 18x2.
∴ p'(x)=−72−36x
⇒x=−7236=−2
Also,
p''(−2)=−36<0
By second derivative test, x=−2x=-2 is the point of local maxima of p.
∴ Maximum profit=p(−2)= 41−72(−2)−18(−2)2=41+144−72=113
Hence, the maximum profit that the company can make is 113 units.
The solution given in the book has some error. The solution is created according to the question given in the book.
Find both the maximum value and the minimum value of
3x4 − 8x3 + 12x2 − 48x + 25 on the interval [0, 3]
Let f(x) = 3x4 − 8x3 + 12x2 − 48x + 25.
Now,
gives x = 2 or x2+ 2 = 0 for which there are no real roots.
Therefore, we consider only x = 2 ∈[0, 3].
Now, we evaluate the value of f at critical point x = 2 and at the end points of the interval [0, 3].
Hence, we can conclude that the absolute maximum value of f on [0, 3] is 25 occurring at x = 0 and the absolute minimum value of f at [0, 3] is − 39 occurring at x = 2.
What is the maximum value of the function sin x + cos x?
Let f(x) = sin x + cos x.
Now, 
will be negative when (sin x + cos x) is positive i.e., when sin x and cos x are both positive. Also, we know that sin x and cos x both are positive in the first quadrant. Then, will be negative when
.
Thus, we consider
.
∴By second derivative test, f will be the maximum atand the maximum value of f is
.
Find the maximum value of 2x3 − 24x + 107 in the interval [1, 3]. Find the maximum value of the same function in [−3, −1].
Let f(x) = 2x3 − 24x + 107.
We first consider the interval [1, 3].
Then, we evaluate the value of f at the critical point x = 2 ∈ [1, 3] and at the end points of the interval [1, 3].
f(2) = 2(8) − 24(2) + 107 = 16 − 48 + 107 = 75
f(1) = 2(1) − 24(1) + 107 = 2 − 24 + 107 = 85
f(3) = 2(27) − 24(3) + 107 = 54 − 72 + 107 = 89
Hence, the absolute maximum value of f(x) in the interval [1, 3] is 89 occurring at x = 3.
Next, we consider the interval [−3, −1].
Evaluate the value of f at the critical point x = −2 ∈ [−3, −1] and at the end points of the interval [1, 3].
f(−3) = 2 (−27) − 24(−3) + 107 = −54 + 72 + 107 = 125
f(−1) = 2(−1) − 24 (−1) + 107 = −2 + 24 + 107 = 129
f(−2) = 2(−8) − 24 (−2) + 107 = −16 + 48 + 107 = 139
Hence, the absolute maximum value of f(x) in the interval [−3, −1] is 139 occurring at x = −2.
It is given that at x = 1, the function x4− 62x2 + ax + 9 attains its maximum value, on the interval [0, 2]. Find the value of a.
Let f(x) = x4 − 62x2 + ax + 9.
It is given that function f attains its maximum value on the interval [0, 2] at x = 1.
Hence, the value of a is 120.
Find the maximum and minimum values of x + sin 2x on [0, 2π].
Let f(x) = x + sin 2x.
Then, we evaluate the value of f at critical points 
and at the end points of the interval [0, 2π].
Hence, we can conclude that the absolute maximum value of f(x) in the interval [0, 2π] is 2π occurring at x = 2π and the absolute minimum value of f(x) in the interval [0, 2π] is 0 occurring at x = 0.
Find two numbers whose sum is 24 and whose product is as large as possible.
Let one number be x. Then, the other number is (24 − x).
Let P(x) denote the product of the two numbers. Thus, we have:
∴By second derivative test, x = 12 is the point of local maxima of P. Hence, the product of the numbers is the maximum when the numbers are 12 and 24 − 12 = 12.
Find two positive numbers x and y such that x + y = 60 and xy3 is maximum.
The two numbers are x and y such that x + y = 60.
⇒ y = 60 − x
Let f(x) = xy3.
∴By second derivative test, x = 15 is a point of local maxima of f. Thus, function xy3 is maximum when x = 15 and y = 60 − 15 = 45.
Hence, the required numbers are 15 and 45.
Find two positive numbers whose sum is 16 and the sum of whose cubes is minimum.
Let one number be x. Then, the other number is (16 − x).
Let the sum of the cubes of these numbers be denoted by S(x). Then,
Now, 
∴ By second derivative test, x = 8 is the point of local minima of S.
Hence, the sum of the cubes of the numbers is the minimum when the numbers are 8 and 16 − 8 = 8.
A square piece of tin of side 18 cm is to made into a box without top, by cutting a square from each corner and folding up the flaps to form the box. What should be the side of the square to be cut off so that the volume of the box is the maximum possible?
Let the side of the square to be cut off be x cm. Then, the length and the breadth of the box will be (18 − 2x) cm each and the height of the box is x cm.
Therefore, the volume V(x) of the box is given by,
V(x) = x(18 − 2x)2
x = 9 or x = 3
If x = 9, then the length and the breadth will become 0.
x ≠ 9.
x = 3.
Now, 
By second derivative test, x = 3 is the point of maxima of V.
