# NCERT Solutions For Class 12 Maths Chapter 6 Exercise 6.5

## Chapter 6 Ex.6.5 Question 1

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

(i) $$f\left( x \right) = {\left( {2x - 1} \right)^2} + 3$$

(ii) $$f\left( x \right) = 9{x^2} + 12x + 2$$

(iii) $$f\left( x \right) = - {\left( {x - 1} \right)^2} + 10$$

(iv) $$g\left( x \right) = {\left( x \right)^3} + 1$$

### Solution

(i) The given function is $$f\left( x \right) = {\left( {2x - 1} \right)^2} + 3$$

It can be observed that $${\left( {2x - 1} \right)^2} \ge 0$$ for every $$x \in {\bf{R}}$$.

Therefore, $$f\left( x \right) = {\left( {2x - 1} \right)^2} + 3 \ge 3$$ for every $$x \in {\bf{R}}$$.

The minimum value of $$f$$ is attained when $$2x - 1 = 0$$

\begin{align}&2x - 1 = 0\\& \Rightarrow \; x = \frac{1}{2}\end{align}

Hence, minimum value of

\begin{align}f &= f\left( {\frac{1}{2}} \right)\\ &= {\left( {2\left( {\frac{1}{2}} \right) - 1} \right)^2} + 3\\& = 3\end{align}

Thus, the function $$f$$ does not have a maximum value.

(ii) The given function is $$f\left( x \right) = 9{x^2} + 12x + 2$$

It can be observed that $${\left( {3x + 2} \right)^2} \ge 0$$ for every $$x \in {\bf{R}}$$.

Therefore, $$f\left( x \right) = {\left( {3x + 2} \right)^2} - 2 \ge - 2$$ for every $$x \in {\bf{R}}$$.

The minimum value of $$f$$ is attained when $$3x + 2 = 0$$

\begin{align}&3x + 2 = 0\\& \Rightarrow \; x = \frac{{ - 2}}{3}\end{align}

Therefore, Minimum value of

\begin{align}f &= f\left( { - \frac{2}{3}} \right)\\ &= {\left( {3\left( { - \frac{2}{3}} \right) + 2} \right)^2} - 2\\ &= - 2\end{align}

Hence, the function $$f$$ does not have a maximum value.

(iii) The given function is $$f\left( x \right) = - {\left( {x - 1} \right)^2} + 10$$

It can be observed that $${\left( {x - 1} \right)^2} \ge 0$$ for every $$x \in {\bf{R}}$$.

Therefore, $$f\left( x \right) = - {\left( {x - 1} \right)^2} + 10 \le 10$$ for every $$x \in {\bf{R}}$$.

The maximum value of f is attained when $$\left( {x - 1} \right) = 0$$

\begin{align}&\left( {x - 1} \right) = 0\\ &\Rightarrow \; x= 1\end{align}

Therefore, Maximum value of

\begin{align}f &= f\left( 1 \right)\\& = - {\left( {1 - 1} \right)^2} + 10\\ &= 10\end{align}

Hence, the function $$f$$ does not have a minimum value.

(iv) The given function is $$g\left( x \right) = {\left( x \right)^3} + 1$$

Hence, function $$g$$ neither has a maximum value nor a minimum value.

## Chapter 6 Ex.6.5 Question 2

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

(i) $$f\left( x \right) = \left| {x + 2} \right| - 1$$

(ii) $$g\left( x \right) = - \left| {x + 1} \right| + 3$$

(iii) $$h\left( x \right) = \sin \left( {2x} \right) + 5$$

(iv) $$f\left( x \right) = \left| {\sin 4x + 3} \right|$$

(v) $$h\left( x \right) = x + 1,x \in \left( { - 1,1} \right)$$

### Solution

(i) The given function is $$f\left( x \right) = \left| {x + 2} \right| - 1$$

It can be observed that $$\left| {x + 2} \right| \ge 0$$ for every $$x \in {\bf{R}}$$.

Therefore, $$f\left( x \right) = \left| {x + 2} \right| - 1 \ge - 1$$ for every $$x \in {\bf{R}}$$.

The minimum value of f is attained when $$\left| {x + 2} \right| = 0$$

\begin{align}&\left| {x + 2} \right| = 0\\ &\Rightarrow \; x = - 2\end{align}

Therefore, Minimum value of

\begin{align}f& = f\left( { - 2} \right)\\ &= \left| { - 2 + 1} \right| - 1\\ &= - 1\end{align}

Hence, the function $$f$$ does not have a maximum value.

(ii) The given function is $$g\left( x \right) = - \left| {x + 1} \right| + 3$$

It can be observed that $$- \left| {x + 1} \right| \le 0$$ for every $$x \in {\bf{R}}$$.

Therefore, $$f\left( x \right) = - \left| {x + 1} \right| + 3 \le 3$$ for every $$x \in {\bf{R}}$$.

The maximum value of g is attained when $$\left| {x + 1} \right| = 0$$

\begin{align}&\left| {x + 1} \right| = 0\\& \Rightarrow \; x = - 1\end{align}

Therefore, maximum value of

\begin{align}g &= g\left( { - 1} \right)\\ &= - \left| { - 1 + 1} \right| + 3\\& = 3\end{align}

Hence, the function $$g$$ does not have a minimum value.

(iii) The given function is $$h\left( x \right) = \sin \left( {2x} \right) + 5$$

We know that $$- 1 \le \sin 2x \le 1$$

Therefore,

\begin{align}& \Rightarrow \; - 1 + 5 \le \sin 2x \le 1 + 5\\ &\Rightarrow \; 4 \le \sin 2x + 5 \le 6\end{align}

Hence, the maximum and minimum values of $$h$$ are 6 and 4, respectively.

(iv) The given function is $$f\left( x \right) = \left| {\sin 4x + 3} \right|$$

We know that $$- 1 \le \sin 4x \le 1$$

Therefore,

\begin{align} &\Rightarrow \; 2 \le \sin 4x + 3 \le 4\\ &\Rightarrow \; 2 \le \left| {\sin 4x + 3} \right| \le 4\end{align}

Hence, the maximum and minimum values of $$f$$ are 4 and 2, respectively.

(v) The given function is $$h\left( x \right) = x + 1,x \in \left( { - 1,1} \right)$$

Here, if a point $${x_0}$$ is closest to $$- 1$$ , then we find $$\frac{{{x_0}}}{2} + 1 > {x_0} + 1$$ for all $${x_0} \in \left( { - 1,1} \right)$$.

Also, if $${x_1}$$ is closest to 1, then find $${x_1} + 1 < \frac{{{x_1} + 1}}{2} + 1$$ for all $${x_1} \in \left( { - 1,1} \right)$$.

Hence, function $$h\left( x \right)$$ has neither maximum nor minimum value in $$\left( { - 1,1} \right)$$.

## Chapter 6 Ex.6.5 Question 3

Find the local maxima and minima, if any, of the following functions. Find also the local maximum and the local minimum values, as the case may be:

(i) $$f\left( x \right) = {x^2}$$

(ii) $$g\left( x \right) = {x^3} - 3x$$

(iii) $$h\left( x \right) = \sin x + \cos x,0 < x < \frac{\pi }{2}$$

(iv) $$f\left( x \right) = \sin x - \cos x,0 < x < 2\pi$$

(v) $$f\left( x \right) = {x^3} - 6{x^2} + 9x + 15$$

(vi) $$g\left( x \right) = \frac{x}{2} + \frac{2}{x},x > 0$$

(vii) $$g\left( x \right) = \frac{1}{{{x^2} + 2}}$$

(viii) $$f\left( x \right) = x\sqrt {1 - x} ,0 < x < 1$$

### Solution

(i) $$f\left( x \right) = {x^2}$$

Therefore,

$$f'\left( x \right) = 2x$$

Now,

\begin{align}&f'\left( x \right) = 0\\ &\Rightarrow \; x = 0\end{align}

Thus, $$x = 0$$ is the only critical point which could possibly be the point of local maxima or local minima of $$f$$.

We have $$f''\left( 0 \right) = 2$$, which is positive.

Therefore, by second derivative test, $$x = 0$$ is a point of local minima and local minimum value of $$f$$ at $$x = 0$$ is $$f\left( 0 \right) = 0$$.

(ii) $$g\left( x \right) = {x^3} - 3x$$

Therefore,

$$g'\left( x \right) = 3{x^2} - 3$$

Now,

\begin{align}&g'\left( x \right) = 0\\ &\Rightarrow \; 3{x^2} - 3 = 0\\ &\Rightarrow \; x = \pm 1\end{align}

Also,

\begin{align}&g''\left( x \right) = 6x\\&g''\left( 1 \right) = 6 > 0\\&g''\left( { - 1} \right) = - 6 < 0\end{align}

By second derivative test, $$x = 1$$ is a point of local minima and local minimum value of $$g$$ at $$x = 1$$ is

\begin{align}g\left( 1 \right) &= {1^3} - 3\\& = 1 - 3\\& = - 2\end{align}

However, $$x = - 1$$ is a point of local maxima and local maximum value of $$g$$ at $$x = - 1$$ is

\begin{align}g\left( { - 1} \right)&= {\left( { - 1} \right)^3} - 3\left( { - 1} \right)\\ &= - 1 + 3\\ &= 2\end{align}

(iii) $$h\left( x \right) = \sin x + \cos x,0 < x < \frac{\pi }{2}$$

Therefore,

$$h'\left( x \right) = \cos x - \sin x$$

Now,

\begin{align}&h'\left( x \right) = 0\\& \Rightarrow \; \cos x - \sin x = 0\\ &\Rightarrow \; \sin x = \cos x\\& \Rightarrow \; \tan x = 1\\& \Rightarrow \; x = \frac{\pi }{4} \in \left( {0,\frac{\pi }{2}} \right)\end{align}

Also,

\begin{align}h''\left( x \right) &= - \sin x - \cos x\\ &= - \left( {\sin x + \cos x} \right)\end{align}

Hence,

\begin{align}h''\left( {\frac{\pi }{4}} \right)& = - \left( {\frac{1}{{\sqrt 2 }} + \frac{1}{{\sqrt 2 }}} \right)\\ &= \frac{{ - 2}}{{\sqrt 2 }}\\& = - \sqrt 2 < 0\end{align}

Therefore, by second derivative test, $$x = \frac{\pi }{4}$$ is a point of local maxima and the local maximum value of $$h$$ at $$x = \frac{\pi }{4}$$ is

\begin{align}h\left( {\frac{\pi }{4}} \right) &= \sin \frac{\pi }{4} + \cos \frac{\pi }{4}\\ &= \frac{1}{{\sqrt 2 }} + \frac{1}{{\sqrt 2 }}\\ &= \sqrt 2 \end{align}

