# EXERCISE 6.5( Application of derivative)

Question 1: Find the maximum and minimum values, if any, of the following functions given by:(class 12 maths ex 6.5 ncert solution)

(i)

(ii)

(iii)

(iv)

Solution: (i) Given function is:

We know that

for all

Adding 3 both sides, we get

The minimum value of is 3

when ,

this function does not have a maximum value.

(ii) Given function is:

As

Subtracting 2 from both sides,

Therefore, minimum value of is and is obtained when

And this function does not have a maximum value.

(iii) Given function is:

As for all

Multiplying both sides by and adding 10 both sides,

Maximum value of is 10 which is obtained when

.

And minimum value of does not exist.

(iv) Given function is:

At

At

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

Question :2)Â Find the maximum and minimum values, if any, of the following functions given by:

(i)

(ii)

(iii)

(iv)

(v)

Solution: (i) Given function is:

We know that

for all

Subtracting 1 from both sides,

Therefore, minimum value of is

when .
.
From equation (1), maximum value of hence it does not exist.

(ii) Given function is:

We know that

for all

Multiplying by both sides and adding 3 both sides,

Maximum value of is 3

when .

From equation (1), minimum value of , does not exist.

(iii) Given function is:

We know that

for all

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

(iv) Given function is:

We know that

for all

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

(v) Given function is:

As

Therefore, neither minimum value not maximum value of exists.

Question: 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)

Solution: (i) Given function is:

and

Now

[Turning point]

Again, when [Positive]

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

(ii) Given function is:

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 function is:

Now

can have values in both 1st and 3rd quadrant.

But,

therefore, is only in I quadrant.

As,

is a point of local maxima and local maxima value

(iv) Given function is:

and

Now

can have values in both 2 nd and 4 th quadrant.

,

So, is a point of local maxima and local maximum value

is a point of local maxima and local maximum value

(v) Given function is:

and

Now .

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 function is:

and

But , therefore is only the turning point.

is a point of local minima and local minimum value is

(vii) Given function is:

and

Now

[Negative]

is a point of local maxima and local maximum value is

(viii) Given function is:

And

Now

is a point of local maxima and local maximum value is

Again

Therefore, has local maximum value at .

Question:4) Prove that the following functions do not have maxima or minima:

(i)

(ii)

(iii)

Solution: (i) Given function is:

Now

But this gives no real value of . Therefore, there is no turning point. does not have maxima or minima.

(ii) Given function is:

(iii) Given function is:

Here, values of are imaginary.

does not have maxima or minima.

Question:5) Find the absolute maximum value and the absolute minimum value of the following functions in the given intervals:

(i)

(ii)

(iii)

(iv)

Solution: (i) Given function is:

Now

,

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

(ii) Given function is:

Now

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

(iii) Given function is:

Now

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

(iv) Given function is:

Now

Therefore, absolute minimum value is 3 and absolute maximum value is 19 .

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

Solution: Profit function

Now

has a local maximum value at .

Question:7) Find both the maximum value and minimum value of on the interval .

Solution: on [0, 3]

Now

As is imaginary, therefore it is rejected.

here is turning point.

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

Question:8) At what points on the interval does the function attain its maximum value?

Solution: Consider

Now

and

Therefore, the required points are and .

Question:9Â  What is the maximum value of the function ?

Solution: Consider

—(i)

Now

Again differentiate with respect to x of (i)

(i) At

f(x) is maximum at and maximum value
is

(ii) At

f(x) is minimum at and minimum value
is

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

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

Solution: Consider

Now

or [Turning points]

For Interval is turning point.

At

At

At

Therefore, maximum value of is 89 .

For Interval is turning point.

Therefore, maximum value of is 139 .

Question:11) It is given that at , the function attains its maximum value, on the interval . Find the value of .

Solution: Consider

As, attains its maximum value at in the interval , therefore

Question:12) Find the maximum and minimum value of on .

Solution: Consider

At

At

At

At

At

At

Therfore max value and minimum value

Question:13) Find two numbers whose sum is 24 and whose product is as large as possible.

Solution: Let two numbers be = x and y

Let Z is the product of two numbers

And

Now for max and minima

is a point of local maxima

From equation (i),

Therefore, the two required numbers are 12 and 12.

Question:14) Find two positive integers and such that and is maximum.

Solution: Given function is:

Let [To be maximized]

Putting from equation (i), in equation (ii),

And

At

Therefore, is maximum when .
Hence, is maximum when and .

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

Solution: Given function is:

Let

From equation (i)

Now

or or

or or

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

Also is rejected because from equation (i), , but is positive.

Therefore, is only the turning point.

