Application of Derivatives: Maxima and Minima

A figure consists of a semicircle with a rectangle on its diameter. Given that the perimeter of the figure is $20 \; m$, find its dimensions in order that its area may be maximum.

Let $r$ be the radius of the semicircle.

Let the dimensions of the rectangle be $b$ and $2r$.

Perimeter of the figure $= \pi r + 2b + 2r$

Given: $\;$ Perimeter $= 20 \; m$

i.e. $\;$ $\pi r + 2b + 2r = 20$

i.e. $\;$ $r \left(2 + \pi\right) + 2b = 20$

i.e. $\;$ $b = 10 - \dfrac{r \left(2 + \pi\right)}{2}$ $\;\;\; \cdots \; (1)$

Area of the figure $= A = \dfrac{\pi r^2}{2} + 2 r b$

i.e. $\;$ $A = \dfrac{\pi r^2}{2} + 2 r \left[10 - \dfrac{r \left(2 + \pi\right)}{2}\right]$ $\;\;\;$ [in view of equation $(1)$]

i.e. $\;$ $A = \dfrac{\pi r^2}{2} + 20 r - r^2 \left(2 + \pi\right)$ $\;\;\; \cdots \; (2)$

For maximum area, $\;$ $\dfrac{dA}{dr} = 0$ $\;\;\; \cdots \; (3)$

$\therefore \;$ Differentiating equation $(2)$ w.r.t $x$ we have,

$\dfrac{dA}{dr} = \dfrac{2 \pi r}{2} + 20 - 2r \left(2 + \pi\right)$

i.e. $\;$ $\dfrac{dA}{dr} = \pi r + 20 - r \left(4 + 2 \pi\right)$

i.e. $\;$ $\dfrac{dA}{dr} = 20 - r \left(\pi + 4\right)$ $\;\;\; \cdots \; (4)$

$\therefore \;$ $\dfrac{dA}{dr} = 0$ $\implies$ $20 - r \left(\pi + 4\right) = 0$

i.e. $\;$ $r = \dfrac{20}{\pi + 4}$

From equation $(4)$, $\;$ $\dfrac{d^2 A}{dr^2} \Big |_{r = \frac{20}{\pi + 4}} = - \left(\pi + 4\right) < 0$

$\implies$ $r = \dfrac{20}{\pi + 4}$ $\;$ gives maximum area.

Substituting the value of $r$ in equation $(1)$ gives

$b = 10 - \dfrac{20 \left(\pi + 2\right)}{2 \left(\pi + 4\right)}$

i.e. $\;$ $b = 10 \left(1 - \dfrac{\pi + 2}{\pi + 4}\right) = \dfrac{20}{\pi + 4}$

$\therefore \;$ For maximum area of the figure, $\;$ radius of semicircle $= \dfrac{20}{\pi + 4} \; m$ $\;$ and the sides of the rectangle are $\;$ $\dfrac{20}{\pi + 4} \; m$, $\;$ $\dfrac{40}{\pi + 4} \; m$

Application of Derivatives: Increasing Decreasing Functions

Find the intervals in which $f\left(x\right) = \dfrac{x}{\log x}$ is increasing or decreasing.

$f\left(x\right) = \dfrac{x}{\log x}$

$\therefore$ $f'\left(x\right) = \dfrac{\log x - x \times \dfrac{1}{x}}{\left(\log x\right)^2}$

i.e. $f'\left(x\right) = \dfrac{\log x -1}{\left(\log x\right)^2}$

Now, $\left(\log x\right)^2 > 0 \;\; \forall \; x $

$\therefore$ $f'\left(x\right) > 0$ $\implies$ $\log x - 1 > 0$

i.e. $\log x > 1$ $\implies$ $\log x > \log e$ $\implies$ $x>e$

$\therefore$ $f\left(x\right)$ is increasing in the interval $\left(e,\infty\right)$

$f'\left(x\right) < 0$ $\implies$ $\log x -1 < 0$

i.e. $\log x < 1$ $\implies$ $x < e$

$\therefore$ $f\left(x\right)$ is decreasing in the interval $\left(0,e\right)$

Application of Derivatives: Increasing Decreasing Functions

Determine the intervals in which $f:R \rightarrow R$, $f\left(x\right) = 2 \cos x + \sin^2 x$ is strictly increasing or strictly decreasing.

