Ncert Solutions For Class 12 Maths Ex 6.4

Ncert Solutions For Class 12 Maths Chapter 6 Ex 6.4

Que.1. For the given functions, find the maximum and minimum values:

(i) f(y) = (2y − 1)2 + 3

Ans. f(y) = (2y − 1)2 + 3

It is observed that (2y – 1)2 \(\geq\) 0 for y \(\epsilon\) R.

Hence, f(y) = (2y – 1)2 + 3 \(\geq\) 3 for y \(\epsilon\) R.

When 2y – 1 = 0, the minimum value of f is obtained.

2y – 1 = 0. Therefore, x = \(\frac{1}{2}\)

Minimum value = f(\(\frac{1}{2}\)) = ( 2. \(\frac{1}{2}\) – 1 )2 + 3 = 3

Therefore, the maximum value of the function f does not exist.

(ii)f(y) = 9y2 + 12y + 2 = (3y + 2)2 − 2.

Ans. It is observed that (3y + 2)2 \(\geq\) 0 for y \(\epsilon\) R.

Hence, f(y) = (3x + 2)2 – 2 \(\geq\) –2 for y \(\epsilon\) R.

When 3y + 2 = 0, the minimum value of f is obtained.

3y + 2 = 0. Therefore, y = \(\frac{-2}{3}\)

Minimum value = f(\(\frac{-2}{3}\)) = (3 (\(\frac{-2}{3}\)) + 2)2 – 2 = –2

Therefore, the maximum value of the function f does not exist.

(iii) f(y) = –(y – 1)2 + 10.

Ans. It is observed that (x – 1)2 \(\geq\) 0 for y \(\epsilon\) R.

Hence, f(y) = –(y – 1)2 + 10 \(\leq\) 10 for y \(\epsilon\) R.

When (y – 1) = 0, the maximum value of f is obtained.

(y – 1) =0. Therefore, y = 0

Maximum value = f(1) = –(1 – 1)2 + 10 = 10

Therefore, the minimum value of the function f does not exist.

(iv) g(y) = y3 + 1

Ans. Therefore, the function g does not have maximum or minimum value.

Que.2. For the given functions, find the maximum and minimum values:

(i) f(y) = \(\left | y+2 \right |\) – 1

Ans. \(\left | y+2 \right |\) \(\geq\) 0 for y \(\epsilon\) R

Hence, f(y) = \(\left | x+2 \right |\) – 1 \(\geq\) – 1 for y \(\epsilon\) R

When \(\left | x+2 \right |\) =0, the minimum value of f is obtained.

\(\Rightarrow\) x = –2

Minimum value = f(–2) = \(\left | -2+2 \right |\) –1 = –1

Therefore, the maximum value of the function f does not exist.

(ii) g(y) = – \(\left | y+1 \right |\) + 3

Ans. – \(\left | y+1 \right |\) \(\leq\) 0 for y \(\epsilon\) R

g(x) = – \(\left | y+1 \right |\) + 3 for y \(\epsilon\) R

When \(\left | y+1 \right |\) = 0, the maximum value of g is obtained.

x = –1

Maximum value = g(-1) = –\(\left | –1+1 \right |\) + 3 = 3

Therefore, the minimum value of the function f does not exist.

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

Ans. – 1 \(\leq\) sin 2y \(\leq\) 1.

–1 + 5 \(\leq\) sin 2y + 5 \(\leq\) 6

Therefore, 6 and 4 are the maximum and minimum values.

(iv) f(y) = \(\left | sin 4y + 3 \right |\)

Ans. –1 \(\leq\) sin 4y \(\leq\) 1

2 \(\leq\) sin 4y + 3 \(\leq\) 4

2 \(\leq\) \(\left | sin 4y + 3 \right |\) \(\leq\) 4

Therefore 4 and 2 are the maximum and minimum values.

(v) h(y) = y + 1, y \( \)(-1,1)

Ans. If point y0 is close to – 1, then \(\frac{y_{0}}{2}\) \(<\) y0 + 1 for y \(\epsilon\) (-1, 1)

If y1 is close to 1, then y1 + 1 < \(\frac{y_{1}+1}{2}\) + 1 for y \(\epsilon\) (-1,1).

Therefore, h(y) neither has a maximum or a minimum value at (-1,1).

Que.3. For the following, find the local maxima and minima and also local maximum and minimum values, if any.

(i) f(y) = y2

\({f}'(y)\)(x) = 2y

\({f}”(y)\) = 0 \(\Rightarrow\) y = 0

Thus, the critical point y = 0 can be the local maxima or local minima of f.

\({f}”(y)\) = 2 which is positive

Thus, by second derivative test, y = 0 is the point of local maxima and minima value of f at y = 0 is f(0) = 0.

(ii) g(y) = y3 – 3y

\( \) \({g}’\)(y) = 3y2 – 3

\({g}’\)(y) = 0

3y2 = 3

y = \(\pm\)1

\({g}”\)(y) = 6y

\({g}”\)(1) = 6 > 0

\({g}’\)(-1) = – 6 < 0

By second derivative,

The point of local minimum and minima at g is y = 1

g(1) = 13 – 3 = 1 – 3 = –2

The point of local maximum and local maxima at g is y = – 1

g(-1) = (-1)3 – 3 (– 1) = – 1 + 3 = 2.

