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

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

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)

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

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

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

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
...

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

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

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

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...

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...

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

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

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

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

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

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

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 ...

- Previous
- Next

Exams

Articles

Questions