(iv) $$f\left( x \right) = \sin x - \cos x,0 < x < 2\pi$$

Therefore,

$$f'\left( x \right) = \cos x + \sin x$$

Now,

\begin{align}&f'\left( x \right) = 0\\ &\Rightarrow \; \cos x + \sin x = 0\\ &\Rightarrow \; \sin x = - \cos x\\ &\Rightarrow \; \tan x = - 1\\& \Rightarrow \; x = \frac{{3\pi }}{4},\frac{{7\pi }}{4} \in \left( {0,2\pi } \right)\end{align}

Also,

$$f''\left( x \right) = - \sin x + \cos x$$

Hence,

\begin{align}f''\left( {\frac{{3\pi }}{4}} \right)& = - \sin \left( {\frac{{3\pi }}{4}} \right) + \cos \left( {\frac{{3\pi }}{4}} \right)\\& = \left( { - \frac{1}{{\sqrt 2 }} - \frac{1}{{\sqrt 2 }}} \right)\\& = - \sqrt 2 < 0\\f''\left( {\frac{{7\pi }}{4}} \right)& = - \sin \left( {\frac{{7\pi }}{4}} \right) + \cos \left( {\frac{{7\pi }}{4}} \right)\\& = \left( {\frac{1}{{\sqrt 2 }} + \frac{1}{{\sqrt 2 }}} \right)\\ &= \sqrt 2 > 0\end{align}

Therefore, by second derivative test, $$x = \left( {\frac{{3\pi }}{4}} \right)$$ is a point of local maxima and the local maximum value of $$f$$ at $$x = \left( {\frac{{3\pi }}{4}} \right)$$ is

\begin{align}f\left( {\frac{{3\pi }}{4}} \right)& = \sin \left( {\frac{{3\pi }}{4}} \right) - \cos \left( {\frac{{3\pi }}{4}} \right)\\ &= \left( {\frac{1}{{\sqrt 2 }} + \frac{1}{{\sqrt 2 }}} \right)\\& = \sqrt 2 \end{align}

However, $$x = \left( {\frac{{7\pi }}{4}} \right)$$ is a point of local minima and the local minimum value of $$f$$ at $$x = \left( {\frac{{7\pi }}{4}} \right)$$ is

\begin{align}f\left( {\frac{{7\pi }}{4}} \right) &= \sin \left( {\frac{{7\pi }}{4}} \right) - \cos \left( {\frac{{7\pi }}{4}} \right)\\ &= \left( { - \frac{1}{{\sqrt 2 }} - \frac{1}{{\sqrt 2 }}} \right)\\ &= - \sqrt 2 \end{align}

(v) $$f\left( x \right) = {x^3} - 6{x^2} + 9x + 15$$

Therefore,

$$f'\left( x \right) = 3{x^2} - 12x + 9$$

Now,

\begin{align}&f'\left( x \right) = 0\\& \Rightarrow \; 3{x^2} - 12x + 9 = 0\\& \Rightarrow \; 3\left( {x - 1} \right)\left( {x - 3} \right) = 0\\ &\Rightarrow \; x = 1,3\end{align}

Also,

\begin{align}f''\left( x \right) &= 6x - 12\\ &= 6\left( {x - 2} \right)\end{align}

Hence,

\begin{align}f''\left( 1 \right)&= 6\left( {1 - 2} \right) = - 6 < 0\\f''\left( 3 \right) &= 6\left( {3 - 2} \right) = 6 > 0\end{align}

Therefore, by second derivative test, $$x = 1$$ is a point of local maxima and the local maximum value of $$f$$ at $$x = 1$$ is

\begin{align}f\left( 1 \right) &= 1 - 6 + 9 + 15\\& = 19\end{align}

However, $$x = 3$$ is a point of local minima and the local minimum value of $$f$$ at $$x = 3$$ is

\begin{align}f\left( 3 \right)&= 27 - 54 + 27 + 15\\& = 15\end{align}

(vi) $$g\left( x \right) = \frac{x}{2} + \frac{2}{x},x > 0$$

Therefore,

$$g'\left( x \right) = \frac{1}{2} - \frac{2}{{{x^2}}}$$

Now,

\begin{align}&g'\left( x \right) = 0\\ &\Rightarrow \; \frac{1}{2} - \frac{2}{{{x^2}}} = 0\\& \Rightarrow \; \frac{2}{{{x^2}}} = \frac{1}{2}\\& \Rightarrow \; {x^2} = 4\\ &\Rightarrow \; x = \pm 2\end{align}

Since, $$x > 0$$, we take $$x = 2$$

Hence,

\begin{align}g''\left( x \right) &= \frac{4}{{{x^3}}}\\g''\left( 2 \right) &= \frac{4}{{{2^3}}} = \frac{1}{2} > 0\end{align}

Therefore, by second derivative test, $$x = 2$$ is a point of local minima and the local minimum value of $$g$$ at $$x = 2$$ is

\begin{align}g\left( 2 \right)& = \frac{2}{2} + \frac{2}{2}\\&= 1 + 1\\ &= 2\end{align}

(vii) $$g\left( x \right) = \frac{1}{{{x^2} + 2}}$$

Therefore,

$$g'\left( x \right) = \frac{{ - \left( {2x} \right)}}{{{{\left( {{x^2} + 2} \right)}^2}}}$$

Now,

\begin{align}&g'\left( x \right) = 0\\& \Rightarrow \; \frac{{ - \left( {2x} \right)}}{{{{\left( {{x^2} + 2} \right)}^2}}} = 0\\ &\Rightarrow \; x = 0\end{align}

Now, for values close to $$x = 0$$ and to the left of 0, $$g'\left( x \right) > 0$$.

Also, for values close to $$x = 0$$ and to the right of 0, $$g'\left( x \right) < 0$$.

Therefore, by first derivative test, $$x = 0$$ is a point of local maxima and the local maximum value of

\begin{align}g\left( 0 \right)& = \frac{1}{{0 + 2}}\\ &= \frac{1}{2}\end{align}

(viii) $$f\left( x \right) = x\sqrt {1 - x} ,0 < x < 1$$

Therefore,

\begin{align}f'\left( x \right) &= \sqrt {1 - x} + x\left( {\frac{1}{{2\sqrt {1 - x} }}} \right)\left( { - 1} \right)\\ &= \sqrt {1 - x} - \frac{x}{{2\sqrt {1 - x} }}\\ &= \frac{{2\left( {1 - x} \right) - x}}{{2\sqrt {1 - x} }}\\& = \frac{{2 - 3x}}{{2\sqrt {1 - x} }}\end{align}

Now,

\begin{align}&f'\left( x \right) = 0\\& \Rightarrow \; \frac{{2 - 3x}}{{2\sqrt {1 - x} }} = 0\\ &\Rightarrow \; 2 - 3x = 0\\& \Rightarrow \; x = \frac{2}{3}\end{align}

Also,

\begin{align}f''\left( x \right) &= \frac{1}{2}\left[ {\frac{{\sqrt {1 - x} \left( { - 3} \right) - \left( {2 - 3x} \right)\left( {\frac{{ - 1}}{{2\sqrt {1 - x} }}} \right)}}{{1 - x}}} \right]\\ &= \frac{{\sqrt {1 - x} \left( { - 3} \right) + \left( {2 - 3x} \right)\left( {\frac{1}{{2\sqrt {1 - x} }}} \right)}}{{2\left( {1 - x} \right)}}\\ &= \frac{{ - 6\left( {1 - x} \right) + \left( {2 - 3x} \right)}}{{4\left( {1 - x} \right)\left( {\sqrt {1 - x} } \right)}}\\ &= \frac{{3x - 4}}{{4{{\left( {1 - x} \right)}^{\frac{3}{2}}}}}\end{align}

Hence,

\begin{align}f''\left( {\frac{2}{3}} \right) &= \frac{{3\left( {\frac{2}{3}} \right) - 4}}{{4{{\left( {1 - \frac{2}{3}} \right)}^{\frac{3}{2}}}}}\\ &= \frac{{2 - 4}}{{4{{\left( {\frac{1}{3}} \right)}^{\frac{3}{2}}}}}\\& = \frac{{ - 1}}{{2{{\left( {\frac{1}{3}} \right)}^{\frac{3}{2}}}}} < 0\end{align}

Therefore, by second derivative test, $$x = \frac{2}{3}$$ is a point of local maxima and the local maximum value of $$f$$ at $$x = \frac{2}{3}$$ is

\begin{align}f\left( {\frac{2}{3}} \right) &= \frac{2}{3}\sqrt {1 - \frac{2}{3}} \\& = \frac{2}{3}\sqrt {\frac{1}{3}} \\ &= \frac{2}{{3\sqrt 3 }}\\ &= \frac{{2\sqrt 3 }}{9}\end{align}

## Chapter 6 Ex.6.5 Question 4

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

(i) $$f\left( x \right) = {e^x}$$

(ii) $$g\left( x \right) = \log x$$

(iii) $$h\left( x \right) = {x^3} + {x^2} + x + 1$$

### Solution

(i) $$f\left( x \right) = {e^x}$$

Therefore,

$$f'\left( x \right) = {e^x}$$

Now,

\begin{align}&f'\left( x \right) = 0\\& \Rightarrow \; {e^x} = 0\end{align}

But the exponential function can never assume 0 for any value of $$x$$ .

Therefore, there does not exist $$c \in {\bf{R}}$$ such that $$f'\left( c \right) = 0$$

Hence, function $$f$$ does not have maxima or minima.

(ii) $$g\left( x \right) = \log x$$

Therefore,

$$g'\left( x \right) = \frac{1}{x}$$

Since $$\log x$$ is defined for a positive number $$x,g'\left( x \right) > 0$$ for any $$x$$.

Therefore, there does not exist $$c \in {\mathbf{R}}$$ such that $$g'\left( c \right) = 0$$.

Hence, function $$g$$ does not have maxima or minima.

(iii) $$h\left( x \right) = {x^3} + {x^2} + x + 1$$

Therefore,

$$h'\left( x \right) = 3{x^2} + 2x + 1$$

Now,

\begin{align}&h'\left( x \right) = 0\\& \Rightarrow \; 3{x^2} + 2x + 1 = 0\\& \Rightarrow \; x = \frac{{ - 2 \pm 2\sqrt 2 i}}{6}\\& \Rightarrow \; x = \frac{{ - 1 \pm \sqrt 2 i}}{3} \ne {\bf{R}}\end{align}

Therefore, there does not exist $$c \in {\bf{R}}$$ such that $$h'\left( c \right) = 0$$.

Hence, function $$h$$ does not have maxima or minima.