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

$\therefore$ $f'\left(x\right) = -2 \sin x + 2 \sin x \cos x$

i.e. $f'\left(x\right) = -2 \sin x \left(1 - \cos x\right) $

i.e. $f'\left(x\right) = -2 \sin x \times 2 \sin^2 \left(\dfrac{x}{2}\right)$

i.e. $f'\left(x\right) = -4 \sin x \sin^2 \left(\dfrac{x}{2}\right)$ $\;\; \cdots$ (1)

Now, $\sin^2 \left(\dfrac{x}{2}\right) > 0 \;\; \forall \; x \in R$

When $f'\left(x\right) > 0$ $\implies$ $f\left(x\right)$ is increasing.

Now, from equation (1), $f'\left(x\right) > 0$ $\implies$ $\sin x < 0$

$\implies$ $- \pi < x < 0$

Generalizing, we have $\left(2 k - 1\right)\pi < x < 2 k \pi$

i.e. $f\left(x\right)$ is strictly increasing in the interval $\left(\left(2 k - 1\right)\pi, 2 k \pi\right), \;\; k \in Z$

When $f'\left(x\right) < 0$ $\implies$ $f\left(x\right)$ is decreasing.

Now, from equation (1), $f'\left(x\right) < 0$ $\implies$ $\sin x > 0$

$\implies$ $0 < x < \pi$

Generalizing, we have $2 k \pi < x < \left(2 k + 1\right) \pi$

i.e. $f\left(x\right)$ is strictly decreasing in the interval $\left(2 k \pi, \left(2 k + 1 \right) \pi \right), \;\; k \in Z$

Application of Derivatives: Increasing Decreasing Functions

Find the intervals in which the function $f\left(x\right) = \dfrac{4x^2 + 1}{x}$ is increasing and the intervals in which it is decreasing.

$f\left(x\right) = \dfrac{4x^2 + 1}{x}$ $\implies$ $f\left(x\right) = 4x + \dfrac{1}{x}$

$\therefore$ $f'\left(x\right) = 4 - \dfrac{1}{x^2} = \dfrac{4x^2 -1}{x^2}$

Now, $x^2 > 0 \;\; \forall \; x, \;\; x \neq 0$

$f'\left(x\right) = 0$ $\implies$ $4x^2 -1 = 0$

i.e. $x^2 = \dfrac{1}{4}$ $\implies$ $x = \dfrac{1}{2}$, $x = -\dfrac{1}{2}$

These points divide the real line into three intervals namely $\left(-\infty, -\dfrac{1}{2}\right)$, $\left(-\dfrac{1}{2}, \dfrac{1}{2}\right)$ and $\left(\dfrac{1}{2},\infty\right)$

Interval	Sign of $\left(4x^2 - 1\right)$	Sign of $f'\left(x\right)$	Nature of function f
$\left(-\infty,-\dfrac{1}{2}\right)$	+ve	+ve	Increasing
$\left(-\dfrac{1}{2},\dfrac{1}{2}\right)$	-ve	-ve	Decreasing
$\left(\dfrac{1}{2},\infty\right)$	+ve	+ve	Increasing

$\therefore$ $f\left(x\right)$ is increasing in the interval $\left\{x : x < \dfrac{-1}{2}\right\} \cup \left\{x : x > \dfrac{1}{2}\right\} $

$f\left(x\right)$ is decreasing in the interval $\left\{x : \dfrac{-1}{2} < x < \dfrac{1}{2}\right\} $

Application of Derivatives: Increasing Decreasing Functions

Determine the intervals in which $f\left(x\right) = \sin^4 x + \cos^4 x$, $x \in \left(0,\dfrac{\pi}{2}\right)$ is increasing and in which it is decreasing.

