NCERT Solutions Class 12 maths Chapter-6 (Applications of Derivatives)Exercise 6.5

NCERT Solutions Class 12 Maths from class 12th Students will get the answers of Chapter-6 (Applications of Derivatives)Exercise 6.5 This chapter will help you to learn the basics and you should expect at least one question in your exam from this chapter.
We have given the answers of all the questions of NCERT Board Mathematics Textbook in very easy language, which will be very easy for the students to understand and remember so that you can pass with good marks in your examination.

Exercise 6.5

Set 1

Question 1. Find the maximum and minimum values, if any, of the following function given by

(i) f(x) = (2x – 1)² + 3                    

Solution:

Given that, f(x) = (2x – 1)² + 3

From the given function we observe that

(2x – 1)² ≥ 0 ∀ x∈ R,

So, 

(2x – 1)² + 3 ≥ 3 ∀ x∈ R,

Now we find the minimum value of function f when 2x-1 = 0

So, x = 1/2

f = f(1/2) = (2(1/2) – 1)² + 3 = 3

Hence, the minimum value of the function is 3 and this function does not contain maximum value.

  (ii) f(x) = 9x² + 12x + 2

Solution:

Given that, f(x) = 9x² + 12x + 2

we can also write as f(x) = (3x + 2)² – 2

From the given function we observe that

(3x + 2)² ≥ 0 ∀ x∈ R,

So, 

(3x + 2)² – 2 ≥ 2 ∀ x∈ R,

Now we find the minimum value of function f when 3x + 2 = 0

So, x = -2/3

f = f(-2/3) = (3(-2/3) + 2)² – 2 = -2

Hence, the minimum value of the function is -2 and this function does not contain maximum value.

(iii) f(x) = -(x – 1)² + 10  

Solution:

Given that, f(x) = -(x – 1)² + 10  

From the given function we observe that

(x – 1)² ≥ 0 ∀ x∈ R,

So, 

-(x – 1)² + 10 ≤ 10 ∀ x∈ R,

Now we find the maximum value of function f when x – 1 = 0

So, x = 1

f = f(1) = -(1 – 1)² +10 = 10

Hence, the maximum value of the function is 10 and this function does not contain minimum value.

(iv) g(x) = x³ + 1

Solution:

Given that, g(x) = x³ + 1

When x —> ∞, then g(x) —> ∞

When x —> -∞, then g(x) —> -∞

So, this function has neither minimum nor maximum value

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

(i) f(x) = |x + 2| – 1

Solution:

Given that, f(x) = |x + 2| – 1

At x = -2 f'(x) change sign from negative to positive, hence by first derivative test, x = -2

is a point of local minima. 

So, the minimum value f = f(-1)= |(-1)+ 2| – 1 = -1

So this function doesn’t contain maximum value

(ii) g(x) = -|x + 1| + 3

Solution:

Given that, g(x) = -|x + 1| + 3

At x = -1 f'(x) change sign from positive to negative, hence by first derivative test, x = -1

is a point of local minima. 

So, the maximum value of f = f(-1) = -|(-1) + 1| + 3 = 3

So, this function doesn’t contain minimum value

(iii) h(x) = sin 2x + 5

Solution:

Given that, h(x) = sin(2x) + 5

On differentiate both side w.r.t x, we get

h'(x) = 2cos2x

Now put h'(x) = 0

2cos 2x = 0

2x = (2x – 1)π/2

x = (2x – 1)π/4

Let’s perform second derivative test,

h”(x) = -4sin2x

h”(π/4) < 0 

& so on.

So, 

of local maxima.

& are points of local minima.

So, the minimum value of the given function is 4 and the maximum value of the given function is 6

(iv) f(x) = |sin(4x + 3)|

Solution:

Given that, f(x) = |sin(4x + 3)|

Now for any value of x, sin4x has the least value as -1. i.e., sin 4x + 3 ≥ 2

So f(x) = |sin(4x + 3)| = sin 4x + 3

On differentiate both side w.r.t x, we get

f'(x) = 4cos4x

Now put f'(x) = 0

4cos 4x = 0

4x = (2x – 1)π/2

x = (2x – 1)π/8

  & son on

Let’s perform second derivative test,

f”(x) = -16 sin4x

f”(π/8) < 0; f”(3π/8) > 0;  f”(5π/8) < 0

So,  …. are points of local maxima. Minimum value = 4.