## Chapter 6 Ex.6.5 Question 5

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

(i) $$f\left( x \right) = {x^3},x \in \left[ { - 2,2} \right]$$

(ii) $$f\left( x \right) = \sin x + \cos x,x \in \left[ {0,\pi } \right]$$

(iii) $$f\left( x \right) = 4x - \frac{1}{2}{x^2},x \in \left[ { - 2,\frac{9}{2}} \right]$$

(iv) $$f\left( x \right) = {\left( {x - 1} \right)^2} + 3,x \in \left[ { - 3,1} \right]$$

### Solution

(i) The given function is $$f\left( x \right) = {x^3}$$

Therefore,

$$f'\left( x \right) = 3{x^2}$$

Now,

\begin{align}&f'\left( x \right) = 0\\ &\Rightarrow \; 3{x^2} = 0\end{align}

Then, we evaluate the value of $$f$$ at critical point $$x = 0$$ and at end points of the interval $$\left[ { - 2,2} \right]$$.

Therefore,

\begin{align}f\left( 0 \right) &= 0\\f\left( { - 2} \right) &= {\left( { - 2} \right)^3}\\ &= - 8\\f\left( 2 \right) = {\left( 2 \right)^3}\\ &= 8\end{align}

Hence, we can conclude that the absolute maximum value of $$f$$ on $$\left[ { - 2,2} \right]$$ is 8 occurring at $$x = 2$$.

Also, the absolute minimum value of $$f$$ on $$\left[ { - 2,2} \right]$$ is $$- 8$$ occurring at $$x = - 2$$.

(ii) The given function is $$f\left( x \right) = \sin x + \cos x$$

Therefore,

$$f'\left( x \right) = \cos x - \sin x$$

Now,

\begin{align}&f'\left( x \right) = 0\\& \Rightarrow \; \cos x - \sin x = 0\\ &\Rightarrow \; \sin x = \cos x\\ &\Rightarrow \; \tan x = 1\\& \Rightarrow \; x = \frac{\pi }{4}\end{align}

Then, we evaluate the value of $$f$$ at critical point $$x = \frac{\pi }{4}$$ and at the end points of the interval $$\left[ {0,\pi } \right]$$.

Therefore,

\begin{align}f\left( {\frac{\pi }{4}} \right) &= \sin \left( {\frac{\pi }{4}} \right) + \cos \left( {\frac{\pi }{4}} \right)\\ &= \frac{1}{{\sqrt 2 }} + \frac{1}{{\sqrt 2 }} = \frac{2}{{\sqrt 2 }}\\ &= \sqrt 2 \end{align}

\begin{align}f\left( 0 \right) &= \sin 0 + \cos 0\\ &= 0 + 1\\& = 1\\f\left( \pi \right) &= \sin \pi + \cos \pi \\& = 0 - 1\\ &= - 1\end{align}

Hence, we can conclude that the absolute maximum value of $$f$$ on $$\left[ {0,\pi } \right]$$ is $$\sqrt 2$$ occurring at $$x = \frac{\pi }{4}$$.

Also, the absolute minimum value of $$f$$ on $$\left[ {0,\pi } \right]$$ is $$- 1$$ occurring at $$x = \pi$$.

(iii) The given function is $$f\left( x \right) = 4x - \frac{1}{2}{x^2}$$

Therefore,

\begin{align}f'\left( x \right) &= 4 - \frac{1}{2}\left( {2x} \right)\\ &= 4 - x\end{align}

Now,

\begin{align}&f'\left( x \right) = 0\\& \Rightarrow \; 4 - x = 0\\ &\Rightarrow \; x = 4\end{align}

Then, we evaluate the value of $$f$$ at critical point $$x = 4$$ and at end points of the interval $$\left[ { - 2,\frac{9}{2}} \right]$$.

Therefore,

\begin{align}f\left( 4 \right) &= 16 - \frac{1}{2}\left( {16} \right)\\ &= 16 - 8\\& = 8\\f\left( { - 2} \right)& = - 8 - \frac{1}{2}\left( 4 \right)\\ &= - 8 - 2\\ &= - 10\\f\left( {\frac{9}{2}} \right) &= 18 - \frac{1}{2}{\left( {\frac{9}{2}} \right)^2}\\ &= 18 - \frac{{81}}{8}\\ &= 18 - 10.125\\ &= 7.875\end{align}

Hence, we can conclude that the absolute maximum value of $$f$$ on $$\left[ { - 2,\frac{9}{2}} \right]$$ is 8 occurring at $$x = 4$$.

Also, the absolute minimum value of $$f$$ on $$\left[ { - 2,\frac{9}{2}} \right]$$ is $$- 10$$ occurring at $$x = - 2$$.

(iv) The given function is $$f\left( x \right) = {\left( {x - 1} \right)^2} + 3$$

Therefore,

$$f'\left( x \right) = 2\left( {x - 1} \right)$$

Now,

\begin{align}&f'\left( x \right) = 0\\& \Rightarrow \; 2\left( {x - 1} \right) = 0\\& \Rightarrow \; x = 1\end{align}

Then, we evaluate the value of $$f$$ at critical point $$x = 1$$ and at end points of the interval $$\left[ { - 3,1} \right]$$.

\begin{align}f\left( 1 \right) &= {\left( {1 - 1} \right)^3} + 3\\ = 3\\f\left( { - 3} \right) &= {\left( { - 3 - 1} \right)^2} + 3\\ &= 16 + 3\\ &= 19\end{align}

Hence, we can conclude that the absolute maximum value of $$f$$ on $$\left[ { - 3,1} \right]$$ is 19 occurring at $$x = - 3$$.

Also, the absolute minimum value of $$f$$ on $$\left[ { - 3,1} \right]$$ is 3 occurring at $$x = 1$$.

## Chapter 6 Ex.6.5 Question 6

Find the absolute maximum profit that a company can make, if the profit function is given by:$$p\left( x \right) = 41 - 24x - 18{x^2}$$

### Solution

The profit function is given as $$p\left( x \right) = 41 - 24x - 18{x^2}$$

Therefore,

$$p'\left( x \right) = - 24 - 36x$$

Now,

\begin{align}&p'\left( x \right) = 0\\ &\Rightarrow \; 24 - 36x = 0\\& \Rightarrow \; x = - \frac{{24}}{{36}}\\& \Rightarrow \; x = \frac{{ - 2}}{3}\end{align}

Also,

$$p''\left( {\frac{{ - 2}}{3}} \right) = - 36 < 0$$

By second derivative test, $$x = - \frac{2}{3}$$ is the point of local maxima of $$p$$.

Therefore, maximum profit

\begin{align}p\left( { - \frac{2}{3}} \right)& = 41 - 24\left( { - \frac{2}{3}} \right) - 18{\left( { - \frac{2}{3}} \right)^2}\\ &= 41 + 16 - 8\\& = 49\end{align}

Hence, the maximum profit that the company can make is 49 units.

## Chapter 6 Ex.6.5 Question 7

Find both the maximum value and the minimum value of $$3{x^4} - 8{x^3} + 12{x^2} - 48x + 25$$ on the interval $$\left[ {0,3} \right]$$.

### Solution

Let $$f\left( x \right) = 3{x^4} - 8{x^3} + 12{x^2} - 48x + 25$$

Therefore,

\begin{align}f'\left( x \right) &= 12{x^3} - 24{x^2} + 24x - 48\\ &= 12\left( {{x^3} - 2{x^2} + 2x - 4} \right)\\ &= 12\left[ {{x^2}\left( {x - 2} \right) + 2\left( {x - 2} \right)} \right]\\ &= 12\left( {x - 2} \right)\left( {{x^2} + 2} \right)\end{align}

Now,

\begin{align}&f'\left( x \right) = 0\\& \Rightarrow \; x - 2 = 0\\& \Rightarrow \; x = 2\end{align}

Now, we evaluate the value of $$f$$ at critical point $$x = 2$$ and at the end points of the interval $$\left[ {0,3} \right]$$.

Therefore,

\begin{align}f\left( 2 \right) &= 3{\left( 2 \right)^4} - 8{\left( 2 \right)^3} + 12{\left( 2 \right)^2} - 48\left( 2 \right) + 25\\ &= 48 - 64 + 48 - 96 + 25\\& = - 39\\f\left( 0 \right) = 3{\left( 0 \right)^4} - 8{\left( 0 \right)^3} + 12{\left( 0 \right)^2} - 48\left( 0 \right) + 25\\ &= 25\\f\left( 3 \right) = 3{\left( 3 \right)^4} - 8{\left( 3 \right)^3} + 12{\left( 3 \right)^2} - 48\left( 3 \right) + 25\\& = 243 - 216 + 108 - 144 + 25\\ &= 16\end{align}

Hence, we can conclude that the absolute maximum value of $$f$$ on $$\left[ {0,3} \right]$$ is 25 occurring at $$x = 0$$ and the absolute minimum value of $$f$$ at $$\left[ {0,3} \right]$$ is $$- 39$$ occurring at $$x = 2$$.

## Chapter 6 Ex.6.5 Question 8

At what points in the interval $$\left[ {0,{\text{2}}\pi } \right]$$ does the function $$\sin 2x$$ attain its maximum value?

### Solution

Let $$f\left( x \right) = \sin 2x$$

Therefore,

$$f'\left( x \right) = 2\cos 2x$$

Now,

\begin{align}&f'\left( x \right) = 0\\& \Rightarrow \; 2\cos 2x = 0\\ &\Rightarrow \; 2x = \frac{\pi }{2},\frac{{3\pi }}{2},\frac{{5\pi }}{2},\frac{{7\pi }}{2}\\& \Rightarrow \; x = \frac{\pi }{4},\frac{{3\pi }}{4},\frac{{5\pi }}{4},\frac{{7\pi }}{4}\end{align}

Now, we evaluate the value of $$f$$ at critical point $$x = \frac{\pi }{4},\frac{{3\pi }}{4},\frac{{5\pi }}{4},\frac{{7\pi }}{4}$$ and at the end points of the interval $$\left[ {0,{\text{2}}\pi } \right]$$.

Therefore,

\begin{align}f\left( {\frac{\pi }{4}} \right) &= \sin \frac{\pi }{2} = 1\\f\left( {\frac{{3\pi }}{4}} \right) &= \sin \frac{{3\pi }}{2} = - 1\\f\left( {\frac{{5\pi }}{4}} \right) &= \sin \frac{{5\pi }}{2} = 1\\f\left( {\frac{{7\pi }}{4}} \right) &= \sin \frac{{7\pi }}{2} = - 1\end{align}

\begin{align}f\left( 0 \right) &= \sin 0 = 0\\f\left( {2\pi } \right) &= \sin 2\pi = 0\end{align}

Hence, we can conclude that the absolute maximum value of $$f$$ on $$\left[ {0,{\text{2}}\pi } \right]$$ is occurring at $$x = \frac{\pi }{4}$$ and $$x = \frac{{5\pi }}{4}$$.