$f\left(x\right) = \sin^4 x + \cos^4 x$

$\therefore$ $f'\left(x\right) = 4 \sin^3 x \cos x - 4 \cos^3 x \sin x$

i.e. $f'\left(x\right) = 4 \sin x \cos x \left(\sin^2 x - \cos^2 x\right)$

i.e. $f'\left(x\right) = -4 \sin x \cos x \cos 2x$ $\;\; \cdots$ (1)

Now, $\sin x > 0 \; \forall \; x \in \left(0,\dfrac{\pi}{2}\right)$; $\cos x > 0 \; \forall \; x \in \left(0,\dfrac{\pi}{2}\right)$

$f\left(x\right)$ is increasing when $f'\left(x\right) > 0$

Since $\sin x > 0$ and $\cos x > 0$, $\therefore$ $\;$ $f'\left(x\right) > 0$ $\implies$ $\cos 2x < 0$

i.e. $\dfrac{\pi}{2} < 2x < \pi$ or $\pi < 2x < \dfrac{3\pi}{2}$

i.e. $\dfrac{\pi}{4} < x < \dfrac{\pi}{2}$ or $\dfrac{\pi}{2} < x < \dfrac{3\pi}{4}$

Since $x \in \left(0,\dfrac{\pi}{2}\right)$, therefore $f\left(x\right)$ is increasing in $\left(\dfrac{\pi}{4}, \dfrac{\pi}{2}\right)$

$f\left(x\right)$ is decreasing when $f'\left(x\right) < 0$

Since $\sin x > 0$ and $\cos x > 0$, $\therefore$ $\;$ $f'\left(x\right) < 0$ $\implies$ $\cos 2x > 0$

i.e. $0 < 2x < \dfrac{\pi}{2}$ or $\dfrac{3\pi}{2} < 2x < 2\pi$

i.e. $0 < x < \dfrac{\pi}{4}$ or $\dfrac{3\pi}{4} < x < \pi $

Since $x \in \left(0,\dfrac{\pi}{2}\right)$, therefore $f\left(x\right)$ is decreasing in $\left(0,\dfrac{\pi}{4}\right)$

Application of Derivatives: Increasing Decreasing Functions

Find the intervals in which the function $f\left(x\right) = \sin x - \cos x$, $0 < x < 2 \pi$ is increasing or decreasing.

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

$\begin{aligned} \therefore \; f'\left(x\right) & = \cos x + \sin x \\ & = \sqrt{2}\left(\dfrac{1}{\sqrt{2}} \cos x + \dfrac{1}{\sqrt{2}} \sin x\right) \\ & = \sqrt{2} \left(\sin \dfrac{\pi}{4} \cos x + \cos \dfrac{\pi}{4} \sin x\right) \\ & = \sqrt{2} \sin \left(\dfrac{\pi}{4} + x\right) \end{aligned}$

$\therefore$ $\;$ $f'\left(x\right) = 0$ $\implies$ $\sin \left(\dfrac{\pi}{4} + x\right) = 0$

i.e. $\sin \left(\dfrac{\pi}{4} + x\right) = \sin 0 \;\; \text{OR} \;\; \sin \pi \;\; \text{OR} \;\; \sin 2 \pi$

$\implies$ $x = - \dfrac{\pi}{4}$ OR $x = \pi - \dfrac{\pi}{4} = \dfrac{3\pi}{4}$ OR $x = 2 \pi - \dfrac{\pi}{4} = \dfrac{7\pi}{4}$

Since $0 < x < 2 \pi$, $\;$ $x = - \dfrac{\pi}{4}$ is not a valid solution.

The points $x = \dfrac{3\pi}{4}$ and $x = \dfrac{7\pi}{4}$ divide the interval $0 < x < 2 \pi$ into three intervals namely $\left(0,\dfrac{3\pi}{4}\right)$, $\left(\dfrac{3\pi}{4},\dfrac{7\pi}{4}\right)$ and $\left(\dfrac{7\pi}{4},2\pi\right)$

Interval	Sign of $\sin \left(\dfrac{\pi}{4} + x\right)$	Nature of function f
$\left(0,\dfrac{3\pi}{4}\right)$	+ve	Increasing
$\left(\dfrac{3\pi}{4},\dfrac{7\pi}{4}\right)$	-ve	Decreasing
$\left(\dfrac{7\pi}{4}, 2 \pi\right)$	+ve	Increasing

$\therefore$ $f\left(x\right)$ is increasing in the interval $\left\{x : 0< x < \dfrac{3\pi}{4}\right\} \cup \left\{x : \dfrac{7\pi}{4}< x < 2\pi\right\} $

$f\left(x\right)$ is decreasing in the interval $\left\{x : \dfrac{7\pi}{4} < x < 2\pi\right\} $

Application of Derivatives: Increasing Decreasing Functions

Find the values of x for which $y = \left[x\left(x-2\right)\right]^2$ is an increasing function.