(iii) h(y) = sin y + cos y, 0 < y < \(\frac{\pi }{2}\)

\({h}’\)(y) = cos y – sin y

\({h}’\)(y) = 0

Sin y = cos y

tan y = 1 \(\Rightarrow\) y = \(\frac{\pi }{4}\)\(\epsilon\) (0, \(\frac{\pi }{2}\))

\({h}”\)(y) = – sin y – cos y = –(sin y + cos y)

\({h}”\) \(\frac{\pi }{4}\) = –(\(\frac{1}{\sqrt{2}}\) + \(\frac{1}{\sqrt{2}}\)) = \(-\frac{2}{\sqrt{2}}\) = \(-\sqrt{2}\) < 0

Thus, by second derivative, the local maximum and maxima at point y = \(\frac{\pi }{4}\) is

h(\(\frac{\pi }{4}\)) = sin \(\frac{\pi }{4}\) + cos \(\frac{\pi }{4}\) = \(\frac{1}{\sqrt{2}}\) +\(\frac{1}{\sqrt{2}}\) = \(\sqrt{2}\).

(iv) f(y) = sin y – cos y, 0 < y < 2n

\({f}’\)(y) = cos y + sin y

\({h}’\)(y) = 0

cos y = – sin y

tan y = – 1 \(\Rightarrow\) y = \(\frac{3\pi }{4}\), \(\frac{7\pi }{4}\) \(\epsilon\) (0,2\(2\pi\))

\({f}”\)(y) = – sin y + cos y

\({f}”\)( \(\frac{3\pi }{4}\) ) = – sin \(\frac{3\pi }{4}\) + cos \(\frac{3\pi }{4}\) = –\(\frac{1}{\sqrt{2}}\) – \(\frac{1}{\sqrt{2}}\) = –\(\sqrt{2}\) > 0

\({f}”(\) \(\frac{7\pi }{4}\)) = – sin \(\frac{7\pi }{4}\) + cos \(\frac{7\pi }{4}\) = \(\frac{1}{\sqrt{2}}\) + \(\frac{1}{\sqrt{2}}\) = \(\sqrt{2}\) > 0

Thus, by second derivative, the local maximum and maxima at point y = \(\frac{3\pi }{4}\) is

f(\(\frac{3\pi }{4}\) = sin \(\frac{3\pi }{4}\) – cos \(\frac{3\pi }{4}\) = \(\frac{1}{\sqrt{2}}\) + \(\frac{1}{\sqrt{2}}\) = \(\sqrt{2}\)

The local minimum and minima at point y = \(\frac{7\pi }{4}\) is

sin \(\frac{7\pi }{4}\) – cos \(\frac{7\pi }{4}\) = –\(\frac{1}{\sqrt{2}}\) – \(\frac{1}{\sqrt{2}}\) = –\(\sqrt{2}\)

(v) f(y) = y3 – 6y2 + 9y + 15

\({f}’\)(y) = 3y3 – 12y + 9

\({f}’\)(y) = 0

3(y2 – 4y + 3) = 0

3(y – 1) (y – 3) = 0

y = 1,3

\({f}”(\)(y) = 6y – 12 = 6(x – 2)

\({f}”(\) = 6(1 – 2) = -6 < 0

\({f}”'(\) = 6(3 – 2) = 6 > 0

Thus, by second derivative, the local maximum and maxima at point y = 1 is

f(1) = 1 – 6 + 9 + 15 = 19

The local minimum and minima at point y = 3 is

f(3) = 27 – 54 + 27 + 15 = 15

(vi) g(y) = \(\frac{y}{2}\) + \(\frac{2}{y}\) , y > 0

\({g}’\)(y) = \(\frac{1}{2}\) – \(\frac{2}{y^3}\)

\({g}’\)(y) = 0

\(\frac{2}{y^3}\) = \(\frac{1}{2}\)

y3 = 4 \(\Rightarrow\) y = \(\pm\)2

As y > 0, y = 2

\({g}”\)(y) = \(\frac{4}{y^3}\)

\({g}”\)(2) = \(\frac{4}{2^3}\) = \(\frac{1}{2}\) > 0

By second derivative, the local minimum and minima at point y =2 is

g(2) = \(\frac{2}{2}\) +\(\frac{2}{2}\) = 1+ 1 = 2.

(vii) g(y) = \(\frac{1}{y^{2}+2}\)

\({g}’\)(y) = \(\frac{-2y}{(y^{2}+2)^{2}}\)

\({g}’\)(y) = 0

\(\frac{-2y}{(y^{2}+2)^{2}}\) = 0

y = 0

By first derivative, the local maximum and maxima at point y = 0 is

g(0) = \(\frac{1}{0+2}\) = \(\frac{1}{2}\).

(viii) f(y) = y\(\sqrt{1-y}\), y > 0

\({f}’\)(y) = \(\sqrt{1-y}\) + y . \(\frac{1}{2\sqrt{1-y}}\) (-1) = \(\sqrt{1-y}\) – \(\frac{y}{2\sqrt{1-y}}\) = \(\frac{2-3y}{2\sqrt{1-y}}\)

\({f}’\)(y) = 0

\(\frac{2-3y}{2\sqrt{1-y}}\) = 0 \(\Rightarrow\) 2 – 3y

y = \(\frac{2}{3}\)

\({f}”\)(y) = \(\frac{1}{2}\) \(\frac{\sqrt{1-y}(-3)-(2-3y)(\frac{-1}{2\sqrt{1-y}})}{1-y}\)

= \(\frac{\sqrt{1-y}(-3)-(2-3y)(\frac{-1}{2\sqrt{1-y}})}{2(1-y)}\)

= \(\frac{-6(1-y)+(2-3y)}{4(1-y)^{\frac{3}{2}}}\)

=\(\frac{3y-2}{4(1-y)^{\frac{3}{2}}}\)

\({f}”\)(y) = \(\frac{2-4}{4(\frac{1}{3})^{\frac{3}{2}}}\) =\(\frac{-1}{2(\frac{1}{3})^{\frac{3}{2}}}\) < 0

By second derivative, the local maximum and maxima at point y = \(\frac{2}{3}\) is

f(\(\frac{2}{3}\)) = \(\frac{2}{3}\) \(\sqrt{1-\frac{2}{3}}\) = \(\frac{2}{3\sqrt{3}}\) = \(\frac{2\sqrt{3}}{9}\).

