# NCERT Solutions class 12 Maths Exercise 6.5

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## NCERT Solutions for Class 12 Maths Application of Derivatives

1. Find the maximum and minimum values, if any, of the following functions given by:

(i)

(ii)

(iii)

(iv)

Ans. (i) Given:

Since  for all  R

Therefore, the minimum value of  is 3 when , i.e.,

This function does not have a maximum value.

(ii) Given:

=

……….(i)

Since  for all  R

Subtracting 2 from both sides,

Therefore, minimum value of  is  and is obtained when , i.e.,

And this function does not have a maximum value.

(iii) Given:    ……….(i)

Since  for all  R

Multiplying both sides by  and adding 10 both sides,

[Using eq. (i)]

Therefore, maximum value of  is 10 which is obtained when  i.e.,

And therefore, minimum value of  does not exist.

(iv) Given:

As

As

Therefore, maximum value and minimum value of  do not exist.

### 2. Find the maximum and minimum values, if any, of the following functions given by:

(i)

(ii)

(iii)

(iv)

(v)

Ans. (i) Given:         ……….(i)

Since  for all  R

Subtracting 1 from both sides,

Therefore, minimum value of  is  which is obtained when  i.e.,

From eq. (i), maximum value of  hence it does not exist.

(ii) Given:

Since  for all  R

Multiplying by  both sides and adding 3 both sides,

Therefore, maximum value of  is  which is obtained when  i.e.,

From eq. (i), minimum value of  hence it does not exist.

(iii) Given:         ……….(i)

Since   for all  R

Therefore, minimum value of  is 4 and maximum value is 6.

(iv) Given:

Since   for all  R

Therefore, minimum value of  is 2 and maximum value is 4.

(v) Given:        ……….(i)

Since

Therefore, neither minimum value not maximum value of  exists.

### 3. 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)

(ii)

(iii)

(iv)

(v)

(vi)

(vii)

(viii)

Ans. (i) Given:

and

Now

[Turning point]

Again, when ,    [Positive]

Therefore,  is a point of local minima and local minimum value =

(ii) Given:

and

Now

or  [Turning points]

Again, when ,

[Negative]

is a point of local maxima and local maximum value

And when    [Positive]

is a point of local minima and local minimum value

(iii) Given:         ……….(i)

and

Now

[Positive]

can have values in both I and III quadrant.

But,  therefore,  is only in I quadrant.

=  [Turning point]

At

=

=  =  =   [Negative]

is a point of local maxima and local maximum value

=

=

(iv) Given:         ……….(i)

and

Now

[Negative]

can have values in both II and IV quadrant.

=

=  or

or

and    [Turning point]

At   =

=

=

=

=  =   [Negative]

is a point of local maxima and local maximum value =

=

=  =

At

=

=

=

=

=  =      [Positive]

is a point of local maxima and local maximum value =

=

=  =

(v) Given:

and

Now

or    [Turning points]

At   [Negative]

is a point of local maxima and local maximum value is

At     [Positive]

is a point of local minima and local minimum value is

(vi) Given:

=

=   and

Now

= 0

or

But  therefore  is only the turning point.

is a point of local minima and local minimum value is

(vii) Given:

and

=

Now

[Turning point]

At    [Negative]

is a point of local maxima and local maximum value is

(viii) Given:

=

=  =

And

=

=

Now

= 0

is a point of local maxima and local maximum value is

Again

=

has local maximum value at

### 4. Prove that the following functions do not have maxima or minima:

(i)

(ii)

(iii)

Ans. (i) Given:

Now

But this gives no real value of  Therefore, there is no turning point.

does not have maxima or minima.

(ii) Given:

Now

1 = 0

#### But this is not possible. Therefore, there is no turning point.

does not have maxima or minima.

(iii) Given:

Now

=

=

Here, values of  are imaginary.

does not have maxima or minima.

### 5. Find the absolute maximum value and the absolute minimum value of the following functions in the given intervals:

(i)

(ii)

(iii)

(iv)

Ans. (i) Given:

Now

At

At

At

Therefore, absolute minimum value of  is  and absolute maximum value is 8.

(ii) Given:

Now

[Positive]

is in I quadrant. [ ]

Therefore, absolute minimum value is  and absolute maximum value is 1.

(iii) Given:

Now

At

At

At

Therefore, absolute minimum value is  and absolute maximum value is 8.

(iv) Given:

Now

At

At

Therefore, absolute minimum value is  and absolute maximum value is 19.