## Chapter 6 Ex.6.5 Question 9

What is the maximum value of the function $$\sin ~x+\cos ~x$$ ?

### Solution

Let $$f\left( x \right)=\text{sin}~x+\text{cos}~x$$

Therefore,

$${f}'\left( x \right)=\text{cos}~x-\sin x$$

Now,

\begin{align}&f'\left( x \right) = 0\\& \Rightarrow \; \cos x - \sin x = 0\\ &\Rightarrow \; \sin x = \cos x\\ &\Rightarrow \; \tan x = 1\\& \Rightarrow \; x = \frac{\pi }{4},\frac{{5\pi }}{4} \ldots \end{align}

Hence,

\begin{align}f''\left( x \right)& = - \sin x - \cos x\\ &= - \left( {\sin x + \cos x} \right)\end{align}

Now, $$f''\left( x \right)$$ will be negative when $$\left( {\sin x + \cos x} \right)$$ is positive i.e., when $$\sin x$$ and $$\cos x$$ are both positive.

Also, we know that $$\sin x$$ and $$\cos x$$ both are positive in the first quadrant.

Then, $$f''\left( x \right)$$ will be negative when $$x \in \left( {0,\frac{\pi }{2}} \right)$$ .

Thus, we consider $$x = \frac{\pi }{4}$$

\begin{align}f''\left( {\frac{\pi }{4}} \right)& = - \left( {\sin \frac{\pi }{4} + \cos \frac{\pi }{4}} \right)\\& = - \left( {\frac{2}{{\sqrt 2 }}} \right)\\& = - \sqrt 2 < 0\end{align}

By second derivative test, $$f$$ will be the maximum at $$x = \frac{\pi }{4}$$ and the maximum value of $$f$$ is

\begin{align}f\left( {\frac{\pi }{4}} \right) &= \sin \frac{\pi }{4} + \cos \frac{\pi }{4}\\ &= \frac{1}{{\sqrt 2 }} \times \frac{1}{{\sqrt 2 }}\\& = \frac{2}{{\sqrt 2 }}\\ &= \sqrt 2 \end{align}

## Chapter 6 Ex.6.5 Question 10

Find the maximum value of $$2{x^3} - 24x + 107$$ in the interval $$\left[ {1,3} \right]$$. Find the maximum value of the same function in $$\left[ { - 3, - 1} \right]$$.

### Solution

Let $$f\left( x \right) = 2{x^3} - 24x + 107$$

Therefore,

\begin{align}f'\left( x \right) &= 6{x^2} - 24\\ &= 6\left( {{x^2} - 4} \right)\end{align}

Now,

\begin{align}&f'\left( x \right) = 0\\ &\Rightarrow \; 6\left( {{x^2} - 4} \right) = 0\\& \Rightarrow \; {x^2} = 4\\& \Rightarrow \; x = \pm 2\end{align}

We first consider the interval $$\left[ {1,3} \right]$$.

Then, we evaluate the value of $$f$$ at the critical point $$x = 2 \in \left[ {1,3} \right]$$ and at the end points of the interval $$\left[ {1,3} \right]$$.

Hence,

\begin{align}f\left( 2 \right)& = 2{\left( 2 \right)^3} - 24\left( 2 \right) + 107\\ &= 16 - 48 + 107\\ = 75\\f\left( 1 \right) &= 2{\left( 1 \right)^3} - 24\left( 1 \right) + 107\\ &= 2 - 24 + 107\\& = 85\\f\left( 3 \right)& = 2{\left( 3 \right)^3} - 24\left( 3 \right) + 107\\ &= 54 - 72 + 107\\& = 89\end{align}

Thus, the absolute maximum value of $$f\left( x \right)$$ in the interval $$\left[ {1,3} \right]$$ is 89 occurring at $$x = 3$$.

Next, we first consider the interval $$\left[ { - 3, - 1} \right]$$.

Then, we evaluate the value of $$f$$ at the critical point $$x = - 2 \in \left[ { - 3, - 1} \right]$$ and at the end points of the interval $$\left[ { - 3, - 1} \right]$$.

Hence,

\begin{align}f\left( { - 3} \right)& = 2{\left( { - 3} \right)^3} - 24\left( { - 3} \right) + 107\\ &= - 54 + 72 + 107\\ = 125\\f\left( { - 1} \right) &= 2{\left( { - 1} \right)^3} - 24\left( { - 1} \right) + 107\\ &= - 2 + 24 + 107\\ &= 129\\f\left( { - 2} \right) &= 2{\left( { - 2} \right)^3} - 24\left( { - 2} \right) + 107\\& = - 16 + 48 + 107\\& = 139\end{align}

Hence, the absolute maximum value of $$f\left( x \right)$$ in the interval $$\left[ { - 3, - 1} \right]$$ is 139 occurring at $$x = - 2$$.

## Chapter 6 Ex.6.5 Question 11

It is given that at $$x = 1$$, the function $${x^4} - 62{x^2} + ax + 9$$ attains its maximum value, on the interval $$\left[ {0,2} \right]$$. Find the value of $$a$$.

### Solution

Let $$f\left( x \right) = {x^4} - 62{x^2} + ax + 9$$

Therefore,

$$f'\left( x \right) = 4{x^3} - 124x + a$$

It is given that function $$f$$ attains its maximum value on the interval $$\left[ {0,2} \right]$$ at $$x = 1$$.

Hence,

\begin{align}&f'\left( 1 \right) = 0\\ &\Rightarrow \; 4{x^3} - 124x + a = 0\\ &\Rightarrow \; 4 - 124 + a = 0\\ &\Rightarrow \; - 120 + a = 0\\ &\Rightarrow \; a = 120\end{align}

Thus, the value of $$a = 120$$.

## Chapter 6 Ex.6.5 Question 12

Find the maximum and minimum values of $$x + \sin 2x$$ on $$\left[ {0,2\pi } \right]$$.

### Solution

Let $$f\left( x \right) = x + \sin 2x$$

Therefore,

$$f'\left( x \right) = 1 + 2\cos 2x$$

Now,

\begin{align}&f'\left( x \right) = 0\\& \Rightarrow \; 1 + 2\cos 2x = 0\\& \Rightarrow \; \cos 2x = \frac{{ - 1}}{2} = - \cos \frac{\pi }{3} = \cos \left( {\pi - \frac{\pi }{3}} \right) = \cos \frac{{2\pi }}{3}\\ &\Rightarrow \; 2x = 2n\pi \pm \frac{{2\pi }}{3}\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\left[ {n \in {\bf{Z}}} \right]\\& \Rightarrow \; x = n\pi \pm \frac{\pi }{3}\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\left[ {n \in {\bf{Z}}} \right]\\& \Rightarrow \; x = \frac{\pi }{3},\frac{{2\pi }}{3},\frac{{4\pi }}{3},\frac{{5\pi }}{3} \in \left[ {0,2\pi } \right]\end{align}

Then, we evaluate the value of $$f$$ at critical points $$x = \frac{\pi }{3},\frac{{2\pi }}{3},\frac{{4\pi }}{3},\frac{{5\pi }}{3}$$ and at the end points of the interval $$\left[ {0,2\pi } \right]$$.

Hence,

\begin{align}f\left( {\frac{\pi }{3}} \right)& = \left( {\frac{\pi }{3}} \right) + \sin 2\left( {\frac{\pi }{3}} \right)\\& = \frac{\pi }{3} + \frac{{\sqrt 3 }}{2}\\f\left( {\frac{{2\pi }}{3}} \right) &= \left( {\frac{{2\pi }}{3}} \right) + \sin 2\left( {\frac{{2\pi }}{3}} \right)\\& = \frac{{2\pi }}{3} - \frac{{\sqrt 3 }}{2}\\f\left( {\frac{{4\pi }}{3}} \right) &= \left( {\frac{{4\pi }}{3}} \right) + \sin 2\left( {\frac{{4\pi }}{3}} \right)\\& = \frac{{4\pi }}{3} + \frac{{\sqrt 3 }}{2}\\f\left( {\frac{{5\pi }}{3}} \right)& = \left( {\frac{{5\pi }}{3}} \right) + \sin 2\left( {\frac{{5\pi }}{3}} \right)\\ &= \frac{{5\pi }}{3} - \frac{{\sqrt 3 }}{2}\end{align}

\begin{align}f\left( 0 \right) &= 0 + \sin 0\\ &= 0\\f\left( {2\pi } \right) &= 2\pi + \sin 4\pi \\& = 2\pi + 0\\ &= 2\pi \end{align}

Hence, we can conclude that the absolute maximum value of $$f\left( x \right)$$ in the interval $$\left[ {0,2\pi } \right]$$ is $$2\pi$$ occurring at $$x = 2\pi$$ and the absolute minimum value of $$f\left( x \right)$$ in the interval $$\left[ {0,2\pi } \right]$$ is 0 occurring at $$x = 0$$.

## Chapter 6 Ex.6.5 Question 13

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

### Solution

Let one number be $$x$$.

Then, the other number be $$\left( {24 - x} \right)$$.

Let $$P\left( x \right)$$ denote the product of the two numbers.

Thus, we have:

\begin{align}P\left( x \right) &= x\left( {24 - x} \right)\\& = 24x - {x^2}\end{align}

Therefore,

\begin{align}P'\left( x \right)& = 24 - 2x\\P''\left( x \right) &= - 2\end{align}

Now,

\begin{align}&P'\left( x \right) = 0\\& \Rightarrow \; 24 - 2x = 0\\ &\Rightarrow \; 24 = 2x\\& \Rightarrow \; x = 12\end{align}

Also,

$$P''\left( {12} \right) = - 2 < 0$$

By second derivative test, $$x = 12$$ is the point of local maxima of $$P$$ .

Thus, the numbers are 12 and $$\left( {24 - 12} \right) = 12$$.

Hence, the product of the numbers is the maximum when the numbers are 12 each.

## Chapter 6 Ex.6.5 Question 14

Find two positive numbers $$x$$ and $$y$$ such that $$x + y = 60$$ and $$x{y^3}$$ is maximum.