& are points of local minima. Minimum value = 2.

(v) h(x) = x + 1, x ∈ (-1, 1)

Solution:

Given that, h(x) = x + 1, x ∈ (-1, 1)

As we can clearly see from the function that h(x) is a strictly increasing function.

So, the minimum value of x will give minimum value of h(x).

Now, x ∈ (-1, 1)       

So, this function has no minimum nor maximum value.

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) f(x) = x²                                                                                         

Solution:

Given that f(x) = x²

On differentiate both side w.r.t x, we get

f'(x) = 2x

Now put f'(x) = 0

2x = 0

x = 0

Let’s do second derivative test,

f”(x) = 2 > 0

At, x = 0, f'(x) = 0 and f”(x) > 0,

So x = 0 is a point of local minima. Local minimum value.

(ii) g(x) = x³ – 3x

Solution:

Given that g(x) = x³ – 3x

On differentiate both side w.r.t x, we get

g'(x) = 3x² – 3

Now put g'(x) = 0

3x² – 3 = 0

x² = 1

x = ±1

Let’s do the second derivative test,

g”(x) = 6x ….(i)

g”(1) = 6 > 0

g”(-1) = -6 > 0

So by second derivatives test, x = 1 is a point of local maxima and the maximum value is 

g(1) = (1)³ – 3(1) = -2

So by second derivatives test, x = -1 is a point of local minima and the minimum value is 

g(-1) = (-1)³ – 3(-1) = 2

Hence, the local minimum value is -2 and the local maximum value is 2

(iii) h(x) = sin x + cos x, 0 < x < π/2 

Solution:

h(x) = sin x + cos x, x∈(0,π/2)

On differentiate both side w.r.t x, we get

h'(x) = cos x – sin x

Now put h'(x) = 0

cos x – sin x = 0

cos x = sin x, x ∈ (0, π/2)

Clearly x = π/4 [both cos x and sin x attain 1/√2 at π/4]

Let’s do second derivative test,

h”(x) = -sin x – cos x

h”(π/4) = 

At  is a point of local maxima and the maximum value is

h(π/4) = sin π/4 + cos π/4

= 1/√2 + 1/√2 = √2

(iv) f(x) = sin x – cos x, 0 < x < 2π

Solution:

Given that, f(x) = sin x – cos x, x ∈ (0, 2π)

On differentiate both side w.r.t x, we get

f'(x) = cos x + sin x

Now put f'(x) = 0

cos x + sin x = 0

x = in (0, 2π)

Now let’s do the second derivative test 

f”(x) = -sin x + cos x

f”(3π​/4) = – √2 > 0

f”(7π​/4) = √2 > 0

So by second derivatives test, x =  is a point of local maxima and the maximum value is 

f() = -sin 3π​/4 + cos 3π​
4 = 1/√2 + 1/√2 = √2 > 0

So by second derivatives test, x =  is a point of local minima and the minimum value is 

f() = -sin 7π​/4 + cos 7π​
4 = -1/√2 – 1/√2 = -√2 > 0

Hence, the local minimum value is -√2 and the local maximum value is √2.

(v) f(x) = x³ – 6x² + 9x + 15 

Solution:

Given that, f(x) = x³ – 6x² + 9x + 15

On differentiate both side w.r.t x, we get

f'(x) = 3x² – 12x + 9

Now put f'(x) = 0

3x² – 12x + 9 = 0

3(x² – 4x + 3) = 0

x = 1, 3

Let’s do the second derivative test,

f”(x) = 6x – 12

f”(1) = -6 < 0

f”(3) = 6 > 0

So by second derivatives test, x = 1 is a point of local maxima and the maximum value is 

f'(1) = 3(1)² – 12(1) + 9 = 19

So by second derivatives test, x = 3 is a point of local minima and the minimum value is 

f'(3) = 3(3)² – 12(3) + 9 = 15

Hence, the local minimum value is 15 and the local maximum value is 19.

(vi) , x > 0

Solution:

Given that, , x > 0

On differentiate both side w.r.t x, we get

g'(x)=

Now put g'(x) = 0

but ‘x > 0’

So, x = 2

Now we will do the second derivative test,

g”(x)=

Hence, x = 2 is a point of local minima.