Que.4. Show that the given equation does not have maxima and minima:

f(y) = ey

Ans.­
\({f}’\)(y) = ey

\({f}’\)(y) = 0

ey = 0. For any values of y, the exponential function cannot assume 0

Thus, the function f do not have maxima or minima.

Que.5. For the given functions find the absolute maximum and minimum values in the given intervals.

(i)f(y) = y3, y \(\epsilon\) (-2,2)

\({f}’\)(y) = 3y3

\({f}’\)(y) = 0 \(\Rightarrow\) y = 0

f(0) = 0

f(-2) = (-2)3 = -8

f(2) = (2)3 = 8

Thus, the absolute maximum is 8 and the absolute minimum is -8 for the given interval.

(ii) f(y) = (y – 1)2 + 3, y \(\epsilon\) (-3,1)

\({f}’\)(y) = 2(y – 1)

\({f}’\)(y) = 0

2(y – 1) = 0

y = 1

The given intervals are (-3,1)

f(1) = (1 – 1)2 + 3 = 0 + 3 = 3

f(-3) = (–3 – 1)2+ 3 = 16 + 3 = 19

Thus, the absolute maximum is 19 and the absolute minimum is -3 for the given interval.

Que.6. The profit function is p(y) = 41 – 34y – 18y2. Find the maximum profit.

Ans. \({p}’\) (y) = -24 – 36y

\({f}”\)(y) = -36

\({p}’\)(y) = 0

y = \(\frac{-24}{36}\) = \(\frac{-2}{3}\)

\({f}”\)(\(\frac{-2}{3}\)) = -36 < 0

By second derivative

The local maxima point is y = \(\frac{-2}{3}\)

Maximum profit = p(= \(\frac{-2}{3}\))

= 41 – 24(\(\frac{-2}{3}\)

) – 18 (\(\frac{-2}{3}\))2

= 41 + 16 – 8

= 49

The maximum profit is 49 units.

Que.7. The maximum value for the function sin 2y can be attained at what point in the interval (0,2n).

Ans. f(y) = sin 2y

\({f}’\)(y) = 2cos 2y

\({f}’\)(y) = 0

cos 2y = 0

2y = \(\frac{\pi }{2}\), \(\frac{3\pi }{2}\), \(\frac{5\pi }{2}\), \(\frac{7\pi }{2}\)

y = \(\frac{\pi }{4}\), \(\frac{3\pi }{4}\), \(\frac{5\pi }{4}\), \(\frac{7\pi }{4}\)

f(\(\frac{\pi }{4}\)) = sin \(\frac{\pi }{2}\) = 1

f(\(\frac{3\pi }{4}\)) = sin \(\frac{3\pi }{2}\) = -1

f(\(\frac{5\pi }{4}\)) = sin \(\frac{5\pi }{2}\) = 1

f(\(\frac{7\pi }{2}\)) = sin \(\frac{7\pi }{2}\) = -1

f(0) = sin 0 = 0, f(2\(\pi\)) = sin 2\(\pi\) = 0

Thus, the absolute maximum is occurring at y = \(\frac{\pi }{4}\) and y = (\(\frac{5\pi }{4}\)).

Que.8. In a function sin y + cos y, find the maximum value.

f(y) = sin y + cos y

Ans. \({f}’\)(y) = cos y – sin y

\({f}’\)(y) = 0

sin y = cos y = – (sin y + cos y) \(\Rightarrow\) tan y = 1

y = \(\frac{\pi }{4}\), \(\frac{5\pi }{4}\)

\({f}”\)(y) = – sin y – cos y = – (sin y + cos y)

When (sin y + cos y) is positive, \({f}”\)(y) will be negative.

\({f}”\)(y) is negative when y \(\epsilon\) (0, \(\frac{\pi }{2}\))

Thus, y = \(\frac{\pi }{4}\)

\({f}”\)( \(\frac{\pi }{4}\)) = – (sin \(\frac{\pi }{4}\) + cos \(\frac{\pi }{4}\)) = – (\(\frac{2}{\sqrt{2}}\)) = –\(\sqrt{2}\) < 0

By second derivative, maximum is at y = \(\frac{\pi }{4}\)

f(\(\frac{\pi }{4}\)) = sin \(\frac{\pi }{4}\)

+ cos \(\frac{\pi }{4}\) = \(\sqrt{2}\)

Que.9. In the interval (1, 3) and (-3, -1), find the maximum value of 2y3 – 24y + 107.

