29) The maximum value of

Given function is
Hence, x = 1/2 is the critical point s0 we need to check the value at x = 1/2 and at the end points of given range i.e. at x = 1 and x = 0
Hence, by this we can say that maximum value of given function is 1 at x = 0 and x = 1
option c is correct

28) For all real values of x, the minimum value of

is

(A) 0 (B) 1 (C) 3 (D) 1/3

Given function is
Hence, x = 1 and x = -1 are the critical points
Now,
Hence, x = 1 is the point of minima and the minimum value is
Hence, x = -1 is the point of maxima
Hence, the minimum value of
is
Hence, (D) is the correct answer

27) The point on the curve which is nearest to the point (0, 5) is

Given curve is
Let the points on curve be
Distance between two points is given by
Hence, x = 0 is the point of maxima
Hence, the point is the point of minima
Hence, the point is the point on the curve which is nearest to the point (0, 5)
Hence, the correct answer is (A)

26) Show that semi-vertical angle of the right circular cone of given surface area and

maximum volume is

Let r, l, and h are the radius, slant height and height of cone respectively
Now,
Now,
we know that
The surface area of the cone (A) =
Now,
Volume of cone(V) =
On differentiate it w.r.t to a and after that
we will get
Now, at
Hence, we can say that is the point if maxima
Hence proved

25) Show that the semi-vertical angle of the cone of the maximum volume and of given slant height is

Let a be the semi-vertical angle of cone
Let r , h , l are the radius , height , slent height of cone
Now,
we know that
Volume of cone (V) =
Now,
Now,
Now, at
Therefore, is the point of maxima
Hence proved

24) Show that the right circular cone of least curved surface and given volume has an altitude equal to time the radius of the base.

Volume of cone(V)
curved surface area(A) =
Now , we can clearly varify that
when
Hence, is the point of minima
Hence proved that the right circular cone of least curved surface and given volume has an altitude equal to time the radius of the base

23) Prove that the volume of the largest cone that can be inscribed in a sphere of radius r is 8/27 of the volume of the sphere.

Volume of cone (V) =
Volume of sphere with radius r =
By pythagoras theorem in we ca say that
V =
Now,
Hence, point is the point of maxima
Hence, the altitude of the right circular cone of maximum volume that can be inscribed in a sphere of radius r is
Volume =
Hence proved

22) 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?

Area of the square (A) =
Area of the circle(S) =
Given the length of wire = 28 m
Let the length of one of the piece is x m
Then the length of the other piece is (28 - x) m
Now,
and
Area of the combined circle and square = A + S
Now,
Hence, is the point of minima
Other length is = 28 - x
...

21) 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 be the radius of base and h be the height of the cylinder
The volume of the cube (V) =
It is given that the volume of cylinder = 100
Surface area of cube(A) =
Hence, is the critical point
Hence, is the point of minima
Hence, and are the dimensions of the can which has the minimum surface area

20. Show that the right circular cylinder of given surface and maximum volume is such that its height is equal to the diameter of the base.

Let r be the radius of the base of cylinder and h be the height of the cylinder
we know that the surface area of the cylinder
Volume of cylinder
Hence, is the critical point
Now,
Hence, is the point of maxima
Hence, the right circular cylinder of given surface and maximum volume is such that its height is equal to the diameter(D = 2r) of the base

19) Show that of all the rectangles inscribed in a given fixed circle, the square has the maximum area.

Let assume that length and breadth of rectangle inscribed in a circle is l and b respectively
and the radius of the circle is r
Now, by Pythagoras theorem
a = 2r
Now, area of reactangle(A) = l b
Now,
Hence, is the point of maxima
Since, l = b we can say that the given rectangle is a square
Hence, of all the rectangles inscribed in a given fixed circle, the square has the maximum area

18) 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 maximum ?

It is given that the sides of the rectangle are 45 cm and 24 cm
Let assume the side of the square to be cut off is x cm
Then,
Volume of cube
But x cannot be equal to 18 because then side (24 - 2x) become negative which is not possible so we neglect value x= 18
Hence, x = 5 is the critical value
Now,
Hence, x = 5 is the point of maxima
Hence, the side of the...

17. A square piece of tin of side 18 cm is to be 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.

It is given that the side of the square is 18 cm
Let assume that the length of the side of the square to be cut off is x cm
So, by this, we can say that the breath of cube is (18-2x) cm and height is x cm
Then,
Volume of cube =
But the value of x can not be 9 because then the value of breath become 0 so we neglect value x = 9
Hence, x = 3 is the critical point
Now,
Hence, x = 3...

16. Find two positive numbers whose sum is 16 and the sum of whose cubes is

minimum.

let x an d y are positive two numbers
It is given that
x + y = 16 , y = 16 - x
and is minimum
Now,
Hence, x = 8 is the only critical point
Now,
Hence, x = 8 is the point of minima
y = 16 - x
= 16 - 8 = 8
Hence, values of x and y are 8 and 8 respectively

15) Find two positive numbers x and y such that their sum is 35 and the product

is a maximum.

It is given that
x + y = 35 , x = 35 - y
and is maximum
Therefore,
Now,
Now,
Hence, y = 35 is the point of minima
Hence, y= 0 is neither point of maxima or minima
Hence, y = 25 is the point of maxima
x = 35 - y
= 35 - 25 = 10
Hence, the value of x and y are 10 and 25 respectively

14) Find two positive numbers x and y such that x + y = 60 and is maximum.

It is given that
x + y = 60 , x = 60 -y
and is maximum
let
Now,
Now,
hence, 0 is niether point of minima or maxima
Hence, y = 45 is point of maxima
x = 60 - y
= 60 - 45 = 15
Hence, values of x and y are 15 and 45 respectively

14. Find two numbers whose sum is 24 and whose product is as large as possible.

Let x and y are two numbers
It is given that
x + y = 24 , y = 24 - x
and product of xy is maximum
let
Hence, x = 12 is the only critical value
Now,
at x= 12
Hence, x = 12 is the point of maxima
Noe, y = 24 - x
= 24 - 12 = 12
Hence, the value of x and y are 12 and 12 respectively

12. Find the maximum and minimum values of

Given function is
So, values of x are
These are the critical points of the function
Now, we need to find the value of the function at and at the end points of given range i.e. at x = 0 and
Hence, at function attains its maximum value and value is in the given range of
and at x= 0 function attains its minimum value and value is 0

11. It is given that at x = 1, the function attains its maximum value,

on the interval [0, 2]. Find the value of a.

Given function is
Function attains maximum value at x = 1 then x must one of the critical point of the given function that means
Now,
Hence, the value of a is 120

10. Find the maximum value of in the interval [1, 3]. Find the

the maximum value of the same function in [–3, –1].

Given function is
we neglect the value x =- 2 because
Hence, x = 2 is the only critical value of function
Now, we need to check the value at x = 2 and at the end points of given range i.e. x = 1 and x = 3
Hence, maximum value of function occurs at x = 3 and vale is 89 when
Now, when
we neglect the value x = 2
Hence, x = -2 is the only critical value ...

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