### 6.  Find the maximum profit that a company can make, if the profit function is given by

Ans. Given: Profit function

and

Now

At ,  [Negative]

has a local maximum value at .

At , Maximum profit

=

= 41 + 16 – 8 = 49

7. Find both the maximum value and minimum value of  on the interval [0, 3].

Ans. Let  on [0, 3]

Now

or

Since  is imaginary, therefore it is rejected.

is turning point.

At  =

At

At

Therefore, absolute minimum value is  and absolute maximum value is 25.

8. At what points on the interval  does the function  attain its maximum value?

Ans. Let

Now

Putting

Now

=

=  =

Putting

Also   and

Since  attains its maximum value 1 at  and

Therefore, the required points are  and

### 9. What is the maximum value of the function

Ans. Let

Now

=

[Turning point]

=

=

=

=

If  is even, then

If  is odd, then

Therefore, maximum value of  is  and minimum value of  is

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

Ans. Let

Now

or   [Turning points]

For Interval [1, 3],  is turning point.

At

At

At

Therefore, maximum value of  is 89.

For Interval    is turning point.

At

At

At

Therefore, maximum value of  is 139.

### 11. It is given that at  the function  attains its maximum value, on the interval [0, 20]. Find the value of

Ans. Let

Since,  attains its maximum value at  in the interval [0, 2], therefore

### 12. Find the maximum and minimum value of  on

Ans. Let

Now

=  =

=

where  Z

For   But  , therefore

For   =  and

For

But , therefore

Therefore, it is clear that the only turning point of  given by  which belong to given closed interval  are,

At

nearly

At

nearly

At

nearly

At

nearly

At

At

nearly

Therefore, Maximum value =  and minimum value = 0

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

Ans. Let the two numbers be  and

According to the question,

…….(i)

And let  is the product of  and

[From eq. (i)]

and

Now to find turning point,

At ,  [Negative]

is a point of local maxima and  is maximum at .

From eq. (i),

Therefore, the two required numbers are 12 and 12.

### 14. Find two positive integers  and  such that  and  is maximum.

Ans. Given:         ……….(i)

Let  P =   [To be maximized]      ……….(ii)

Putting from eq. (i),  in eq. (ii),

P =

…..(iii)

Now

It is clear that  changes sign from positive to negative as  increases through 45.

Therefore, P is maximum when

Hence,  is maximum when  and

### 15. Find two positive integers  and  such that their sum is 35 and the product  is a maximum.

Ans. Given:

……….(i)

Let

[From eq. (i)]

……….(ii)

Now

or  or

or  or

Now  is rejected because according to question,  is a positive number.

Also  is rejected because from eq. (i),  = 35 – 35 = 0, but  is positive.

Therefore,  is only the turning point.

At ,

=

By second derivative test,  will be maximum at  when .

Therefore, the required numbers are 10 and 25.

### 16. Find two positive integers whose sum is 16 and sum of whose cubes is minimum.

Ans. Let the two positive numbers are  and

……….(i)

Let

[From eq. (i)]

=

and

Now  = 0

At    is positive.

is a point of local minima and  is minimum when .

Therefore, the required numbers are 8 and 8.

### 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?

Ans. Given: Each side of square piece of tin is 18 cm.

Let  cm be the side of each of the four squares cut off from each corner.

Then dimensions of the open box formed by folding the flaps after cutting off squares are  and  cm.

Let  denotes the volume of the open box.

=

and

Now  = 0

or

is rejected because at  length =  which is impossible.

is the turning point.

At  [Negative]

·  is minimum at  i.e., side of each square to be cut off from each corner for maximum volume is 3 cm.

### 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?

Ans. Given: Dimensions of rectangular sheet are 45 cm and 24 cm.

Let  cm be the side of each of the four squares cut off from each corner.

Then dimensions of the open box formed by folding the flaps after cutting off squares are  and  cm.

Let  denotes the volume of the open box.

=

and

Now  = 0

or

is rejected because at  length =  which is impossible.

is the turning point.

At  [Negative]

·  is minimum at  i.e., side of each square to be cut off from each corner for maximum volume is 5 cm.

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

Ans. Let PQRS be the rectangle inscribed in a given circle with centre O and radius

Let  and  be the length and breadth of the rectangle, i.e.,  and

In right angled triangle PQR, using Pythagoras theorem,

PQ2 + QR2 = PR2

…..(i)

Let A be the area of the rectangle, then A =  =

=

And

=

=

Now

= 0

At ,   [Negative]

At , area of rectangle is maximum.