Ans. f(y) = 2y3 – 24y + 107

\({f}’\)(y) = 6y2 – 24 = 6(y2 – 4)

\({f}’\)(y) = 0

6(y2 – 4) = 0

y = \(\pm\)2

If we consider the interval (1, 3)

f(2) = 2(8) – 24(2) + 107 = 16 – 48 + 107 = 75

f(1) = 2(1) – 24(1) + 107 = 2 – 24 + 107 = 85

f(3) = 2(27) – 24(3) + 107 = 54 – 72 + 107 = 89

Therefore, the absolute maximum value of f(y) in interval (1, 3) is 89 at y = 3

Now, if we consider the interval (-3, -1)

f(-3) = 2 (-27) – 24 (-3) + 107 = -54 + 72 + 107 = 125

f(-1) = 2 (-1) – 24 (-1) + 107 = -2 + 24 + 107 = 129

f(-2) = 2 (-8) – 24 (-2) + 107 = -16 + 28 + 107 = 139

Therefore, in the interval (-3, -1), the maximum value is 139 at y = -2.

Que.10. For the function y4 – 62y2 + ay + 9, the maximum value is attained at y = 1 in the interval (0, 2). Find the value of a.

Ans. f(y) = y4 – 62y2 + ay + 9

\({f}’\)(y) = 4y3 – 124y + a

Maximum value is attained at y = 1

\({f}’\)(y) = 0

4 – 124 + a = 0

a = 120

Therefore, 120 is the maximum value

Que.11. For the function y + sin 2y, find the maximum and minimum value at the interval (0, 2n).

Ans. f(y) = y + sin 2y

\({f}’\)(y) = 1 + 2cos 2y

\({f}’\)(y) = 0

cos 2y = –\(\frac{1}{2}\) = – cos \(\frac{\pi }{2}\) = cos(\(\pi\) – \(\frac{\pi }{3}\)) = cos \(\frac{2\pi }{3}\)

2y = 2n \(\pm\) \(\frac{2\pi }{3}\), n \(\epsilon\) Z

y = n\(\pi\) \(\pm\) \(\frac{\pi }{3}\), n \(\epsilon\) Z

y = \(\frac{\pi }{3}\), \(\frac{2\pi }{3}\), \(\frac{4\pi }{3}\), \(\frac{5\pi }{3}\) \(\epsilon\) (0, 2\(\pi\))

f(\(\frac{\pi }{3}\)) = \(\frac{\pi }{3}\) + sin \(\frac{2\pi }{3}\) = \(\frac{\pi }{3}\) + \(\frac{\sqrt{3}}{2}\)

f(\(\frac{2\pi }{3}\)) = \(\frac{2\pi }{3}\) + sin \(\frac{4\pi }{3}\) = \(\frac{2\pi }{3}\) – \(\frac{\sqrt{3}}{2}\)

f(\(\frac{4\pi }{3}\)) = \(\frac{4\pi }{3}\) + sin \(\frac{8\pi }{3}\) = \(\frac{4\pi }{3}\) + \(\frac{\sqrt{3}}{2}\)

f(\(\frac{5\pi }{3}\)) = \(\frac{5\pi }{3}\) + sin \(\frac{10\pi }{3}\) =

\(\frac{5\pi }{3}\) – \(\frac{\sqrt{3}}{2}\)

f(0) = 0 + sin 0 = 0

f(2\(\pi\)) = 2\(\pi\) + sin 4\(\pi\) = 2\(\pi\) + 0 = 2\(\pi\)

Therefore, the absolute maximum and minimum value is 2n and 0 in the interval (0, 2n)

Que.12. A sum of two numbers in 24 whose product is as large as possible. Find the two numbers.

Ans. Let one number be y. Then, the other number is (24 – y)

P(y) is the product of two numbers.

P(y) = y(24 – y) = 24y – y2

\({P}’\)(y) = 24 – 2y

\({P}”\)(y) = -2

\({P}’\)(y) = 0

y = 12

\({P}”\)(12) = -2 < 0

By second derivative, the local maxima of P is at point y = 12. Thus, the product of two numbers is maximum when there are 12 and 24.

Que.13. If y + z = 60 and yz3 is maximum, find the two positive numbers.

Ans. y + z = 60

y = 60 – y

f(y) = yz3

f(y) = x(60 – y)3

\({f}’\)(y) = (60 – y)3 – 3y(60 – y)2

= (60 – y)2 (60 – y – 3y)

= (60 – y)2 (60 – 4y)

\({f}”\)(y) = -2(60 – y) ( 60 – 4y) – 4(60 – y)2

= -2 (60 – y) (60 – 4y + 2(60 – y))

When y = 60, \({f}”\)(y) = 0.

When y = 15, \({f}”\)(y) = -12(60 – 15) (30 – 15) = -12 x 45 x 15 < 0

\({f}’\)(y) = 0

y = 60 or y = 15

By second derivative, the local maxima is at the point y = 15.

Thus the required numbers are y =15 and z = 60 – 15 = 45.

Que.14. The sum of two positive numbers is 16 and whose cube is minimum. Find the two numbers.

Ans. Let y be a number. The other number is (16 – y)

S(y) is the sum of the cubes.

S(y) = y3 + (16 – y)3

\({s}’\)(y) = 3y2 – 3(16 – y)2 = 0

y2 – (165 – y)2 = 0

y2 – 256 – y2 + 32y = 0

y = \(\frac{256}{32}\) = 8

\({S}”\)(8) = 6(8) + 6(16 – 8) = 48 + 48 = 96 > 0

By second derivative, the local minima is at point y = 8.

Thus, the numbers are 8 and 16 − 8 = 8.

Que.15. A box is made without the top from a square piece of tin whose sides are 18 cm by cutting a square from each corner and folding up the flaps. The volume of box should be maximum, so what length of sides are to be cut?

Ans. Let y be the length to be cut. So, the length and breadth are (18 – 2y) cm and the height is y cm.