Local maximum value = g(2) = 2

(vii)  

Solution:

Given that, 

On differentiate both side w.r.t x, we get

Now put g'(x) = 0

Now, let’s perform the second derivative test,

= -8/16 = -1/2 < 0

At x = 0, g'(x) = 0 and g”(x) < 0

Hence, ‘x = 0’ is a point of local maxima.

Now the domain of g(x) is (-∞, ∞).

Value of g(x) at the extreme values of x is 0

So the global maxima of g(x)=is at x = 0.

The maximum value is g(0) = 1/2

(viii) , x > 0

Solution:

Given that, 

Now put f'(x) = 0

2(1 – x) = x

2 – 2x = x

3x = 2

x = 2/3

Now let’s do the second derivative test,

x = 2/3 is a point of local maxima f(2/3) = 

Now, f (x) = x

For domain, 1 – x ≥ 0 or x ≤ 1

So x ∈ [0, 1]

Local maxima is at x = 2/3 and the local maximum value is 

Question 4. Prove that the following function do not have maxima or minima:

(i) f(x) = e^x                     

Solution:

Given that, f(x) = e^x

f'(x) = e^x

Now e^x > 0, f'(x) > 0

Hence, f(x) is a strictly increasing function with no maxima or minima.  

(ii) g(x) = log x               

Solution:

Given that, g(x) = log x

g'(x) = 1/x

Now the domain of log x is x > 0

So, 1/x > 0, i.e., g'(x) > 0

Hence, g(x) is a strictly increasing function with no maxima or minima.  

(iii) h(x) = x³ + x² + x + 1  

Solution:

Given that, h(x) = x³ + x² + x + 1

h'(x) = 3x² + 2x + 1

Now for this quadratic expression 3x² + 2x + 1,

Its discriminant 0 = 2² – 4(3)(1) = -8 < 0

So, 3x² + 2x + 1 > 0

Hence, h(x) is a strictly increasing function with no maxima or minima.    

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

(i) f(x) = x³, x ∈ [-2, 2]                             

Solution:

Given that, f(x) = x³, x ∈ [-2, 2]  

f'(x) = 3x²

f'(x) = 0 at x = 0 

f”(x) = 6x

f”(0) = 0, second derivative failure

Now f'(3⁺) > 0 and f'(3^–) > 0

f'(x) does not change sign at x = 0.

x = 0 is neither maxima nor minima

f(x) = x³ is a strictly increasing function.

(ii) f(x) = sin x + cos x, x ∈ [0, π]

Solution:

Given that, f(x) = sin x + cos x, x ∈ [0, π]

First derivative

f'(x) = cos x – sin x

Now put f'(x) = 0

cos x = sin x

x = π/4

On applying second derivative test, 

f”(x) = -sin x – cos x

f”(π/4) = 

Hence, x = π/4 is pof local maxima . f(π/4)=

Now, for global maxima = max{f(0), f(π/4), f(π)}

= max{1, √2, -1}

For global maxima is at x = π/4 and the global maximum value is √2.

Now, for global minima = max{f(0), f(π/4), f(π)}

= max{1, √2, -1}

Global minima is at x = π and the global minimum value is -1.

(iii)  

Solution:

Given that, 

f'(x) = 4 – x

Now put f'(x) = 0

4 – x = 0

x = 4

Now applying second derivative test f”(x) = -1 < 0

Hence, x = 4 is a pt. of local maxima.

f(4) = 

Global maxima = max{f(-2), f(4), f(9/2)}

= max{-10, 8, 7.8}

= 8

Global maxima occur at x = 9/2 and global maximum value is f(9/2) = 8

Global minima = min{f(-2), f(4), f(9/2)}

= max{-10, 8, 16.9}

= -10

Global minima occur at x = -2 and the global minimum value is f(-2) = -10.