$y = \left[x\left(x-2\right)\right]^2$

$\therefore$ $\dfrac{dy}{dx}= 2 x \left(x-2\right) \left(x+x-2\right)$

i.e. $\dfrac{dy}{dx}= 4x \left(x-2\right) \left(x-1\right)$

$\dfrac{dy}{dx}=0$ $\implies$ $x=0$, $x=2$, $x=1$

These points divide the real line into four disjoint intervals namely $\left(-\infty,0\right)$, $\left(0,1\right)$, $\left(1,2\right)$ and $\left(2,\infty\right)$

Interval	Sign of $\dfrac{dy}{dx}$	Nature of function y
$\left(-\infty,0\right)$	-ve	Decreasing
$\left(0,1\right)$	+ve	Increasing
$\left(1,2\right)$	-ve	Decreasing
$\left(2,\infty\right)$	+ve	Increasing

$\therefore$ y is increasing in the intervals $\left(0,1\right)$ and $\left(2,\infty\right)$.

Application of Derivatives: Increasing Decreasing Functions

Find the intervals in which the function $f\left(x\right) = \left(x+1\right)^2 \left(x-3\right)^3$ is increasing or decreasing.

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

$\therefore$ $f'\left(x\right) = 3 \left(x+1\right)^2 \left(x-3\right)^2 + 2 \left(x+1\right) \left(x-3\right)^3$

i.e. $f'\left(x\right) = \left(x+1\right) \left(x-3\right)^2 \left[3\left(x+1\right) +2 \left(x-3\right)\right]$

i.e. $f'\left(x\right) = \left(x+1\right) \left(x-3\right)^2 \left(5x-3\right)$

Now, $\left(x-3\right)^2 > 0 \; \forall \; x$

For critical points, $f'\left(x\right) = 0$

$\implies$ $x+1 = 0$ $\implies$ $x=-1$

or $\left(x-3\right)^2 =0$ $\implies$ $x=3$

or $5x-3=0$ $\implies$ $x=\dfrac{3}{5}$

These points divide the real line into four intervals namely $\left(-\infty, -1\right)$, $\left(-1, \dfrac{3}{5}\right)$, $\left(\dfrac{3}{5},3\right)$ and $\left(3,\infty\right)$

Interval	Sign of $\left(x+1\right)$	Sign of $\left(5x-3\right)$	Sign of $f'\left(x\right)$	Nature of function f
$\left(-\infty,-1\right)$	-ve	-ve	+ve	Increasing
$\left(-1,\dfrac{3}{5}\right)$	+ve	-ve	-ve	Decreasing
$\left(\dfrac{3}{5},3\right)$	+ve	+ve	+ve	Increasing
$\left(3,\infty\right)$	+ve	+ve	+ve	Increasing

$\therefore$ $f\left(x\right)$ is increasing in the interval $\left\{x : x<-1\right\} \cup \left\{x : x>\dfrac{3}{5}\right\} $

$f\left(x\right)$ is decreasing in the interval $\left\{x : -1 < x < \dfrac{3}{5}\right\} $

Application of Derivatives: Increasing Decreasing Functions

Find the least value of $a$ such that the function f given by $f\left(x\right) = x^2 +ax + 1$ is strictly increasing on $\left(1,2\right)$.