V(y) = y(18 – 2y)2

\({V}’\)(y) = (18 – 2y)2 – 4y(18 – 2y)

= (18 – 2y) (18 – 6y)

= 12(9 – y) (3 – y)

\({V}”\)(y) = 12(– (9 – y) – (3 – y) )

= –12(9 – y + 3 – y)

= – 12(12 – 2y)

= – 24(6 – y)

\({V}’\)(y) = 0

y = 9 or y = 3

If y = 9, then the length and breadth will become 0

y \(\neq\) 9

y = 3

\({V}”\)(y) = -24 (6 – 3) = –72 < 0

By second derivative, the point of maxima is y = 3

Thus, 3cm should be removed from the sides.

Que.16. Prove that the all the rectangles inscribed inside a circle, square has the maximum area

5.19.1

Ans. Let the radius be a.

The diagonal passes through the center whose length is 2a cm.

By Pythagoras theorem,

(2a)2 = l2 + b2

b2 = 4a2 – l2

b = \(\sqrt{4a^{2}-l^{2}}\)

Area A = l\(\sqrt{4a^{2}-l^{2}}\)

\(\frac{dA}{dl}\) = \(\sqrt{4a^{2}-l^{2}}\) + l \(\frac{1}{2\sqrt{4a^{2}-l^{2}}}\) (-2l) = \(\sqrt{4a^{2}-l^{2}}\) – \(\frac{l^{2}}{\sqrt{4a^{2}-l^{2}}}\)

= \(\frac{4a^{2}-2l^{2}}{\sqrt{4a^{2}-l^{2}}}\)

\(\frac{d^2A}{dl^2}\) = \(\frac{\sqrt{4a^{2}-l^{2}}(-4l)-(4a^{2}-2l^{2})\frac{-2l}{2\sqrt{4a^{2}-l^{2}}}}{4a^{2}-l^{2}}\)

= \(\frac{(4a^{2}-l^{2})(-4l)+(4a^{2}-2l^{2})}{(4a^{2}-l^{2})^{\frac{3}{2}}}\)

=\(\frac{-2l(6a^{2}-l^{2})}{(4a^{2}-l^{2})^{\frac{3}{2}}}\)

Now, \(\frac{dA}{dl}\) = 0

4a2 = 2l2

l = \(\sqrt{2}a\)

b = \(\sqrt{4a^{2}-2a^{2}}\) = \(\sqrt{2}a\)

When l = \(\sqrt{2}a\)

\(\frac{d^2A}{dl^2}\) = \(\frac{-2(\sqrt{2}a)(6a^{2}-2a^{2})}{2\sqrt{2}a^{3}}\) = \(\frac{-8\sqrt{2}a^{3}}{2\sqrt{2}a^{3}}\) = -4 < 0

By second derivative, when l = \(\sqrt{2}a\)

Area of the rectangle is maximum.

Since l = b = \(\sqrt{2}a\), the rectangle is a square.

Hence, it has been proved that of all the rectangles inscribed in the given fixed circle, the square has the maximum area.

Que.17. Show that the height of the right circular cylinder of given surface and maximum volume is equal to the diameter of the base.

Ans. Let a and h be the radius and height of the cylinder.

S = 2\(\pi\)a2 + 2\(\pi\)ah

h = \(\frac{S-2\pi a^{2}}{2\pi a}\)

= \(\frac{S}{2\pi }\frac{1}{a}\) – a

V = \(\pi a^{2}h\) = \(\pi\)a2(\(\frac{S}{2\pi }\frac{1}{a}\) – a) = \(\frac{Sa}{2}-\pi a^{3}\)

\(\frac{dV}{da}\) = \(\frac{S}{2}\) – \(3\pi a^{2}\)

\(\frac{d^2V}{da^2}\) = -6\(\pi\)a

\(\frac{dV}{da}\) = 0

\(\frac{S}{2}\) = \(3\pi a^{2}\)

a2 = \(\frac{S}{6\pi }\)

When, a2 = \(\frac{S}{6\pi }\), then \(\frac{d^2V}{da^2}\) = -6\(\pi\) \(\sqrt{}\frac{S}{6\pi }\) < 0

By second derivative, when a3 = \(\frac{S}{6\pi }\) it has the maximum value.

When a2 = \(\frac{S}{6\pi }\), then h =\(\frac{6\pi a^{2}}{2\pi }\frac{1}{a}\) – a = 2a.

Thus, when the height is twice the radius, the volume is maximum.

Que.18. For all right circular closed cylinder of volume 100 cubic cm, find the dimension of can which has the minimum surface area.

Ans. Let a and h be the radius and height

V = \(\pi a^{2}h\) = 100

h = \(\frac{100}{\pi a^{2}}\)

S = \(\pi\)a2h + 2\(\pi\)ah = 2\(\pi\)a2 + \(\frac{200}{a}\)

\(\frac{dS}{da}\) = 4\(\pi\)a + \(\frac{200}{a^{2}}\)

\(\frac{d^2S}{da^2}\) = 4\(\pi\) + \(\frac{400}{a^{3}}\)

\(\frac{dS}{da}\) = 0

4\(\pi\)a = \(\frac{200}{a^{2}}\)

a3 = \(\frac{200}{4\pi }\) = \(\frac{50}{\pi }\)

a = \((\frac{50}{\pi })^{\frac{1}{3}}\)

Since a = \((\frac{50}{\pi })^{\frac{1}{3}}\), \(\frac{d^2S}{da^2}\) > 0

By second derivative, When the radius is \((\frac{50}{\pi })^{\frac{1}{3}}\) cm, the surface area is minimum.