(iv) f(x) = (x – 1)² + 3, x ∈ [-3, 1]

Solution:

Given that, f(x) = (x – 1)2 + 3, x ∈ [-3, 1]

f'(x) = 2(x – 1)

Now put f'(x) = 0

2(x – 1) = 0

x = 1

Now applying second order derivative test,

f”(x) = 2 > 0

Hence, x = 1 is a point of local minima. f(1) = 3

Global maxima = max{f(-3), f(1)} 

= max{19, 3}

= 19

The global or absolute maxima occurs at x = -3 and the absolute maximum value is f(-3) = 19

Global minima = min{f(-3), f(1)}

= min{19, 3]

= 3

The global or absolute minima occurs at x = 1 and the absolute value is f(1) = 3

Question 6. Find the maximum profit that a company can make, if the profit function is given by p(x) = 41 – 24x – 18x²

Solution:

Given that p(x) = 41 – 24x – 18x²

p'(x) = -24 – 36x

Now put p'(x) = 0

-24 – 36x = 0

x = -24/36

x = -2/3

Now, doing the second order derivative test,

p”(x) = -36 < 0

Hence, x = -2/3 is point of local maxima.

Now in quadratic function with domain R, if there is a local maxima, it is the global maxima also. BC₃ p(-∞)⇢ -∞ and p(+∞)⇢ -∞

The maximum profit is p(-2/3) = 49

If negative units (x) do not exist, then maximum profit is p(0) = 41.

Question 7. Find both the maximum value and the minimum value of 3x⁴ – 8x³ + 12x² – 48x + 25 on the interval [0, 3].

Solution:

Given that f(x) = 3x⁴ – 8x³ + 12x² – 48x + 25, x ∈ [0, 3]

f'(x) = 12x² – 24x² + 24x – 48

Now put f'(x) = 0

12x³ – 24x² + 24x – 48 = 0

12(x² – 2x² + 2x – 4) = 0

12(x²(x – 2) + 2(x – 2)) = 0

12(x² + 2)(x – 2) = 0

x = 2 because x² + 2 ≠ 0

Now applying second derivative test,

f”(x) = 12(3x² – 4x + 2)

f”(2) = 12(3.2² – 4.2 + 2)

f”(2) = 12.6 = 72 > 0

Hence, x = 2 is point of local minima.

f(2) = -39

Global maxima = max{f(0), f(2), f(3)}

= max{25, -39, 16}

= 25

Global maxima occur at x = 0 and the global maximum is 25.

Global minima = min{f(0), f(2), f(3)}

= min{25, -39, 16}

= -39

Global minima occur at x = 2 andthe global minimum value is -39.

Question 8. At what points in the interval[0, 2π], does the function sin 2x attain its maximum value?

Solution:

Given that, f(x) = sin 2x                     ,  x ∈ [0, 2π]

f'(x) = 2 cos 2x

Now put f'(x) = 0

2cos2x = 0

2x = (2x – 1)π/2

x = (2x – 1)π/4

x = π/4, 3π/4, 5π/4, 7π/4

Now let’s do second order derivative test.

f”(x) = -4 sin2x

f”(π/4) = 

x = π/4 and x = 5π/4 are point of local maxima.

x = 3π/4 and x = 7π/4 are point of local minima.

f(π/4) = f(5π/4) = 1 and f(3π/4) = f(7π/4) = -1

Now, 

Global maxima = max{f(0), f(π/4), f(3π/4), f{5π/4}, f(7π/4), f(2π)}

= max{0, 1, -1, 1, -1, 0}

= 1

Global maxima occur at the points x = π/4 and x = 5π/4 and the absolute maximum value is 1.

Question 9. What is the maximum value of the function sin x + cos x?

Solution:

Given that, f(x) = sin x + cos x

f'(x) = cos x – sin x

Now put f'(x) = 0

cos x = sin x

= {-√2, √2, +√2, -√2, -√2}

Now, second order derivative test,

f”(x) = -sin x – cos x

f”(π/4) = f”(9π/4) = f”(17π/4)……….. = -√2 < 0

A liter ⇢ f(x) = sin x + cos x =  

= 

Question 10. Find the maximum value of 2x³ – 24x + 107 in the interval [1, 3]. Find the maximum value of the same function in [-3, -1].

Solution:

Given that f(x) = 2x³ – 24x + 107 

On differentiating w.r.t. x we get

f'(x) = 6x² – 24

Now, put f'(x) = 0

6x² = 24

x² = 4

x = ±2                    

Now second order test

f”(x) = 12x

f”(2) = 12.2 = 24 > 0

x = 2 is a pt. of local minima 

f(2) = 75

f”(-2) = 12(-2) = -24 < 0

x = -2 is point of local maxima. f(-2) = 139

Now, in the interval [1, 3]

Global maxima = max{f(1), f(2), f(3)}

= max{85, 75, 89}

= 89

Now, in the interval [-3,-1]

Global maxima = max{f(-3), f(-2), f(-1)}

= max{125, 139, 129}

= 139

Question 11. 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.