And from eq. (i), ,

i.e.,

Therefore, the area of inscribed rectangle is maximum when it is square.

### 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.

Ans. Let  be the radius of the circular base and  be the height of closed right circular cylinder.

Total surface area (S) =

=  (say)

…..(i)

Volume of cylinder

=  [From eq. (i)]

and

Now

At   [Negative]

is maximum at .

From eq. (i),

=  =

Height = Diameter

Therefore, the volume of cylinder is maximum when its height is equal to the diameter of its base.

### 21. Of all the closed cylindrical cans (right circular), of a given volume of 100 cubic centimeters, find the dimensions of the can which has the minimum surface area.

Ans. Let  be the radius of the circular base and  be the height of closed right circular cylinder.

According to the question, Volume of the cylinder

…(i)

Total surface area (S) =

=

=   [From eq. (i)]

S =

=

and

Now

At

=  =    [Positive]

S is minimum when radius  cm

From eq. (i)

=  =

### 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?

Ans. Let  meters be the side of square and  meters be the radius of the circle.

Length of the wire = Perimeter of square + Circumference of circle

= 28

……….(i)

Area of square =  and Area of circle =

Combined area (A) =  +  =  +

=

and

Now

And   [Positive]

A is minimum when

Therefore, the wire should be cut at a distance  from one end.

### 23. 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.

Ans. Let O be the centre and R be the radius of the given sphere, BM =  and AM =

In right angled triangle OMB, using Pythagoras theorem,

OM2 + BM2 = OB2

……….(i)

Volume of a cone inscribed in the given sphere

=

……….(ii)

and

Now

At

=

=   [Negative]

is maximum at

From eq. (i)

=

Maximum volume of the cone

=

=  (Volume of the sphere)

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

Ans. Let  be the radius and  be the height of the cone.

Volume of the cone (V) =

(say) ……….(i)

And Surface area of the cone (S) =

(say) ….(ii)

=

=

and

Now

……….(iii)

At   [Positive]

is minimum when

From eq. (i),

=  [From eq. (iii)]

Therefore, Surface area is minimum when height =  (radius of base)

### 25. Show that the semi-vertical angle of the cone of the maximum value and of given slant height is

Ans. Let  be the radius,  be the height,  be the slant height of given cone and  be the semi-vertical angle of cone.

……….(i)

Volume of the cone (V) =        ……….(ii)

V =

=

and

Now

At

=    [Negative]

V is maximum at

From eq. (i),

Semi-vertical angle,

=

### 26. Show that the semi-vertical angle of the right circular cone of given surface area and maximum volume is

Ans. Let  be the radius and  be the height of the cone and semi-vertical angle be

Total Surface area of cone (S) =

(say)

……….(i)

Volume of cone (V) =

=

=

=   [Using quotient rule]

……….(ii)

Now

[height can’t be negative]

is the turning point.

Since, , therefore, Volume is maximum at

From eq. (i),

Now Semi-vertical angle of the cone =

Choose the correct answer in the Exercises 27 to 29.

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

(A)

(B)

(C) (0, 0)

(D) (2, 2)

Ans. Equation of the curve is                                                                                 ……….(i)

Let P be any point on the curve (i), then according to question,

Distance between given point (0, 5) and P =   (say)

=             [From eq. (i)]

= Z (say)

and

Now

At

[Positive]

Z is minimum and  is minimum at

From eq. (i)

and  are two points on curve (i) which are nearest to (0, 5).

Therefore, option (A) is correct.

28. For all real values of  the minimum value of  is:

(A) 0

(B) 1

(C) 3

(D)

Ans. Given:  ……….(i)

=

Now

= 0

and  [Turning points]

At ,

from eq. (i),

At ,

from eq. (i),

[Minimum value]

Therefore, option (D) is correct.

29. The maximum value of  is:

(A)

(B)

(C) 1

(D)

Ans. Let

= ,      ……….(i)

=

Now

= 0

[Turning point] and it belongs to the given enclosed interval  i.e., [0, 1].

At , from eq. (i),

At   from eq. (i),

At ,  from eq. (i),

Maximum value of  is 1.

Therefore, option (C) is correct.

## NCERT Solutions class 12 Maths Exercise 6.5

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### 1 thought on “NCERT Solutions class 12 Maths Exercise 6.5”

1. In solution 17th, it should be maximum not minimum.