When a = \((\frac{50}{\pi })^{\frac{1}{3}}\), h = \(\frac{100}{\pi (\frac{50}{\pi })^{\frac{2}{3}}}\) = 2 \((\frac{50}{\pi })^{\frac{1}{3}}\) cm.

Therefore, the dimensions are a = \((\frac{50}{\pi })^{\frac{1}{3}}\) cm

h = 2 \((\frac{50}{\pi })^{\frac{1}{3}}\) cm

Que.19. A wire is cut into two pieces whose length is 28 m. A circle and a square are made from those pieces. What length should be the pieces such that the combined area of the circle and the square is minimum?

Ans. Let l be the length of one piece.

So, the length of the other piece is (28 – l) m

Side of square = \(\frac{l}{4}\)

Radius of the circle is a

\(2\pi a\) = 28 – l

a = \(\frac{1}{2\pi }(28-l)\)

Total area A = \(\frac{l^{2}}{16}\) + \(\pi (\frac{1}{2\pi }(28-l))^{2}\)

=\(\frac{l^{2}}{16}\) + \(\frac{1}{4\pi }(28-l)^{2}\)

\(\frac{dA}{dl}\) = \(\frac{2l}{16}+\frac{2}{4\pi }(28-l)(-1)\) = \(\frac{l}{8}-\frac{1}{2\pi }(28-l)\)

\(\frac{d^2A}{dl^2}\) = \(\frac{1}{8}+\frac{1}{2\pi }\) > 0

Que.20. Prove that the volume of the largest cone that can be inscribed in a sphere of radius R is \(\frac{8}{27}\) of thevolume of the sphere.

5.23.1

Ans:

Let r and h be the radius and height of the cone respectively inscribed in a sphere of radius R.

Let V be the volume of the cone.

Then, \(\frac{1}{3}\Pi r^{2} h\)

Height of the cone is given by,

h = R + AB = R + \(\sqrt{R^{2}-r^{2}}\)

\( V=\frac{1}{3}\Pi r^{2}\left ( R +\sqrt{R^{2}-r^{2}} \right )\)

=\(\frac{1}{3}\Pi r^{2} R +\frac{1}{3}\Pi r^{2}\sqrt{R^{2}-r^{2}}\)

\( \frac{dV}{dr}=\frac{2}{3}\Pi rR +\frac{2}{3}\Pi r\sqrt{R^{2}-r^{2}}+\frac{1}{3} \Pi r^{2}.\frac{-2r}{2\sqrt{R^{2}-r^{2}}}\) \(=\frac{2}{3}\Pi rR +\frac{2}{3}\Pi r\sqrt{R^{2}-r^{2}}-\frac{1}{3} \Pi \frac{r^{3}}{{R^{2}-r^{2}}}\) \(=\frac{2}{3}\Pi rR +\frac{2\Pi r(R^{2}-r^{2})-\Pi r^{3}}{3\sqrt{R^{2}-r^{2}}}\) \(=\frac{2}{3}\Pi rR +\frac{2\Pi rR^{2}-3\Pi r^{3}}{3\sqrt{R^{2}-r^{2}}}\) \(\frac{d^{2}V}{dr^{2}}=\frac{2\Pi R}{3}+\frac{3\sqrt{R^{2}-r^{2}}(2\Pi R^{2}-9\Pi r^{2})-(2\Pi rR^{2}-3\Pi r^{3}).\frac{(-2r)}{6\sqrt{R^{2}-r^{2}}}}{9(R^{2}-r^{2})}\) \(\Rightarrow 2R=\frac{3r^{2}-2R^{2}}{\sqrt{R^{2}-r^{2}}}\Rightarrow 2R\sqrt{R^{2}-r^{2}}=3r^{2}-2R^{2}\) \(\Rightarrow 4R^{2}\left ( R^{2} -r^{2}\right )=(3r^{2}-2R^{2})^{2}\) \(\Rightarrow 4R^{4}-4 R^{2} r^{2}=9r^{4}+4R^{4}-12r^{2}R^{2}\) \(\Rightarrow 9r^{4}=8R^{2}r^{2}\) \(\Rightarrow r^{2}=\frac{8}{9}R^{2}\)

When \(r^{2}=\frac{8}{9}R^{2}\), then \(\frac{d^{2}V}{dr^{2}}\)<0.

\( \) by second derivative test, the volume of thecone is the maximum when \(\Rightarrow r^{2}=\frac{8}{9}R^{2}\).

\(when r^{2}=\frac{8}{9}R^{2},h=R+\sqrt{R^{2}-\frac{8}{9}R^{2}}=R+\sqrt{\frac{1}{9}R^{2}}=R+\frac{R}{3}=\frac{4}{3}R\)

Therefore,

= \(\frac{1}{3}\Pi \left ( \frac{8}{9} R^{2}\right )\left ( \frac{4}{3} R\right )\)

= \(\frac{8}{27} \left ( \frac{4}{3} \Pi R^{3}\right )\)

=\(\frac{8}{27}\) x (volume of the sphere)

Hence, the volume of the largest cone that can be inscribed in a sphere is \(\frac{8}{27}\) the volume of the sphere.

Que.21. Show that the right circular cone of least curved surface and given volume has an altitude equal to \(\sqrt{2}\) time the radius of the base.