Hence, if we remove a square of side 3 cm from each corner of the square tin and make a box from the remaining sheet, then the volume of the box obtained is the largest possible.
A rectangular sheet of tin 45 cm by 24 cm is to be made into a box without top, by cutting off square from each corner and folding up the flaps. What should be the side of the square to be cut off so that the volume of the box is the maximum possible?
Let the side of the square to be cut off be x cm. Then, the height of the box is x, the length is 45 − 2x, and the breadth is 24 − 2x.
Therefore, the volume V(x) of the box is given by,
Now,
x = 18 and x = 5
It is not possible to cut off a square of side 18 cm from each corner of the rectangular sheet. Thus, x cannot be equal to 18.
∴x = 5
Now,
By second derivative test, x = 5 is the point of maxima.
Hence, the side of the square to be cut off to make the volume of the box maximum possible is 5 cm.
Show that of all the rectangles inscribed in a given fixed circle, the square has the maximum area.
Let a rectangle of length l and breadth b be inscribed in the given circle of radius a.
Then, the diagonal passes through the centre and is of length 2a cm.
Now, by applying the Pythagoras theorem, we have:
∴Area of the rectangle, 
By the second derivative test, when
, then the area of the rectangle is the maximum.
Since
, the rectangle is a square.
Hence, it has been proved that of all the rectangles inscribed in the given fixed circle, the square has the maximum area.
Show that the right circular cylinder of given surface and maximum volume is such that is heights is equal to the diameter of the base.
Let r and h be the radius and height of the cylinder respectively.
Then, the surface area (S) of the cylinder is given by,
Let V be the volume of the cylinder. Then,
∴ By second derivative test, the volume is the maximum when
.
Hence, the volume is the maximum when the height is twice the radius i.e., when the height is equal to the diameter.
Of all the closed cylindrical cans (right circular), of a given volume of 100 cubic centimetres, find the dimensions of the can which has the minimum surface area?
Let r and h be the radius and height of the cylinder respectively.
Then, volume (V) of the cylinder is given by,
Surface area (S) of the cylinder is given by,
∴By second derivative test, the surface area is the minimum when the radius of the cylinder is
.
Hence, the required dimensions of the can which has the minimum surface area is given by radius = 
and height = 
A wire of length 28 m is to be cut into two pieces. One of the pieces is to be made into a square and the other into a circle. What should be the length of the two pieces so that the combined area of the square and the circle is minimum?
Let a piece of length l be cut from the given wire to make a square.
Then, the other piece of wire to be made into a circle is of length (28 − l) m.
Now, side of square =
.
Let r be the radius of the circle. Then,
The combined areas of the square and the circle (A) is given by,
Thus, when
By second derivative test, the area (A) is the minimum when
.
Hence, the combined area is the minimum when the length of the wire in making the square is
cm while the length of the wire in making the circle is
.
Prove that the volume of the largest cone that can be inscribed in a sphere of radius R is 
of the volume of the sphere.
Let r and h be the radius and height of the cone respectively inscribed in a sphere of radius R.
Let V be the volume of the cone.
Then, 
Height of the cone is given by,
h = R + AB
∴ By second derivative test, the volume of the cone is the maximum when
Show that the right circular cone of least curved surface and given volume has an altitude equal to 
time the radius of the base.
Let r and h be the radius and the height (altitude) of the cone respectively.
Then, the volume (V) of the cone is given as:
V=13πr2h⇒h=3Vπr2V=13πr2h⇒h=3Vπr2
The surface area (S) of the cone is given by,
S = πrl (where l is the slant height)
Thus, it can be easily verified that when
∴ By second derivative test, the surface area of the cone is the least when
Hence, for a given volume, the right circular cone of the least curved surface has an altitude equal to times the radius of the base.
Show that the semi-vertical angle of the cone of the maximum volume and of given slant height is
.
Let θ be the semi-vertical angle of the cone.
It is clear that 
Let r, h, and l be the radius, height, and the slant height of the cone respectively.
The slant height of the cone is given as constant.
Now, r = l sin θ and h = l cos θ
The volume (V) of the cone is given by,
∴By second derivative test, the volume (V) is the maximum when
.
Hence, for a given slant height, the semi-vertical angle of the cone of the maximum volume is.
Show that semi-vertical angle of right circular cone of given surface area and maximum volume is 
.
Let r be the radius, l be the slant height and h be the height of the cone of given surface area, S.
Also, let α be the semi-vertical angle of the cone.
Then S = πrl + πr2
Let V be the volume of the cone.
Differentiating (2) with respect to r, we get
For maximum or minimum, put 
Differentiating again with respect to r, we get
Thus, V is maximum when S= 4πr2
As S = πrl + πr2
⇒ 4πr2 = πrl + πr2
⇒ 3πr2 = πrl
⇒ l = 3r
Now, in ΔCOB,
The maximum value of 
is
(A) 
(B) 
(C) 1 (D) 0
Let
Then, we evaluate the value of f at critical point
and at the end points of the interval [0, 1] {i.e., at x = 0 and x = 1}.
Hence, we can conclude that the maximum value of f in the interval [0, 1] is 1.
The correct answer is C.
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