Solution:

Give that, f(x) = x⁴ – 62x² + ax + 9

On differentiating w.r.t. x we get

f'(x) = 4x³ – 124x + a

The maximum value is attained at x = 1, and 1 lies between 0 and 2. 

So, at x = 1, there must be a local maxima

That means, f'(1) = 0

f'(1) = 4(1)³ – 124(1) + a = 0

-120 + a = 0

a = 120

Question 12. Find the maximum and minimum values of x + sin2x on [0, 2π].

Solution:

Give that f(x) = x + sin2x, x ∈ [0, 2π]

On differentiating w.r.t. x we get

f'(x) = 1 + 2cos2x

Now put f'(x) = 0, we get

1 + 2cos2x = 0

cos2x = -1/2

 ∈ [0, 2π]

Now, 

For global maxima = max{f(0), f(π/3), f(4π/3), f(2π)}

= max{0, π/3, }

= 2π

Global maxima occur at x = 2π and the maximum value is f(2π) = 2π.

For global minima = min{f(0), f(2π/3), f(5π/3), f(2π)}

= min{0, }

= 0

Global minima occur at x = 0 and the minimum value is 0.

Exercise 6.5  

Set 2

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

Solution:

Let us assume two numbers x and y.

Given that x + y = 24 ⇒ y = 24 – x

Now lets take a product function P(x) = x.y

Now put the value of y

P(x) = x.(24 – x)

P(x) = 24x – x2

Now, we have to maximize our product P(x).

Differentiating P(x) with respect to x, we get,  

Now put P'(x)=0

24 – 2x = 0

x = 12

y = 24 – x = 24-12

y = 12

Hence, the two numbers are 12 and 12 and the maximum product is 144.

Question 14. Find two positive number x and y such that x + y = 60 and xy³ is maximum.

Solution:

Given that, 

x + y = 60 ⇒ x = 60 – y, x > 0, y > 0

Let’s take a function P(y) = xy³

P(x) = (60 – y)y³

P(x) = 60y³ – y⁹

On differentiating P(y) with respect to y, we get

P'(y) = 180y² – 4y³

Now put P'(y) = 0

180y² – 4y³ = 0

4y²(45 – y) = 0

y = 0 or y = 45

We doesn’t accept y = 0 because y > 0

So, y = 45

x = 60 – y = 60 – 45

x = 15

Question 15. Find two positive number x and y such that their sum is 35 and the product x²y⁵ is a maximum.

Solution:

Given that,

x + y = 35 ⇒ y = 35 – x 

Let us considered P(y) = x²y⁵  

Put the value of y from the above equation

P(x) = x²(35 – x)⁵

On differentiating P(x) with respect to y, we get,

P'(x) = x²5(35 – x)⁴(-1) +(35 – x)⁵2x

P'(y) = x(35 – x)⁴[-5x + (35 -x)2]

P'(y) = 7x(35 – x)⁴(10 – x)

Now put P'(x) = 0

7x(35 – x)⁴(10 – x) = 0

x = 0 or x = 35 or x = 10

Here, x = 0 is rejected because x is positive, x = 35 is also rejected because 

y = 35 – 35 = 0, but y is positive. So, x = 10 is the turning point.

Now we do second derivative test

P”(x) = 7(35 – x)³(6x² – 120x +350)

Now put x = 10

we get

P”(x) = 7(35 – 10)³(6 x 100 – 120 x 10 +350)

= 7(25)³(250) < 0

So by second derivative test p'(x) will be maximum at x = 10. So, y = 25

Question 16. Find two positive numbers whose sum is 61 and the sum of whose cubes is minimum.