By second derivative test, will be maximum at when .

Therefore, the required numbers are 10 and 25.

Question:16) Find two positive integers whose sum is 16 and sum of whose cubes is minimum.

Solution: Consider the two positive numbers are and .

Let

[From equation (i)]

Differentiate with respect to x

And

Now

At is positive.

is a point of local minima

Therefore, the required numbers are 8 and 8.

Question:17) A square piece of tin of side 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?

Solution: Each side of square piece of tin is .

Let 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 .

Let denotes the volume of the open box.

Differentiating with respect to x

is not possible

At

Volume of the box is maximum when . Hence side of the square cut off from each corner

Question:18) A rectangula sheet of tin 45 cm by 24 cm is to be made into a box without top, by cutting of square from each corner and folding up the flaps. What should be the side of square to be cutt of so that the volume of the box is maximum?

Solution: Length and breadth of rectangular sheets 45 cm and 24 cm respectively.

Let side of each of the four squares cut off from each corner = x cm

Let Z denote the volume of the box

Now differentiating with respect to x

and

is rejected because at length which is impossible.

Here is the turning point.

is maximum at that is, side of each square to be cut off from each corner for maximum volume is .

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

Solution: LetPQRS be the rectangle inscribed in a given circle with centre and radius a.

Let and be the length and breadth of the rectangle, that is, and
In right angled triangle , using Pythagoras theorem,

–(i)

Let be the area of the rectangle,

then

Squaring both side

Differentiating with respect to x

—–(ii)

For max and minima

or

or

Again differentiate with respect to x of (ii)

At

Hence the area of the square is max when

From eq (i)

Hence area of inscribed rectangle is maximum when it is a square

Question: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.

Solution:Â  Let be the radius of the circular base and be the height of closed right circular cylinder.

Formula for Total surface area

–(i)

Volume of cylinder

Differentiating with respect r

And

For max and minima

.

Since

Hence volume of cylinder is max when

Value of S putting in equation (i)

Height of cylinder = Diametre of cylinder

Hence volume of cylindr is max when

Question:21) Of all the close cylindrical cans(right circular), of a given volume of 100 cubic centimeters, find the dimension of the can which has the minimum surface areas.

Solution : Let r be the radius and h be the height of the circular cylinder

Volume of the cylinder

— (i)

Total surface area

Differentiate with respect to r

—(ii)

For max and minima

Again differentiate with respect to r of eq (ii)

At

Total suface area is minimum when

Hence from eq (i)

Question:22) A wire of length 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?

Solution: 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

Therefore, the length of one piece

Lenghth of second piece

Question:23) Prove that the volume of the largest cone that can be inscribed in a sphere of radius is of the volume of the sphere.

Solution: Consider be the centre and be the radius of the given sphere,

In Î”OBM, using pythagoras theorem

BM is the radius of cone

Volume of cone

Differentiating with respect to x

—(ii)

For max and minima

Hence or

or

Not possible

Agian differentiate with respect to x of eq (ii)

At

Hence volume of cone is max when

Now From eq (i)

Volume of cone volume of sphere

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

Solution: Let r be the radius of cone and h be the height of cone

Volume of cone

–(i)

Curve surface area of cone

Squaring both side

Differentiating with respect to r

For max and minima

Again differentiate with respect to r of (ii)

AT

Hence curve surface area is minimum at

putting in (i)

therefore height of the cone is times the radius of base

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

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

Formula for Volume of the cone

and

Which implies,

Question:26) Show that the semi-vertical angle of the right circular cone of given surface area and maximum volume is

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

Since, slant height

And, Total Surface area of cone (S)

Volume of cone

Squaring both side

Differentiating with respect to r

And

For max and minima

At

Hence Volume of cone is max when

Putting in eq (i)

In

Choose the correct answer in the Exercises 27 to 29.

Question:27) The point on the curve which is nearest to the point is:

(A)

(B)

(C)

(D)

Solution: Option (A) is correct.

Equation of the curve is
Let be any point on the curve (1), then according to question,

Distance between given point and (say)

is minimum and is minimum at

From equation (1), we have

and are two points on curve (1) which are nearest to .

Question:28) For all real values of , the minimum value of is:

(A) 0

(B) 1

(C) 3

(D)

Solution: Option (D) is correct.

Given function is:

Question:29) The maximum value of is:

(A)

(B)

(C) 1

(D)

Solution: Option (C) is correct.

Now

Here is a turning point and it belongs to the given enclosed interval that is, . At , from equation (i),

At , from equation (i),

At , from equation (i),

Maximum value of is 1.

class 12 maths ex 6.3 ncert solution