Ans:

Let r and h be the radius and the height(altitude) of the cone respectively. Then, the volume(V) of the cone is given as:

V=\(\frac{1}{3\prod }\prod r^{2}h\Rightarrow h=\frac{3V}{v^{2}}\)

The surface area(S) of the cone is given by,

S=nrl (where I is the slant height)

=\(\Pi r\sqrt{r^{2}+h^{2}}\)

=\(\Pi r\sqrt{r^{2}+\frac{9V^{2}}{\Pi ^{2}r^{4}}}=\Pi \frac{r\sqrt{9^{2}r^{6}+V^{2}}}{\Pi r^{2}}\)

\(\frac{1}{r}\sqrt{\Pi 2r^{6}+9V^{2}}\) \( \frac{ds}{dr} =\frac{\frac r.{6\Pi ^{2}r^{5}}{2\Pi \sqrt{r^{6}9V^{2}}}-\sqrt{\Pi 2}r^{6}+9V^{2}}{r^{2}}\) \(=\frac{3\Pi ^{2}r^{6}-\Pi ^{2}r^{6}-9V^{2}}{r^{2}\sqrt{\Pi ^{2}r^{6}+9V^{2}}}\)

= \(=\frac{2\Pi ^{2}r^{6}-9V^{2}}{r^{2}\sqrt{\Pi ^{2}r^{6}+9V^{2}}}\)

Now, \(\frac{ds}{dr}=0\Rightarrow 2\Pi ^\frac{}{}{2}r^{6}=9V^{2}\Rightarrow r^{6}=\frac{9V^{2}}{2\Pi ^{2}}\)

Thus, it can be easily verified that when \(r^{6}=\frac{9V^{2}}{2\Pi ^{2}},\frac{d^{2}S}{dr^{2}}>0.\)

By second derivative test, the surface area of the cone is the least when

\(\frac{3V}{\Pi r^{2}}=\frac{3}{\Pi r^{2}}(\frac{2\Pi ^{2}r^{6}}{9})^{\frac{1}{2}}=\frac{3}{\Pi r^{2}}.\frac{\sqrt{2\Pi r^{3}}}{3}=\sqrt{2r}\)
Hence, for a given volume, the right circular cone of the least curved surface has an altitude equal to \(\sqrt{2}\) time the radius of the base.

\(\frac{dA}{dl}\) = 0

(\(\pi\) + 4)l – 112 = 0

l = \(\frac{112}{\pi + 4}\)

Thus, when l = \(\frac{112}{\pi + 4}\), \(\frac{d^2A}{dl^2}\) > 0.

By second derivative when l = \(\frac{112}{\pi + 4}\), the area is minimum.

The length of the wire used in making the circle is 28 – \(\frac{112}{\pi + 4}\) = \(\frac{28\pi }{\pi +4}\) cm.

Que.22. Show that the semi vertical angle of the cone of the maximum volume and of given slant height is \(tan^{-1}\sqrt{2}\).

5.25.1

Ans: Let \(\Theta\) be the semi-vertical angle of the cone.

It is clear that \(\Theta\) \(\in\) \(\left [ 0,\frac{\prod }{2} \right ]\).

Let r, h and I be the radius height and the slant height of the cone respectively.

The slant height of the cone is given as constant.

Now, r = I sin \(\Theta\) and h = I cos \(\Theta\).

The volume(v) of the cone is given by,

\(v=\frac{1}{3}\prod r^{2}h\)

= \(\frac{1}{3}\prod (l^{2}sin^{2}\theta)(l cos \theta )\)

= \(\frac{1}{3}\prod l^{3}sin^{2}\theta cos \theta\)

\( \frac{dv}{d\theta }=\frac{l^{3}\prod }{3}\left [ sin^{2} \theta (-sin\theta )+cos\theta (2sin\theta cos\theta )\right ]\)

=\(\frac{l^{3}\prod }{3}\left [ -sin^{3}+2sin\theta cos^{2}\theta \right ]\)

\(\frac{d^{2}v}{d\theta ^{2}}=\frac{l^{3}\prod }{3}\left [ -3sin^{2}\theta cos\theta +2cos^{3}\theta -4sin^{2}\theta cos\theta \right ]\)

= \(=\frac{l^{3}\prod }{3}\left [ 2cos^{3}\theta -7sin^{2}\theta cos\theta \right ]\)

Now, \(\frac{dv}{d\theta }=0\)

\(\Rightarrow sin ^{3}\theta =2sin\theta cos^{2}\theta\) \(\Rightarrow tan ^{2}\theta =2\) \(\Rightarrow tan \theta =\sqrt{2}\) \(\Rightarrow \theta =tan^{-1}\sqrt{2}\)

Now, when \( \theta =tan^{-1}\sqrt{2}\), then \(\Rightarrow tan ^{2}\theta =2\)

Or \(\Rightarrow sin ^{2}\theta =2 cos^{2}\theta\).

Then we have:

\(\frac{dv^{2}}{d\theta ^{2}} = 2y which is nearest to the point =\frac{l^{3}\prod }{3}\left [ 2cos ^{3}\theta -14cos^{3}\theta \right ]=-4\prod l^{3}cos^{3}\theta <0for\theta \varepsilon [0,\frac{\prod }{2}]\)

By second derivative test, the volume(v) is the maximum when \(\Rightarrow \theta =tan^{-1}\sqrt{2}\).

Hence, for a given slant height, the semi- vertical angle of the cone of the maximum volume is \(tan^{-1}\sqrt{2}\).

Que.23. The point on the curve x2= 2y ehich is nearest to the point (0 , 5) is

(A) (2\(\sqrt{2},4)\)

(B) (2\(\sqrt{2},0)\)

(C) (0,0)

(D) (2,2)

Ans:

The given curve is x2= 2y.