Solution:

Let us considered two numbers a and y

x + y = 16 ⇒ y = 16 – x

Sum of cubes = x³ + y³ 

So let us considered s(x) = x³ + y³ 

S(x) = x³ + (16 – x)³

S(x) = x³ + 16³ – x³ – 3.16.x(16 – x)

S(x) = 16³ – x.16².3 + 3.16.x²

S'(x) = 2.3.16x – 3.16²

Now put S'(x) = 0

2.3.16.x = 3.16²

2.x = 16

x = 8

So at x = 8 S”(x) = 96(positive), so x = 8 is a local minima point. Hence

y = 16 – x = 8

The two numbers are 8 and 8 and the sum of cubes is 8³ + 8³ = 1024

Question 17. A square piece of tin of side 18cm 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:

The dotted square will be the base of the open cuboid.

Side will be 18 – 2x

The cuboid will now have dimensions 18 – 2x, 18 – 2x, x

Volume of the open box,

V(x) = (18 – 2x).(18 – 2x).x

V(x) = x(18 – 2x)²

Now we have to maximize V(x),

Now put V'(x) = 0

V'(x) = 2(18 – 2x).x(-2) + (18 – 2x)² = 0

(18 – 2x)[-4x + 18 – 2x] = 0

(18 – 2x)(18 – 6x) = 0

x = 9 and x = 3

Now put x = 9

V(x) = (18 – 2x)².x

V(9) = (18 – 2 x 9)².9

V(9) = 0

It is impossible so we doesn’t accept x = 9

Now put x = 3

V(x) = (18 – 2x)².x

V(3) = (18 – 2 x 3)².3

V(3) = 432

So, x = 9 is the turning point

Hence at x = 9, V”(x) = -72(negative), So, volume is minimum at x = 3. 

Now, the sides of the square for maximum volume is 3.   

Question 18. A rectangular sheet of tin 45cm by 24cm is to be made into a box without top, by cutting off square form 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? 

Solution:

Given,

A rectangular sheet of dimension 45 x 24. Lets the side of the wall square be x,

Then the cuboid have its base as the blue rectangle and height equal to side of the square.

The dimensions of the cuboid will be 45 – 2x, 24 – 2x and x.

Let V(x) = Volume function, so

V(x) = (45 – 2x)(24 – 2x)(x)

V(x) = (45x – 2x²)(24 – 2x)

V'(x) = (45x – 2x²)(-2) + (24 – 2x)(45 – 4x)

Now put V'(x) = 0

(45x – 2x²).2 = 2(12 – x)(45 – 4x)

45x – 4x² = 12.45 – 48x – 45x + 4x²

6x² – 138x + 12.45 = 0

x² – 23x + 90 = 0

(x – 18)(x – 5) = 0

x = 18 or x = 5

x can’t be equal to 18, 24 – 2x will become negative. Hence x = 5 is the turning point. 

Now, volume is V(x) = (45 – 2x)(24 – 2x)(x)

Now put x = 5, V = (45 – 10).(24 – 10).(5)

V = 35.12.5

V = 2100

Side of small square is 5cm

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

Solution:

Given, a variable rectangle inside a fixed circle.

Let the diameter of the fixed circle be equal to d and the 

sides of the rectangle be equal to x and y 

Now, the diameter d is fixed.

Let the area function be A(x) = x.y

From the triangle ABC,

AB2 + BC2 = AC2

x2 + y2 = (2a)2

y2 = (2a)2 – x2

y = 

Now, A(x) = x.y = x.

A'(x) = x.

A'(x) = 

Now we find the second derivative

A”(x) = 

Now put A'(x) = 0 for maximum and minimum values.

A'(x) =  = 0

x = √2a

When x = √2a,  = -4(negative)

So, at x = √2a the area of rectangle is maximum hence, y = √2a 

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:

Given,

A right circular cylinder of given surface area.

Now, let the total surface are be A.

A = 2πr(r + h)

A/2π = r² + rh

Let us assume A/2π = M

h = M – r²/r

Now the volume of the cylinder V(r) = πr²h

So, V(r) = πr²(M – r²/r)

= π(rM – r³)

V'(r) = π(M – 3r²)

and V”(r) = -π(6r)

Now put V'(r) = 0

π(rM – r³) =0

r = √M/3

At r = √M/3, V”(r) = -π(6√M/3) is negative, so r is maximum at √M/3. 

Hence, h = 2r

Clearly, the height is equal to the diameter of the base.

Question 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 maximum surface area? 

Solution:

Given: cylindrical can of fixed volume = 100 cc.