$f\left(x\right) = x^2 + ax + 1$

i.e. $f'\left(x\right) = 2x + a$

Since $f\left(x\right)$ is a strictly increasing function $\implies$ $f'\left(x\right) > 0$

i.e. $2 x + a > 0$ $\implies$ $x > -\dfrac{a}{2}$

i.e. To find $x > \dfrac{-a}{2}$ when x is in the interval $\left(1,2\right)$

i.e. To find $x > \dfrac{-a}{2}$ when $1 < x < 2$

Least value of x is 1.

$\therefore$ Least value of a is obtained when $x = 1$

$\therefore$ We have, $1 = \dfrac{-a}{2}$ $\implies$ $a = -2$

$\therefore$ The least value of $a = -2$.

Application of Derivatives: Increasing Decreasing Functions

Find the intervals in which the function $f:R \rightarrow R$, $f\left(x\right) = \dfrac{4 \sin x -2x - x \cos x}{2+\cos x}$ is increasing and in which it is decreasing.

$f\left(x\right) = \dfrac{4 \sin x -2x - x \cos x}{2 + \cos x}$

i.e. $f\left(x\right) = \dfrac{4\sin x}{2+\cos x} - \dfrac{x \left(2+\cos x\right)}{2+\cos x}$

i.e. $f\left(x\right) = \dfrac{4\sin x}{2+\cos x} - x$

$\therefore$ $f'\left(x\right) = \dfrac{\left(2+\cos x\right)\times 4 \cos x -4 \sin x \times \left(-\sin x\right)}{\left(2+\cos x\right)^2} -1$

i.e. $f'\left(x\right) = \dfrac{8 \cos x + 4 \cos ^2 x + 4 \sin ^2 x}{\left(2+\cos x\right)^2}-1$

i.e. $f'\left(x\right) = \dfrac{8 \cos x + 4 -4 - 4\cos x -\cos^2 x}{\left(2+\cos x\right)^2}$

i.e. $f'\left(x\right) = \dfrac{4 \cos x - \cos^2 x}{\left(2+\cos x\right)^2}$

i.e. $f'\left(x\right) = \dfrac{\cos x \left(4- \cos x\right)}{\left(2+\cos x\right)^2}$ $\;\; \cdots$ (1)

In equation (1), $\left(2+\cos x\right)^2 > 0 \; \forall \; x$ $\;\; \cdots$ (2)

Also, $4 - \cos x > 0 \; \forall \; x$ since $\cos x$ varies between $\pm 1$ $\;\; \cdots$ (3)

Case (i): $\mathbf{f\left(x\right)}$ is increasing:

$f\left(x\right)$ is increasing when $f'\left(x\right) > 0$

$\therefore$ From equation (1), for $f'\left(x\right) > 0$, $\cos x > 0$

Now, $\cos x > 0$ i.e. positive in the First and Fourth quadrants.

$\therefore$ $\cos x > 0$ $\implies$ $0 < x < \dfrac{\pi}{2}$ (First Quadrant) and $\dfrac{3\pi}{2} < x < 2\pi$ (Fourth Quadrant)

$\therefore$ Generalizing,

$\cos x > 0$ $\implies$ $2k\pi < x < \left(4k+1\right)\dfrac{\pi}{2}$ (First Quadrant) and $\left(4k+3\right) \dfrac{\pi}{2} < x < \left(2k+2\right)\pi$ (Fourth Quadrant), $k \in Z$

$\therefore$ $f\left(x\right)$ is increasing in the interval $\left(2k\pi, \left(4k+1\right)\dfrac{\pi}{2}\right)$ and $\left(\left(4k+3\right)\dfrac{\pi}{2}, \left(2k+2\right)\pi\right)$, $k \in Z$

Case (ii): $\mathbf{f\left(x\right)}$ is decreasing:

$f\left(x\right)$ is decreasing when $f'\left(x\right) < 0$

$\therefore$ In view of equations (2) and (3), we have from equation (1), for $f'\left(x\right) < 0$, $\cos x < 0$

Now, $\cos x < 0$ i.e. negative in the Second and Third quadrants.