For each value of x, the position of the point will be \((x,\frac{x^{2}}{2}).\)

The distance d(x) between the points \((x,\frac{x^{2}}{2}).\) and (0,5) is given by,

\(d(x)=\sqrt{(x-0)^{2}+(\frac{x^{2}}{2}-5)^{2})}=\sqrt{x^{2}+\frac{x^{4}}{4}+25-5x^{2}}=\sqrt{\frac{x^{4}}{4}-4x^{2}+25}\).

\( d'(x)=\frac{(x^{3}-8x)}{2\sqrt{\frac{x^{4}}{4}-4x^{2}+25}}=\frac{(x^{3}-8x)}{\sqrt{x^{4}-16x^{2}+100}}\)

Now, d’’(x) = 0\(\Rightarrow\) x3-8x=0

\(\Rightarrow\) x(x2-8)=0

\(\Rightarrow\) x=0, \(\pm\)2 \(\sqrt{2}\).

\(And,d”(x)=\frac{\sqrt{x^{4}-16x^{2}+100}(3x^{2}-8)-(x^{3}-8x).\frac{4x^{3}-32x}{2\sqrt{x^{4}-16x^{2}+100}}}{(x^{4}-16x^{2}+100)}\)

= \(\frac{{(x^{4}-16x^{2}+100)}(3x^{2}-8)-2(x^{3}-8x)(x^{3}-8x)}{(x^{4}-16x^{2}+100)^{\frac{3}{2}}}\)

\(=\frac{{(x^{4}-16x^{2}+100)}(3x^{2}-8)-2(x^{3}-8x)^{2}}{(x^{4}-16x^{2}+100)^{\frac{3}{2}}}\)

When, x = 0, then d’’(x)= \(\frac{(36(-8))}{6^{3}}<0\)

When, x = \(\pm\) 2 \(\sqrt{2}\),d’’(X)>0.

\( \) by second derivative test,d(x) is the minimum at x = \(\pm\)2 \(\sqrt{2}\)

When x=\(\pm\)2 \(\sqrt{2}\),y=\(\frac{(2\sqrt{2})^{2}}{2}=4\).

Hence, the point on the curve x2=2y which is nearest to the point(0,5) is (+2 \(\sqrt{2}\),4).

The correct answer is A.

Que.24. For all real values of x, the minimum value of \(\frac{1-x+x^{2}}{1+x+x^{2}}is\)

(A) 0

(B) 1

(C) 3

(D) \(\frac{1}{3}\)

Ans:

Let f(x)= \(\frac{1-x+x^{2}}{1+x+x^{2}}is\)

\( f'(x)=\frac{(1+x+x^{2})(-1+2x)-(1-x+x^{2})(1+2x)}{(1+x+x^{2})^{2}}\)

= \(\frac{-1+2x-x+2x^{2}-x^{2}+2x^{3}-1-2x+x+2x^{2}-x^{2}-2x^{3}}{(1+x+x^{2})^{2}}\)

= \(= \frac{2x^{2}-2}{(1+x+x^{2})^{2}}= \frac{2(x^{2}-1)}{(1+x+x^{2})^{2}}\)

\( f'(x)=0\Rightarrow x^{2}=1\Rightarrow x=\pm 1\)

Now, f’’(x) = \(\frac{2\left [ (1+x+x^{2})^{2}(2x)-(x^{2}-1)(2)(1+x+x^{2})(1+2x) \right ]}{(1+x+x^{2})^{4}}\)

\(=\frac{4(1+x+x^{2})[(1+x+x^{2})x-(x^{2}-1)(1+2x)]}{(1+x+x^{2})^{4}}\) \(=\frac{4[x+x^{2}+x^{3}-x^{2}-2x^{3}+1+2x]}{(1+x+x^{2})^{3}}\) \(=\frac{4[1+3x-x^{3}]}{(1+x+x^{2})^{3}}\)

And, f’’(1)= \(\frac{4(1+3-1)}{(1+1+1)^{3}}=\frac{4(3)}{(3)^{3}}=\frac{4}{9}>0\)

Also, f’’(-1)= \(\frac{4(1-3+1)}{(1-1+1)^{3}}=4(-1)=-4<0\)

\( \) by second derivative test, f is the minimum at x = 1 and the minimum value is given by f(1)= \(\frac{1-1+1}{1+1+1}\)= \(\frac{1}{3}\)

The correct ans is D.

 

Que.25. The maximum value of \(\left [ x(x-1)+1 \right ]^{\frac{1}{3}},0\leq x\leq 1\) is

(A) \(\frac({1}{3})^{\frac{1}{3}}\)

(B) \(\frac{1}{2}\)

(C) 1

(D) 0

Ans:

Let f(x)= \(\left [ x(x-1)+1 \right ]^{\frac{1}{3}}\)

\(f'(x)=\left [\frac{2x-1}{3[x(x-1)+1 ]} \right ]^{\frac{2}{3}}\)

Now, f’(x)=0\(\Rightarrow x=\frac{1}{2}\)

Then, we evaluate the value of f at critical point \(x=\frac{1}{2}\) and at the end points of the interval [0,1] {i.e. at x=0 and x=1).

\(f(0)=[0(0-1)+1]^{\frac{1}{3}}=1\) \(f(1)=[1(1-1)+1]^{\frac{1}{3}}=1\) \(f(\frac{1}{2})=\left [ \frac{1}{2}(\frac{-1}{2})+1 \right ]^{\frac{}{3}}=(\frac{3}{4})^{\frac{1}{3}}\)

Hence, we can conclude that the maximum value of f in the interval[0,1] is 1.

The correct answer is C.