### Solution

The two numbers are $$x$$ and $$y$$ such that $$x + y = 60$$

Therefore,

$$\Rightarrow \; y = 60 - x$$

Let, $$f\left( x \right) = x{y^3}$$

$$f\left( x \right) = x{\left( {60 - x} \right)^3}$$

Therefore,

\begin{align}f'\left( x \right) &= {\left( {60 - x} \right)^3} - 3x{\left( {60 - x} \right)^2}\\ &= {\left( {60 - x} \right)^2}\left[ {60 - x - 3x} \right]\\& = {\left( {60 - x} \right)^2}\left( {60 - 4x} \right)\\f''\left( x \right) &= - 2\left( {60 - x} \right)\left( {60 - 4x} \right) - 4{\left( {60 - x} \right)^2}\\ &= - 2\left( {60 - x} \right)\left[ {60 - 4x + 2\left( {60 - x} \right)} \right]\\ &= - 2\left( {60 - x} \right)\left( {180 - 6x} \right)\\ &= - 12\left( {60 - x} \right)\left( {30 - x} \right)\end{align}

Now,

\begin{align}&f'\left( x \right) = 0\\ &\Rightarrow \; x = 60{\text{ or }}x = 15\end{align}

When, $$x = 60$$

Then,

$$f''\left( x \right) = 0$$

When, $$x = 15$$

Then,

\begin{align}f''\left( x \right) &= - 12\left( {60 - 15} \right)\left( {30 - 15} \right)\\ &= - 12 \times 45 \times 15 < 0\end{align}

By second derivative test, $$x = 15$$ is a point of local maxima of $$f$$.

Thus, function $$x{y^3}$$ is maximum when $$x = 15$$ and $$y = 60 - 15 = 45$$.

Hence, the required numbers are 15 and 45.

## Chapter 6 Ex.6.5 Question 15

Find two positive numbers $$x$$ and $$y$$ such that their sum is 35 and the product $${x^2}{y^5}$$ is a maximum.

### Solution

Let one number be $$x$$.

Then, the other number is $$y = \left( {35 - x} \right)$$.

Let $$P\left( x \right) = {x^2}{y^5}$$

Then we have, $$P(x)=x^{2}(35-x)^{5}$$

Therefore,

\begin{align}P'\left( x \right) &= 2x{\left( {35 - x} \right)^5} - 5{x^2}{\left( {35 - x} \right)^4}\\ &= x{\left( {35 - x} \right)^4}\left[ {2\left( {35 - x} \right) - 5x} \right]\\ &= x{\left( {35 - x} \right)^4}\left( {70 - 7x} \right)\\ &= 7x{\left( {35 - x} \right)^4}\left( {10 - x} \right)\\P''\left( x \right) &= 7{\left( {35 - x} \right)^4}\left( {10 - x} \right) + 7x\left[ { - {{\left( {35 - x} \right)}^4} - 4{{\left( {35 - x} \right)}^3}\left( {10 - x} \right)} \right]\\ &= 7{\left( {35 - x} \right)^4}\left( {10 - x} \right) - 7x{\left( {35 - x} \right)^4} - 28x{\left( {35 - x} \right)^3}\left( {10 - x} \right)\\ &= 7{\left( {35 - x} \right)^3}\left[ {\left( {35 - x} \right)\left( {10 - x} \right) - x\left( {35 - x} \right) - 4x\left( {10 - x} \right)} \right]\\ &= 7{\left( {35 - x} \right)^3}\left[ {350 - 45x + {x^2} - 35x + {x^2} - 40x + 4{x^2}} \right]\\& = 7{\left( {35 - x} \right)^3}\left( {6{x^2} - 120x + 350} \right)\end{align}

Now,

\begin{align}&P'\left( x \right) = 0\\ &\Rightarrow \; x = 0,x = 35,x = 10\end{align}

When, $$x = 35$$

Then,

\begin{align}&P'\left( x \right) = P\left( x \right) = 0\\ &\Rightarrow \; y = 35 - 35 = 0\end{align}

This will make the product $${x^2}{y^5}$$ equal to 0.

When, $$x = 0$$

Then,

$$\Rightarrow \; y = 35 - 0 = 35$$

This will make the product $${x^2}{y^5}$$ equal to 0.

Hence, $$x = 0$$ and $$x = 35$$ cannot be the possible values of $$x$$.

When, $$x = 10$$

Then,

\begin{align}P''\left( x \right)& = 7{\left( {35 - 10} \right)^3}\left( {6 \times 100 - 120 \times 10 + 350} \right)\\& = 7{\left( {25} \right)^3}\left( { - 250} \right) < 0\end{align}

By second derivative test, $$P\left( x \right)$$ will be the maximum when $$x = 10$$ and $$y = 35 - 10 = 25$$.

Hence, the required numbers are 10 and 25.

## Chapter 6 Ex.6.5 Question 16

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

### Solution

Let one number be $$x$$.

Then, the other number be $$\left( {16 - x} \right)$$.

Let the sum of the cubes of these numbers be denoted by $$S\left( x \right)$$.

Then,

$$S\left( x \right) = {x^3} + {\left( {16 - x} \right)^3}$$

Therefore,

\begin{align}S'\left( x \right) &= 3{x^2} - 3{\left( {16 - x} \right)^2}\\S''\left( x \right) &= 6x + 6\left( {16 - x} \right)\end{align}

Now,

\begin{align}&S'\left( x \right) = 0\\ &\Rightarrow \; 3{x^2} - 3{\left( {16 - x} \right)^2} = 0\\& \Rightarrow \; {x^2} - {\left( {16 - x} \right)^2} = 0\\& \Rightarrow \; {x^2} - 256 - {x^2} + 32x = 0\\& \Rightarrow \; x = \frac{{256}}{{32}}\\& \Rightarrow \; x = 8\end{align}

Also,

\begin{align}S''\left( 8 \right)& = 6\left( 8 \right) + 6\left( {16 - 8} \right)\\ &= 48 + 48\\ &= 96 > 0\end{align}

By second derivative test, $$x = 8$$ is the point of local minima of $$S$$.

Thus, the numbers are $$8$$ and $$\left( {16 - 8} \right) = 8$$.

Hence, the sum of the cubes of the numbers is the minimum when the numbers are 8 each.

## Chapter 6 Ex.6.5 Question 17

A square piece of tin of side $$18 \rm{cm}$$ 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?

### Solution

Let the side of the square to be cut off be $$x{\text{ cm}}$$.

Then, the length and the breadth of the box will be $$\left( {18 - 2x} \right) \rm{cm}$$ each and the height of the box be $$x{\text{ cm}}$$.

Therefore, the volume $$V\left( x \right)$$ of the box is given by,

$$V\left( x \right) = x{\left( {18 - 2x} \right)^2}$$

Hence,

\begin{align}V'\left( x \right) &= {\left( {18 - 2x} \right)^2} - 4x\left( {18 - 2x} \right)\\ &= \left( {18 - 2x} \right)\left[ {18 - 2x - 4x} \right]\\ &= \left( {18 - 2x} \right)\left( {18 - 6x} \right)\\& = 6 \times 2\left( {9 - x} \right)\left( {3 - x} \right)\\ &= 12\left( {9 - x} \right)\left( {3 - x} \right)\\V''\left( x \right) &= 12\left( { - 9 - x} \right)\left( {3 - x} \right)\\ &= - 12\left( {9 - x + 3 - x} \right)\\ &= - 12\left( {12 - 2x} \right)\\ &= - 24\left( {6 - x} \right)\end{align}

Now,

\begin{align}&V'\left( x \right) = 0\\ &\Rightarrow \; x = 9,x = 3\end{align}

If, $$x = 9$$ then the length and the breadth will become 0.

Hence, $$x \ne 9$$

When, $$x = 3$$

Then,

\begin{align}V''\left( 3 \right) &= - 24\left( {6 - 3} \right)\\& = - 72 < 0\end{align}

By second derivative test, $$x = 3$$ is the point of local maxima of $$V$$.

Hence, if we remove a square of side $$3 \,\rm{cm}$$ from each corner of the square tin and make a box from the remaining sheet, then the volume of the box obtained is the largest possible.

## Chapter 6 Ex.6.5 Question 18

A rectangular sheet of tin $$45cm$$ by $$24cm$$ 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 the maximum possible?

### Solution

Let the side of the square to be cut off be $$x{\text{ }}cm$$.

Then, the height of the box is $$x{\text{ }}cm$$, the length is $$\left( {45 - 2x} \right)cm$$ and the breadth is $$\left( {24 - 2x} \right)cm$$.

Therefore, the volume $$V\left( x \right)$$ of the box is given by,

\begin{align}V\left( x \right) &= x\left( {45 - 2x} \right)\left( {24 - 2x} \right)\\ &= x\left( {1080 - 90x - 48x + 4{x^2}} \right)\\ &= 4{x^3} - 138{x^2} + 1080x\end{align}

Hence,

\begin{align}V'\left( x \right)& = 12{x^2} - 276x + 1080\\ &= 12\left( {{x^2} - 23x + 90} \right)\\ &= 12\left( {x - 18} \right)\left( {x - 5} \right)\\V''\left( x \right) &= 24x - 276\\& = 12\left( {2x - 23} \right)\end{align}

Now,

\begin{align}&V'\left( x \right) = 0\\ &\Rightarrow \; x = 18,x = 5\end{align}

It is not possible to cut off a square of side $$18cm$$ from each corner of the rectangular sheet.

Thus, $$x$$ cannot be equal to $$18$$.

When, $$x = 5$$

Then,

\begin{align}V''\left( 5 \right) &= 12\left( {2\left( 5 \right) - 23} \right)\\& = 12\left( {10 - 23} \right)\\ &= 12\left( { - 13} \right)\\& = - 156 < 0\end{align}

By second derivative test, $$x = 5$$ is the point of local maxima.

Hence, the side of the square to be cut off to make the volume of the box maximum possible is $$5 \rm{cm}$$.

## Chapter 6 Ex.6.5 Question 19

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

### Solution

Let a rectangle of length $$l$$ and breadth $$b$$ be inscribed in the given circle of radius $$a$$.

Then, the diagonal passes through the centre and is of length $$2a{\text{ }}cm$$.