$\therefore$ $\cos x < 0$ $\implies$ $\dfrac{\pi}{2} < x < \pi$ (Second Quadrant) and $\pi < x < \dfrac{3\pi}{2}$ (Third Quadrant)

$\therefore$ Generalizing,

$\cos x < 0$ $\implies$ $\left(4k+1\right)\dfrac{\pi}{2} < x < \left(2k+1\right)\pi$ (Second Quadrant) and $\left(2k+1\right)\pi < x < \left(4k+3\right) \dfrac{\pi}{2}$ (Third Quadrant), $k \in Z$

$\therefore$ $f\left(x\right)$ is decreasing in the interval $\left(\left(4k+1\right)\dfrac{\pi}{2}, \left(2k+1\right)\pi \right)$ and $\left(\left(2k+1\right) \pi, \left(4k+3\right)\dfrac{\pi}{2}\right)$, $k \in Z$

Application of Derivatives: Increasing Decreasing Functions

Discuss the nature of the function f given by $f\left(x\right) = \log \left[\sin \left(x\right)\right]$ in the intervals $\left(0,\dfrac{\pi}{2}\right)$ and $\left(\dfrac{\pi}{2}, \pi\right)$.

$f\left(x\right) = \log \left[\sin \left(x\right)\right]$

$\therefore$ $f'\left(x\right) = \dfrac{\cos x}{\sin x} = \cot x$

For $0 < x < \dfrac{\pi}{2}$, $\cot x > 0$

i.e. $f'\left(x\right) > 0$ for $0 < x < \dfrac{\pi}{2}$

i.e. $f\left(x\right)$ is strictly increasing in the interval $\left(0,\dfrac{\pi}{2}\right)$

For $\dfrac{\pi}{2} < x < \pi$, $\cot x < 0$

i.e. $f'\left(x\right) < 0$ for $\dfrac{\pi}{2} < x < \pi$

i.e. $f\left(x\right)$ is strictly decreasing in the interval $\left(\dfrac{\pi}{2}, \pi\right)$

Application of Derivatives: Increasing Decreasing Functions

Find the intervals in which $f\left(x\right) = \dfrac{3}{10}x^4-\dfrac{4}{5}x^3 -3x^2 + \dfrac{36}{5}x + 11$ is increasing and in which it is decreasing.

$f\left(x\right) = \dfrac{3}{10}x^4 - \dfrac{4}{5}x^3 - 3x^2 + \dfrac{36}{5}x + 11$

$\therefore$ $f'\left(x\right) = \dfrac{6}{5}x^3-\dfrac{12}{5}x^2-6x+\dfrac{36}{5}$

i.e. $f'\left(x\right) = \dfrac{6}{5}\left(x^3 -2x^2-5x+6\right)$

i.e. $f'\left(x\right) = \dfrac{6}{5}\left(x-1\right) \left(x^2 -x-6\right)$

i.e. $f'\left(x\right) = \dfrac{6}{5}\left(x-1\right) \left(x-3\right) \left(x+2\right)$

$\therefore$ For critical points, $f'\left(x\right) = 0$

$\implies$ $\left(x-1\right) \left(x-3\right) \left(x+2\right) = 0$

$\implies$ $x = -2$, $x=1$, $x=3$

These points divide the real line into four disjoint intervals namely $\left(-\infty,-2\right)$, $\left(-2,1\right)$, $\left(1,3\right)$ and $\left(3,\infty\right)$

Interval	Sign of $f'\left(x\right)$	Nature of function f
$\left(-\infty,-2\right)$	$< 0$	Decreasing
$\left(-2,1\right)$	$> 0$	Increasing
$\left(1,3\right)$	$ < 0$	Decreasing
$\left(3,\infty\right)$	$ > 0$	Increasing

$\therefore$ $f\left(x\right)$ is increasing in the intervals $\left(-2,1\right)$ and $\left(3,\infty\right)$

$f\left(x\right)$ is decreasing in the intervals $\left(-\infty,-2\right)$ and $\left(1,3\right)$

Application of Derivatives: Increasing Decreasing Functions

Prove that the function given by $f\left(x\right) = x^3 - 3x^2 +3x - 100$ is increasing in $\mathbb{R}$.