Find: the dimensions of the can 

As we know that

Volume = πr²h = 100cc

h = 100/πr²             ————-(1)

Now, the total surface area(A) = 2πr² + 2πrh

A = 2πr² + 2πr(100/πr²)

A = 2πr² + (200/r)

A'(r) = 4πr + (200/r²)

Now we find

A”(r) = 4π+ (200/r³)

Now put A'(r) = 0

A'(r) = 4πr + (200/r²) = 0

r = (50/π)^1/3

At r = (50/π)^1/3, A”(r) = 4π+ (200/((50/π)^1/3)³) = 12π is positive. 

So, area is minimum when r = (50/π)^1/3.

Hence, the height is (h) = 100/π((50/π)^1/3)² = 2.(50/π)^1/3 = 2r

Question 22. A wire of length 28m is to be cut into two pieces. One of the pieces is to 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:

Given that the length of the wire is 8 and cut into two pieces. From one 

piece a square is formed and form another piece a circle is formed. so let 

us assume x be the side of square and y be the radius of circle. 

So, Length of the wire = Perimeter(of square) + circumference(of circle)

28 = 4x + 2πy

y = (14 – 2x)/π

As we know that the area of circle is πy² and the area of square is x²

So, total area(A) = πy² + x²

Now put the value of y we get

A = π((14 – 2x)/π)² + x²

A'(x) = 2x – 8/π(7 – x)

and 

A”(x) = 2 + 8/π

Now put A'(x) = 0

2x – 8/π(7 – x) = 0

x = 28/π + 4

So, at point x = 28/π + 4, area is minimum because A”(x) is positive. 

Hence, the wire cut at distance 28/π + 4 

Question 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.

Solution:

According to the question it is given that the sphere contains a cone. 

So let us assume that the center of the sphere is O, the radius of the sphere is R and UV = x and SV = y.

So in triangle OVU

OV² + VU ²= OU²

(y – R)² + x² = R²

x² = 2Ry – y² …(1)

So, the volume of the cone(V) = 1/3πx²y

Now put the value of x from eq(1)

= 1/3π(2Ry – y²)y

V = 1/3π(2Ry² – y³)

V'(y) = 1/3π(4Ry – 3y²)

and 

V”(y) = 1/3π(4R – 6y)

Now put V'(y) = 0

1/3π(4Ry – 3y²) = 0

y = 4R/3

So, at y = 4R/3, V”(y) = 1/3π(4R – 6y) = 1/3π(4R – 6(4R/3)) = -4R/3 is negative. 

So, Volume is maximum at y = 4R/3.

Now put the value of y in eq(1), we get

x² = 2R(4R/3) – (4R/3)² 

x = 8R²/9

So the maximum volume of the cone is 

V = 1/3πx²y

V = 1/3π(8R²/9)²(4R/3) = 8/27

Question 24. Show that the right circular cone of least curved surface and given volume has an altitude equal to √2 time the radius of the base.

Solution:

Let us considered r be the radius of the cone and h be the height of the cone.

So, the volume of the cone(V) = 1/3πr2h

r2h = 3V/π

r2 = M/h [3V/π = M] ….(1)

As we know that the surface area of the cone:

(A) = 

(A)2 = π2r2 (r2 + h2)

Now put the value of r2 from above

(A)2 = S = π2M/h (M/h + h2)

= π2M(M/h2 + h)

So, S'(h) = π2M(-2Mh-3 + 1) and

S”(h) = π2M(6Mh-4) 

Now put S'(h) = 0

π2M(-2Mh-3 + 1) = 0

h = (2M)1/3

So at h = (2M)1/3, S”(h) = π2M(6Mh-4) = 6π2M2/(2M)4/3 is positive. 

So, area is minimum at h = (2M)1/3

So, r = M/(2M)1/3

From eq(1), we get

h = √2r

Question 25. Show that the semi-vertical angle of the cone of the maximum volume and of given slant height is tan^-1 √2.

Solution:

Let us assume r be the radius, h be the height, l be the slant height, and θ be the semi vertical angle of cone. 