Now, by applying the Pythagoras theorem, we have:

\begin{align}&{\left( {2a} \right)^2} = {l^2} + {b^2}\\& \Rightarrow \; {b^2} = 4{a^2} - {l^2}\\& \Rightarrow \; b = \sqrt {4{a^2} - {l^2}} \end{align}

Area of triangle, $$A = l\sqrt {4{a^2} - {l^2}}$$

Therefore,

\begin{align}\frac{{dA}}{{dl}}& = \sqrt {4{a^2} - {l^2}} + l\frac{1}{{2\sqrt {4{a^2} - {l^2}} }}\left( { - 2l} \right)\\& = \sqrt {4{a^2} - {l^2}} - \frac{{{l^2}}}{{\sqrt {4{a^2} - {l^2}} }}\\ &= \frac{{4{a^2} - 2{l^2}}}{{\sqrt {4{a^2} - {l^2}} }}\\\frac{{{d^2}A}}{{d{l^2}}} &= \frac{{\sqrt {4{a^2} - {l^2}} \left( { - 4l} \right) - \left( {4{a^2} - 2{l^2}} \right)\frac{{\left( { - 2l} \right)}}{{2\sqrt {4{a^2} - {l^2}} }}}}{{4{a^2} - {l^2}}}\\ &= \frac{{\left( {4{a^2} - {l^2}} \right)\left( { - 4l} \right) + l\left( {4{a^2} - {l^2}} \right)}}{{{{\left( {4{a^2} - {l^2}} \right)}^{\frac{3}{2}}}}}\\ &= \frac{{ - 12{a^2}l + 2{l^3}}}{{{{\left( {4{a^2} - {l^2}} \right)}^{\frac{3}{2}}}}}\\ &= \frac{{ - 2l\left( {6{a^2} - {l^2}} \right)}}{{{{\left( {4{a^2} - {l^2}} \right)}^{\frac{3}{2}}}}}\end{align}

Now,

$$\frac{{dA}}{{dl}} = 0$$

Hence,

\begin{align} &\Rightarrow \; \frac{{4{a^2} - 2{l^2}}}{{\sqrt {4{a^2} - {l^2}} }} = 0\\& \Rightarrow \; 4{a^2} = 2{l^2}\\ &\Rightarrow \; l = \sqrt 2 a\end{align}

Thus,

\begin{align}b &= \sqrt {4{a^2} - 2{a^2}} \\& = \sqrt {2{a^2}} \\ &= \sqrt 2 a\end{align}

When, $$l = \sqrt 2 a$$

Then,

\begin{align}\frac{{{d^2}A}}{{d{l^2}}} &= \frac{{ - 2\left( {\sqrt {2a} } \right)\left( {6{a^2} - 2{a^2}} \right)}}{{2\sqrt 2 {a^3}}}\\ &= \frac{{ - 8\sqrt 2 {a^3}}}{{2\sqrt 2 {a^3}}}\\& = - 4 < 0\end{align}

By the second derivative test, when $$l = \sqrt 2 a$$, then the area of the rectangle is the 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.

## Chapter 6 Ex.6.5 Question 20

Show that the right circular cylinder of given surface and maximum volume is such that is heights is equal to the diameter of the base.

### Solution

Let $$r$$ and $$h$$ be the radius and height of the cylinder respectively.

Then, the surface area $$\left( S \right)$$ of the cylinder is given by,

$$S = 2\pi {r^2} + 2\pi rh$$

Therefore,

\begin{align}h &= \frac{{S - 2\pi {r^2}}}{{2\pi r}}\\& = \frac{S}{{2\pi }}\left( {\frac{1}{r}} \right) - r\end{align}

Let $$V$$ be the volume of the cylinder.

Then,

\begin{align}V &= \pi {r^2}h\\ &= \pi {r^2}\left[ {\frac{S}{{2\pi }}\left( {\frac{1}{r}} \right) - r} \right]\\ &= \frac{{Sr}}{2} - \pi {r^3}\\\frac{{dV}}{{dr}} &= \frac{S}{2} - 3\pi {r^2}\\\frac{{{d^2}V}}{{d{r^2}}}& = - 6\pi r\end{align}

Now,

\begin{align}&\frac{{dV}}{{dr}} = 0\\& \Rightarrow \; \frac{S}{2} - 3\pi {r^2} = 0\\& \Rightarrow \; \frac{S}{2} = 3\pi {r^2}\\& \Rightarrow \; {r^2} = \frac{S}{{6\pi }}\end{align}

When, $${r^2} = \frac{S}{{6\pi }}$$

Then,

$$\frac{{{d^2}V}}{{d{r^2}}} = - 6\pi \left( {\sqrt {\frac{S}{{6\pi }}} } \right) < 0$$

By second derivative test, the volume is the maximum when $${r^2} = \frac{S}{{6\pi }}$$.

Now, when $${r^2} = \frac{S}{{6\pi }}$$,

Then,

\begin{align}h &= \frac{{6\pi {r^2}}}{{2\pi }}\left( {\frac{1}{r}} \right) - r\\ &= 3r - r\\ &= 2r\end{align}

Hence, the volume is the maximum when the height is twice the radius i.e., when the height is equal to the diameter.

## Chapter 6 Ex.6.5 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 minimum surface area?

### Solution

Let $$r$$ and $$h$$ be the radius and height of the cylinder respectively.

Then, volume $$\left( V \right)$$ of the cylinder is given by,

\begin{align}&V = \pi {r^2}h = 100\\& \Rightarrow \; h = \frac{{100}}{{\pi {r^2}}}\end{align}

Surface area $$\left( S \right)$$ is given by:

\begin{align}S &= 2\pi {r^2} + 2\pi rh\\ &= 2\pi {r^2} + \frac{{200}}{r}\end{align}

Hence,

\begin{align}\frac{{dS}}{{dr}} &= 4\pi r - \frac{{200}}{{{r^2}}}\\\frac{{{d^2}S}}{{d{r^2}}} &= 4\pi + \frac{{400}}{{{r^3}}}\end{align}

Now,

\begin{align}&\frac{{dS}}{{dr}} = 0\\& \Rightarrow \; 4\pi r - \frac{{200}}{{{r^2}}} = 0\\& \Rightarrow \; 4\pi r = \frac{{200}}{{{r^2}}}\\& \Rightarrow \; {r^3} = \frac{{200}}{{4\pi }} = \frac{{50}}{\pi }\\ &\Rightarrow \; r = {\left( {\frac{{50}}{\pi }} \right)^{\frac{1}{3}}}\end{align}

When, $$r = {\left( {\frac{{50}}{\pi }} \right)^{\frac{1}{3}}}$$

Then,

$$\frac{{{d^2}S}}{{d{r^2}}} > 0$$

By second derivative test, the surface area is the minimum when the radius of the cylinder is $${\left( {\frac{{50}}{\pi }} \right)^{\frac{1}{3}}}cm$$.

When, $$r = {\left( {\frac{{50}}{\pi }} \right)^{\frac{1}{3}}}$$

Then,

\begin{align}h &= \frac{{100}}{{\pi {{\left( {\frac{{50}}{\pi }} \right)}^{\frac{2}{3}}}}}\\& = \frac{{2 \times 50}}{{{{\left( {50} \right)}^{\frac{2}{3}}}{{\left( \pi \right)}^{1 - \frac{2}{3}}}}}\\ &= 2{\left( {\frac{{50}}{\pi }} \right)^{\frac{1}{3}}}\end{align}

Hence, the required dimensions of the can which has the minimum surface area is given by radius $${\left( {\frac{{50}}{\pi }} \right)^{\frac{1}{3}}}cm$$ and height $$2{\left( {\frac{{50}}{\pi }} \right)^{\frac{1}{3}}}cm$$.

## Chapter 6 Ex.6.5 Question 22

A wire of length $$28 \rm{m}$$ 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?

### Solution

Let a piece of length $$l$$ be cut from the given wire to make a square.

Then, the other piece of wire to be made into a circle is of length $$\left( {28 - l} \right)m$$.

Now, side of square is $$\frac{l}{4}$$

Let $$r$$ be the radius of the circle.

Then,

\begin{align}&2\pi r = 28 - l\\ &\Rightarrow \; r = \frac{l}{{2\pi }}\left( {28 - l} \right)\end{align}

The combined areas of the square and the circle, $$\left( A \right)$$ is given by,

\begin{align}A &= {\left( {side{\text{ }}of{\text{ }}the{\text{ }}square} \right)^2} + \pi {r^2}\\ &= \frac{{{l^2}}}{{16}} + \pi {\left[ {\frac{1}{{2\pi }}\left( {28 - l} \right)} \right]^2}\\ &= \frac{{{l^2}}}{{16}} + \frac{1}{{4\pi }}{\left( {28 - l} \right)^2}\end{align}

Hence,

\begin{align}\frac{{dA}}{{dl}} &= \frac{{2l}}{{16}} + \frac{2}{{4\pi }}\left( {28 - l} \right)\left( { - 1} \right)\\ &= \frac{1}{8} - \frac{1}{{2\pi }}\left( {28 - l} \right)\\\frac{{{d^2}A}}{{d{l^2}}}& = \frac{1}{8} + \frac{1}{{2\pi }} > 0\end{align}

Now,

\begin{align}&\frac{{{d^2}A}}{{d{l^2}}} = 0\\ &\Rightarrow \; \frac{l}{8} - \frac{l}{{2\pi }}\left( {28 - l} \right) = 0\\& \Rightarrow \; \frac{{\pi l - 4\left( {28 - l} \right)}}{{8\pi }} = 0\\ &\Rightarrow \; \left( {\pi + 4} \right)l - 112 = 0\\ &\Rightarrow \; l = \frac{{112}}{{\pi + 4}}\end{align}

When, $$l = \frac{{112}}{{\pi + 4}}$$

Then,

$$\frac{{{d^2}A}}{{d{l^2}}} > 0$$

By second derivative test, the area $$\left( A \right)$$ is the minimum when $$l = \frac{{112}}{{\pi + 4}}m$$ .

Hence, the combined area is the minimum when the length of the wire in making the square is $$\frac{{112}}{{\pi + 4}}m$$ while the length of the wire in making the circle is $$\left( {28 - \frac{{112}}{{\pi + 4}}} \right) = \frac{{28\pi }}{{\pi + 4}}m$$.

## Chapter 6 Ex.6.5 Question 23

Prove that the volume of the largest cone that can be inscribed in a sphere of radius R is $$\frac{8}{{27}}$$ of the volume of the sphere.