So, 

l ² = r²+ h²

r² = l² – h²  ….(1)

As we know that the volume of the cone is 

V = 1/3πr²h

Now put the value of r² from eq(1)

V = 1/3π(l² – h²)h

V = 1/3π(hl² – h³)

So, V'(h) = 1/3π(l² – 3h²) and 

V”(h) = 1/3π(- 6h) = -2πh

Now put V'(h) = 0

1/3π(l² – 3h²) = 0

h = l/√3

So, at h = l/√3, V”(h) = -2πh = -2π(l/√3) is negative so, V is maximum at h = l/√3

So from eq(1), we get

r² = l² – (l/√3)²

r = √2(l/√3)

As we know that the semi vertical angle is tan θ = r/h = √2(l/√3)/l/√3 = √2. So the value of θ = tan^-1√2

Question 26. Show that semi-vertical angle of right circular cone of given surface area and maximum volume is sin^-1(1/3).

Solution:

Let us assume r be the radius, h be the height, and θ be the semi vertical angle of cone. 

So, the total surface area (A) = 

= A/π

= S

 = S – r2

On squaring on both side we get

r2(r2 + h2) = (S – r2)2

r2 = S2/h2 + 2S2 ….(1)

As we know that the volume of the cone is 

V = 1/3πr2h

Now put the value of r, we get

V = 1/3π(S2/h2 + 2S2)h

V = 1/3πS2(h/h2 + 2S)

Now 

V'(h) = 1/3πS2(2S – h2)/(h + 2S)2

Now put V'(h) = 0

 1/3πS2(2S – h2)/(h + 2S)2 = 0

h = +-√2S

Here, the h = √2S is a valid value and h = -√2S is not valid because height can’t be negative. So, h = √2S is the turning point. 

V'(h) > 0, so the volume is maximum at h = √2S. 

So put the value of h = √2S in eq(1), we get

r = √S/2

and as we know that the semi vertical angle is sin θ = r/√r 2+ h2. So the value of θ = tan-11/3

Question 27. The point on the curve x² = 2y which is nearest to point (0,5) is 

(A)(2, √2, 4)           (B) (2, √2, 0)            (C) (0, 0)                  (D) (2, 2)

Solution:

Given that, x2 = 2y

Let any random point on this curve be (h, k)

So, h2 = 2k; k = h2/2

The random point is (h, h2/2)

Minimizing the distance

d2 = (h – 0)2 + () 

d2 = f(x) = h2 + 

f'(x) = 2h + 2.h

Now put f'(h) = 0

2h[1+] = 0

h = 0 or 

h = 0 or h = 2√2​, h =−2√2​

If h = 0, d=

If h = ±2√2,

So, h = ±2√2       [giving minimum distance]

k = h2/2 = 8/2 = 4

Points are (±2√2, 4)

So the correct option is A

Question 28. For all real values of x, the minimum value of

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

Solution:

Given that, 

 Now put f'(x) = 0

1 + x + x2 – 2x2 – x = 0

x2 = 1

x = ±1 

So, x = -1, x = 1 are the turning points

Now let us find the value of f(x) at points x = -1, x = 1

Put x = 1

f(1) = 1/3

Put x = -1

f(-1) = 3

Hence, the maximum value of  is 1/3

So the correct option is D

Question 29. The maximum value of [x(x – 1) + 1]^1/3, 0 ≤ x ≤ 1 is 

(A) (1/3)^1/3                    (B) 1/2                         (C) 1                   (D) 0

Solution:

Given that,

f(x) = [x(x – 1) + 1]1/3 = (x2 – x + 1)1/3, x ϵ [0, 1]

f′(x) = 1/3​(x2 − x + 1)-2/3 ​.(2x−1)

or we can write as

 

Now put f'(x) = 0

2x – 1 = 0

So, x = 1/2 is the turning point which belongs to the given closed interval 0 ≤ x ≤ 1. 

Now let us find the value of f(x) at points x = 1/2, x = 0, x = 1

Put x = 1/2

f(1/2) = (1/4 – 1/2 + 1)1/3

f(1/2) = (3/4)1/3 < 1

Put x = 1

f(1) = (1 – 1 + 1)1/3

f(1) = 1

Put x = 0

f(0) = (0 – 0 + 1)1/3

f(0) = 1

Hence, the maximum value of [x(x – 1) + 1]1/3 is 1

So the correct option is C 

Chapter-6 (Applications of Derivatives)

• Exercise 6.1
• Exercise 6.2
• Exercise 6.3
• Exercise 6.4