### Solution

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, $$V = \frac{1}{3}\pi {r^2}h$$

Height of the cone is given by,

\begin{align}h &= R + AB\\& = R + \sqrt {{R^2} - {r^2}} \;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\left[ {ABC{\text{ is a right triangle}}} \right]\end{align}

Hence,

\begin{align}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}} \end{align}

Therefore,

\begin{align}\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{{\left( { - 2r} \right)}}{{\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}}}{{\sqrt {{R^2} - {r^2}} }}\\& = \frac{2}{3}\pi rR + \frac{{2\pi r\left( {{R^2} - {r^2}} \right) - \pi {r^3}}}{{3\sqrt {{R^2} - {r^2}} }}\\& = \frac{2}{3}\pi rR + \frac{{2\pi r{R^2} - 3\pi {r^3}}}{{3\sqrt {{R^2} - {r^2}} }}\\\frac{{{d^2}V}}{{d{r^2}}}& = \frac{2}{3}\pi R + \frac{{3\sqrt {{R^2} - {r^2}} \left( {2\pi {R^2} - 9\pi {r^2}} \right) - \left( {2\pi r{R^2} - 3\pi {r^3}} \right).\frac{{\left( { - 2r} \right)}}{{6\sqrt {{R^2} - {r^2}} }}}}{{9\left( {{R^2} - {r^2}} \right)}}\\& = \frac{2}{3}\pi R + \frac{{9\left( {{R^2} - {r^2}} \right)\left( {2\pi {R^2} - 9\pi {r^2}} \right) + 2\pi {r^2}{R^2} + 3\pi {r^4}}}{{27{{\left( {{R^2} - {r^2}} \right)}^{\frac{3}{2}}}}}\end{align}

Now,

\begin{align}&\frac{{dV}}{{dr}} = 0\\& \Rightarrow \; \frac{2}{3}\pi rR + \frac{{2\pi r{R^2} - 3\pi {r^3}}}{{3\sqrt {{R^2} - {r^2}} }} = 0\\& \Rightarrow \; \frac{2}{3}\pi rR = \frac{{3\pi {r^3} - 2\pi r{R^2}}}{{3\sqrt {{R^2} - {r^2}} }}\\& \Rightarrow \; 2R = \frac{{3{r^2} - 2{R^2}}}{{\sqrt {{R^2} - {r^2}} }}\\ &\Rightarrow \; 2R\sqrt {{R^2} - {r^2}} = 3{r^2} - 2{R^2}\\& \Rightarrow \; 4{R^2}\left( {{R^2} - {r^2}} \right) = {\left( {3{r^2} - 2{R^2}} \right)^2}\\& \Rightarrow \; 4{R^4} - 4{R^2}{r^2} = 9{r^4} + 4{R^4} - 12{r^2}{R^2}\\ &\Rightarrow \; 9{r^4} = 8{R^2}{r^2}\\& \Rightarrow \; {r^2} = \frac{8}{9}{R^2}\end{align}

When, $${r^2} = \frac{8}{9}{R^2}$$

Then, $$\frac{{dV}}{{d{r^2}}} < 0$$

By second derivative test, the volume of the cone is the maximum when $${r^2} = \frac{8}{9}{R^2}$$.

When, $${r^2} = \frac{8}{9}{R^2}$$

Then,

\begin{align}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\end{align}

Therefore,

\begin{align}V &= \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}} \times \left( {{\text{Volume of sphere}}} \right)\end{align}

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

## Chapter 6 Ex.6.5 Question 24

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.

### Solution

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

Then, the volume $$\left( V \right)$$ of the cone is given by,

\begin{align}&V = \frac{1}{3}\pi {r^2}h\\ &\Rightarrow \; h = \frac{{3V}}{{\pi {r^2}}}\end{align}

The surface area $$\left( S \right)$$ of the cone is given by,

$$S = \pi rl$$, where $$l$$ is the slant height

Hence,

\begin{align} & S~=\pi r\sqrt{{{r}^{2}}+{{h}^{2}}} \\ & =\pi r\sqrt{{{r}^{2}}+\frac{9{{V}^{2}}}{{{\pi }^{2}}{{r}^{4}}}} \\ & =\frac{\pi r\sqrt{{{\pi }^{2}}{{r}^{6}}+9{{V}^{2}}}}{\pi {{r}^{2}}} \\ & =\frac{1}{r}\sqrt{{{\pi }^{2}}{{r}^{6}}+9{{V}^{2}}} \end{align}

Therefore,

\begin{align}\frac{{dS}}{{dr}}& = \frac{{r.\frac{{6{\pi ^2}{r^5}}}{{2\sqrt {{\pi ^2}{r^6} + 9{V^2}} }} - \sqrt {{\pi ^2}{r^6} + 9{V^2}} }}{{{r^2}}}\\& = \frac{{3{\pi ^2}{r^6} - {\pi ^2}{r^6} - 9{V^2}}}{{{r^2}\sqrt {{\pi ^2}{r^6} + 9{V^2}} }}\\& = \frac{{2{\pi ^2}{r^6} - 9{V^2}}}{{{r^2}\sqrt {{\pi ^2}{r^6} + 9{V^2}} }}\end{align}

Now,

\begin{align}&\frac{{dS}}{{dr}} = 0\\& \Rightarrow \; \frac{{2{\pi ^2}{r^6} - 9{V^2}}}{{{r^2}\sqrt {{\pi ^2}{r^6} + 9{V^2}} }} = 0\\& \Rightarrow \; 2{\pi ^2}{r^6} = 9{V^2}\\& \Rightarrow \; {r^6} = \frac{{9{V^2}}}{{2{\pi ^2}}}\end{align}

Thus, it can be easily verified that when$${r^6} = \frac{{9{V^2}}}{{2{\pi ^2}}}$$,$$\Rightarrow \; \frac{{{d^2}S}}{{d{r^2}}} > 0$$

By second derivative test, the surface area of the cone is the least when$${r^6} = \frac{{9{V^2}}}{{2{\pi ^2}}}$$.

When, $${r^6} = \frac{{9{V^2}}}{{2{\pi ^2}}}$$

Then,

\begin{align}h& = \frac{{3V}}{{\pi {r^2}}}\\& = \frac{{3V}}{{\pi {r^2}}}{\left( {\frac{{2{\pi ^2}{r^6}}}{9}2} \right)^{\frac{1}{2}}}\\ &= \frac{3}{{\pi {r^2}}}.\frac{{\sqrt 2 \pi {r^3}}}{3}\\ &= \sqrt 2 r\end{align}

Hence, for a given volume, the right circular cone of the least curved surface has an altitude equal to $$\sqrt 2$$ times the radius of the base.

## Chapter 6 Ex.6.5 Question 25

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

### Solution

Let $$θ$$ be the semi-vertical angle of the cone.

It is clear that $$\theta \in \left[ {0,\frac{\pi }{2}} \right]$$ .

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

The slant height of the cone is given as constant.

Now, $$r~=~l~\text{sin}~\theta ~$$ and $$h~=~l~\text{cos}~\theta$$

The volume $$\left( V \right)$$ of the cone is given by,

\begin{align} & V=\frac{1}{3}\pi {{r}^{2}}h \\ & =\frac{1}{3}\pi \left( {{l}^{2}}~{{\sin }^{2}}~\theta \right)\left( l~\cos ~\theta \right) \\ & =\frac{1}{3}\pi {{l}^{3}}{{\sin }^{2}}~\theta \cos ~\theta \end{align}

Therefore,

\begin{align} & \frac{dV}{d\theta }=\frac{\pi {{l}^{3}}}{3}\left[ \text{si}{{\text{n}}^{2}}~\theta \left( -\text{sin}~\theta \right)+\cos \theta \left( \text{2sin}~\theta \cos ~\theta \right) \right] \\ & =\frac{\pi {{l}^{3}}}{3}\left[ -\text{si}{{\text{n}}^{3}}~\theta +\text{2sin}~\theta \cos {{~}^{3}}\theta \right] \\ & \frac{{{d}^{2}}V}{d{{\theta }^{2}}}=\frac{\pi {{l}^{3}}}{3}\left[ -3\text{sin}{{~}^{2}}\theta \cos ~\theta +2\cos {{~}^{3}}\theta -4\text{sin}{{~}^{2}}\theta \cos ~\theta \right] \\ & =\frac{\pi {{l}^{3}}}{3}\left[ 2\cos {{~}^{3}}\theta -7\text{sin}{{~}^{2}}\theta \cos ~\theta \right] \end{align}

Now,

\begin{align} & \frac{dV}{d\theta }=0 \\ & \Rightarrow \; \frac{\pi {{l}^{3}}}{3}\left[ -{{\sin }^{3}}~\theta +2\sin ~\theta \cos {{~}^{3}}\theta \right]=0 \\ & \Rightarrow \; {{\sin }^{3}}~\theta =2\sin ~\theta \cos {{~}^{3}}\theta \\ & \Rightarrow \; {{\tan }^{2}}\theta =2 \\ & \Rightarrow \; \tan \theta =\sqrt{2} \\ & \Rightarrow \; \theta ={{\tan }^{-1}}\sqrt{2} \\ \end{align}

When, $${\text{\theta }} = {\tan ^{ - 1}}\sqrt 2$$

Then, $${\tan ^2}\theta = {\text{2}}$$ or $$\text{si}{{\text{n}}^{2}}~\theta =\text{2cos}{{~}^{2}}\theta$$ .

Hence, we have:

\begin{align} & \frac{{{d}^{2}}V}{d{{\theta }^{2}}}=\frac{\pi {{l}^{3}}}{3}\left[ 2\cos {{~}^{3}}\theta -14{{\cos }^{3}}~\theta \right] \\ & =\frac{\pi {{l}^{3}}}{3}\left[ -12\cos {{~}^{3}}\theta \right] \\ & =-4\pi {{l}^{3}}\cos {{~}^{3}}\theta <0\text{ }\forall \text{ }\theta \in \left[ 0,\frac{\pi }{2} \right] \end{align}

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

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

## Chapter 6 Ex.6.5 Question 26

Show that semi-vertical angle of right circular cone of given surface area and maximum volume is$${\sin ^{ - 1}}\left( {\frac{1}{3}} \right)$$.

### Solution

Let $$r$$ be the radius, $$l$$ be the slant height and $$h$$ be the height of the cone of given surface area $$S$$.

Also, let $$\alpha$$ be the semi-vertical angle of the cone.

Then,

$$S~=\pi rl~+\pi {{r}^{\text{2}}}$$

$$\Rightarrow \; l = \frac{{S - \pi {r^2}}}{{\pi r}}\;\;\;\;\;\;\;\;\;\; \ldots \left( 1 \right)$$

Let $$V$$ be the volume of the cone.

Then

\begin{align} V &= \frac{1}{3}\pi {r^2}h \\ {V^2} &= \frac{1}{9}{\pi ^2}{r^4}{h^2} \\ &= \frac{1}{9}{\pi ^2}{r^4}\left( {{l^2} - {r^2}} \right)\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\left[ {\because {l^2} = {r^2} + {h^2}} \right] \\& = \frac{1}{9}{\pi ^2}{r^4}\left[ {{{\left( {\frac{{S - \pi {r^2}}}{{\pi r}}} \right)}^2} - {r^2}} \right] \\ &= \frac{1}{9}{\pi ^2}{r^4}\left[ {\frac{{{{\left( {S - \pi {r^2}} \right)}^2} - {\pi ^2}{r^4}}}{{{\pi ^2}{r^2}}}} \right] \\ &= \frac{1}{9}{r^2}\left( {{S^2} - 2S\pi {r^2}} \right) \\ {V^2} &= \frac{1}{9}S{r^2}\left( {{S^2} - 2\pi {r^2}} \right)\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\; \ldots \left( 2 \right) \\ \end{align}

Differentiating ($$2$$) with respect to $$r$$, we get

$$2V\frac{{dV}}{{dr}} = \frac{1}{9}S\left( {2Sr - 8\pi {r^3}} \right)$$

For maximum or minimum, put $$\frac{{dV}}{{dr}} = 0$$

\begin{align} &\Rightarrow \; \frac{1}{9}S\left( {2Sr - 8\pi {r^3}} \right) = 0 \hfill \\ &\Rightarrow \; 2Sr - 8\pi {r^3} = 0\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\left[ {\because S \ne 0} \right] \hfill \\ &\Rightarrow \; S = 4\pi {r^2}\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\left[ {\because r \ne 0} \right] \hfill \\& \Rightarrow \; {r^2} = \frac{S}{{4\pi }} \hfill \\ \end{align}

Differentiating again with respect to $$r$$, we get

\begin{align} &2V\frac{{{d^2}V}}{{d{r^2}}} + 2{\left( {\frac{{dV}}{{dr}}} \right)^2} = \frac{1}{9}S\left( {2S - 24\pi {r^2}} \right) \hfill \\ &\Rightarrow \; 2V\frac{{{d^2}V}}{{d{r^2}}} = \frac{1}{9}S\left( {2S - 24\pi \times \frac{S}{{4\pi }}} \right)\;\;\;\;\;\;\;\;\;\;\;\;\;\;\left[ {\because \frac{{dV}}{{dr}} = 0{\text{ and }}{r^2} = \frac{S}{{4\pi }}} \right] \hfill \\ &\Rightarrow \; 2V\frac{{{d^2}V}}{{d{r^2}}} = \frac{1}{9}S\left( {2S - 6S} \right) \hfill \\& \Rightarrow \; 2V\frac{{{d^2}V}}{{d{r^2}}} = - \frac{4}{9}{S^2} < 0 \hfill \\ \end{align}

Thus, $$V$$ is maximum when $$S=~\text{4}\pi {{r}^{\text{2}}}$$.

Therefore,

\begin{align} & S~=~\pi rl~+\pi {{r}^{2}} \\ & \Rightarrow \; 4\pi {{r}^{2}}=\pi rl+\pi {{r}^{2}} \\ & \Rightarrow \; 3\pi {{r}^{2}}=\pi rl \\ & \Rightarrow \; l=3r \\ \end{align}

Now, in $$\Delta C O B$$,

\begin{align}\sin \alpha& = \frac{{OB}}{{BC}}\\ &= \frac{r}{l}\\& = \frac{r}{{3r}} = \frac{1}{3}\\\alpha &= {\sin ^{ - 1}}\left( {\frac{1}{3}} \right)\end{align}

## Chapter 6 Ex.6.5 Question 27

The point on the curve $${x^2} = 2y$$ which is nearest to the point $$\left( {0,{\text{5}}} \right)$$ is

(A) $$\left( {2\sqrt 2 ,4} \right)$$

(B) $$\left( {2\sqrt 2 ,0} \right)$$

(C) $$\left( {0,0} \right)$$

(D) $$\left( {2,2} \right)$$

### Solution

The given curve is $${x^2} = 2y$$.

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

Let $$P\left( {x,\frac{{{x^2}}}{2}} \right)$$ and $$A\left( {0,5} \right)$$ are the given points.

Now distance between the points P and A is given by,

\begin{align} &\Rightarrow \; PA = \sqrt {{{\left( {x - 0} \right)}^2} + {{\left( {\frac{{{x^2}}}{2} - 5} \right)}^2}} \hfill \\ &\Rightarrow \; P{A^2} = {\left( {x - 0} \right)^2} + {\left( {\frac{{{x^2}}}{2} - 5} \right)^2} \hfill \\& \Rightarrow \; P{A^2} = {x^2} + \frac{{{x^4}}}{4} + 25 - 5{x^2} \hfill \\& \Rightarrow \; P{A^2} = \frac{{{x^4}}}{4} - 4{x^2} + 25 \hfill \\ &\Rightarrow \; P{A^2} = {y^2} - 8y + 25\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\left( {\because {x^2} = 2y} \right) \hfill \\ \end{align}

Let us denote $$P{A^2}$$ by $$Z$$

Then, $$Z = {y^2} - 8y + 25$$

Differentiating both sides with respect to $$y$$, we get

$$\frac{{dZ}}{{dy}} = 2y - 8$$

For maxima or minima, we have

\begin{align}&\frac{{dZ}}{{dy}} = 0\\ &\Rightarrow \; 2y - 8 = 0\\ &\Rightarrow \; 2y = 8\\& \Rightarrow \; y = 4\end{align}

Also,

$$\frac{{{d^2}Z}}{{d{y^2}}} = 2$$

Now,

\begin{align}&{\left[ {\frac{{{d^2}Z}}{{d{y^2}}}} \right]_{y = 4}} = 2 > 0\\ &\Rightarrow \; {x^2} = 2y\\& \Rightarrow \; {x^2} = 2 \times 4\\& \Rightarrow \; {x^2} = 8\\& \Rightarrow \; x = \pm 2\sqrt 2 \end{align}

So, $$Z$$ is minimum at $$\left( {2\sqrt 2 ,4} \right)$$or$$\left( { - 2\sqrt 2 ,4} \right)$$.

Or, $$P A^{2}$$ is minimum at $$\left( {2\sqrt 2 ,4} \right)$$or$$\left( { - 2\sqrt 2 ,4} \right)$$.

$$P A$$ Or, is minimum at $$\left( {2\sqrt 2 ,4} \right)$$or$$\left( { - 2\sqrt 2 ,4} \right)$$.

So, distance between the points $$P\left( {x,\frac{{{x^2}}}{2}} \right)$$ and $$A\left( {0,5} \right)$$ is minimum at $$\left( {2\sqrt 2 ,4} \right)$$or$$\left( { - 2\sqrt 2 ,4} \right)$$.

Thus, the correct option is A.

## Chapter 6 Ex.6.5 Question 28

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)$$\frac{1}{3}$$

### Solution

Let $$f\left( x \right) = \frac{{1 - x + {x^2}}}{{1 + x + {x^2}}}$$

Therefore,

\begin{align}f'\left( x \right) &= \frac{{\left( {1 + x + {x^2}} \right)\left( { - 1 + 2x} \right) - \left( {1 - x + {x^2}} \right)\left( {1 + 2x} \right)}}{{{{\left( {1 + x + {x^2}} \right)}^2}}}\\& = \frac{{ - 1 + 2x - x + 2{x^2} - {x^2} + 2{x^3} - 1 - 2x + x + 2{x^2} - {x^2} - 2{x^3}}}{{{{\left( {1 + x + {x^2}} \right)}^2}}}\\ &= \frac{{2{x^2} - 2}}{{{{\left( {1 + x + {x^2}} \right)}^2}}}\\ &= \frac{{2\left( {{x^2} - 1} \right)}}{{{{\left( {1 + x + {x^2}} \right)}^2}}}\end{align}

Now,

\begin{align}&f'\left( x \right) = 0\\& \Rightarrow \; {x^2} = 1\\& \Rightarrow \; x = \pm 1\end{align}

Also,

\begin{align}f''\left( x \right) &= \frac{{2\left[ {{{\left( {1 + x + {x^2}} \right)}^2}\left( {2x} \right) - \left( {{x^2} - 1} \right)\left( 2 \right)\left( {1 + x + {x^2}} \right)\left( {1 + 2x} \right)} \right]}}{{{{\left( {1 + x + {x^2}} \right)}^4}}}\\ &= \frac{{4\left( {1 + x + {x^2}} \right)\left[ {\left( {1 + x + {x^2}} \right)x - \left( {{x^2} - 1} \right)\left( {1 + 2x} \right)} \right]}}{{{{\left( {1 + x + {x^2}} \right)}^4}}}\\ &= \frac{{4\left[ {x + {x^2} + {x^3} - {x^2} - 2{x^3} + 1 + 2x} \right]}}{{{{\left( {1 + x + {x^2}} \right)}^3}}}\\& = \frac{{4\left( {1 + 3x - {x^3}} \right)}}{{{{\left( {1 + x + {x^2}} \right)}^3}}}\end{align}

Hence,

\begin{align}f''\left( 1 \right) &= \frac{{4\left( {1 + 3 - 1} \right)}}{{{{\left( {1 + 1 + 1} \right)}^3}}}\\& = \frac{{4\left( 3 \right)}}{{{{\left( 3 \right)}^3}}}\\ &= \frac{{12}}{{27}}\\ &= \frac{4}{9} > 0\end{align}

Also,

\begin{align}f''\left( { - 1} \right) &= \frac{{4\left( {1 - 3 - 1} \right)}}{{{{\left( {1 - 1 + 1} \right)}^3}}}\\& = 4\left( { - 1} \right)\\& = - 4 < 0\end{align}

By second derivative test, $$f$$ is the minimum at $$x = 1$$ and the minimum value is given by,

\begin{align}f\left( 1 \right) = \frac{{1 - 1 + 1}}{{1 + 1 + 1}}\\ &= \frac{1}{3}\end{align}

Thus, the correct option is D.

## Chapter 6 Ex.6.5 Question 29

The maximum value of $${\left[ {x\left( {x - 1} \right) + 1} \right]^{\frac{1}{3}}},0 \le x \le 1$$ is

(A)$${\left( {\frac{1}{3}} \right)^{\frac{1}{3}}}$$

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

(C)$$1$$

(D)$$0$$

### Solution

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

Therefore,

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

Now,

\begin{align}&f'\left( x \right) = 0\\& \Rightarrow \; x = \frac{1}{2}\end{align}

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

\begin{align}f\left( 0 \right) &= {\left[ {0\left( {0 - 1} \right) + 1} \right]^{\frac{1}{3}}}\\ &= 1\\f\left( 1 \right) &= {\left[ {1\left( {1 - 1} \right) + 1} \right]^{\frac{1}{3}}}\\ &= 1\end{align}

\begin{align}f\left( {\frac{1}{2}} \right) &= {\left[ {\frac{1}{2}\left( { - \frac{1}{2}} \right) + 1} \right]^{\frac{1}{3}}}\\ &= {\left( {\frac{3}{4}} \right)^{\frac{1}{3}}}\end{align}

Hence, we can conclude that the maximum value of $$f$$ in the interval $$\left[ {0,{\text{1}}} \right]$$ is $$1$$.

Thus, the correct option is $$C$$.

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