$f\left(x\right) = x^3 -3x^2 + 3x -100$

$\begin{aligned} \therefore f'\left(x\right) & = 3x^2 -6x +3 \\ & = 3 \left(x^2 -2x +1\right) \\ & = 3 \left(x-1\right)^2 \end{aligned}$

Now, $\left(x-1\right)^2 > 0 \; \forall \; \mathbb{R}$

$\therefore$ $3 \left(x-1\right)^2 > 0 \; \forall \; \mathbb{R}$

$\implies$ $f'\left(x\right) > 0 \; \forall \; \mathbb{R}$

$\implies$ $f\left(x\right)$ is strictly increasing in $\mathbb{R}$.

Application of Derivatives: Maxima and Minima

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

Let the two positive numbers be x and y.

Given: $x+y=16$

$\implies$ $y = 16-x$ $\;\; \cdots$ (1)

Sum of cubes of the numbers $= S = x^3 + y^3$

i.e. $S = x^3 + \left(16-x\right)^3$ [From equation (1)]

For minimum sum, $\dfrac{dS}{dx}=0$

Now, $\dfrac{dS}{dx}= 3x^2 - 3 \left(16-x\right)^2$ $\;\; \cdots$ (2)

$\therefore$ $\dfrac{dS}{dx}=0$ $\implies$ $3x^2 - 3 \left(16-x\right)^2 = 0$

i.e. $x^2 = \left(16-x\right)^2$

i.e. $x^2 = 256 -32x + x^2$

i.e. $32x = 256$ $\implies$ $x = 8$

From equation (2), $\dfrac{d^2 S}{dx^2} = 6x +6 \left(16-x\right) = 96 > 0 \; \forall \; x$

$\implies$ $x = 8$ gives a minimum for the sum of cubes of the numbers.

From equation (1), when $x = 8$, $y=16-8=8$

$\therefore$ The two numbers are 8, 8.

Application of Derivatives: Maxima and Minima

What fraction of the volume of a sphere is taken up by the largest cylinder that can be fit inside the sphere?

Let radius of sphere $=R$

Let radius of cylinder $=r$

Let height of cylinder $=2h$

From the figure, $R^2 = h^2 + r^2$

$\therefore$ $r^2 = R^2 - h^2$ $\;\; \cdots$ (1)

Volume of sphere $=V_s = \dfrac{4}{3}\pi R^3$ $\;\; \cdots$ (2)

Volume of cylinder $= V_c = 2\pi r^2 h$ $\;\; \cdots$ (3)

In view of equation (1), equation (3) becomes

$V_c = 2 \pi h \left(R^2 - h^2\right)$

i.e. $V_c = 2\pi R^2 h - 2 \pi h^3$ $\;\; \cdots$ (4)

For cylinder with maximum volume, $\dfrac{dV_c}{dh} = 0$

Now, from equation (4), $\dfrac{dV_c}{dh} = 2 \pi R^2 - 6\pi h^2$ $\;\; \cdots$ (5)

$\therefore$ $\dfrac{dV_c}{dh} = 0$ $\implies$ $2\pi R^2 - 6\pi h^2 = 0$

i.e. $3h^2 = R^2$ $\implies$ $h = \dfrac{R}{\sqrt{3}}$ $\;\; \cdots$ (6)

Now, from equation (5), $\dfrac{d^2 V_c}{dh^2} = -12 \pi h$

$\therefore$ $\dfrac{d^2 V_c}{dh^2} \bigg |_{h=\frac{R}{\sqrt{3}}} = \dfrac{-12\pi R}{\sqrt{3}} < 0$

$\implies$ $h = \dfrac{R}{\sqrt{3}}$ gives a cylinder with maximum volume.

Now, from equation (1), when $h = \dfrac{R}{\sqrt{3}}$,

$r = \sqrt{R^2 - \dfrac{R^2}{3}} = \dfrac{R\sqrt{2}}{\sqrt{3}}$ $\;\; \cdots$ (7)

$\therefore$ From equation (2), volume of cylinder $=V_c = 2 \pi \times \dfrac{2R^2}{3} \times \dfrac{R}{\sqrt{3}}$

i.e. $V_c = \dfrac{4\pi R^3}{3\sqrt{3}}$ $\;\; \cdots$ (8)

$\therefore$ From equations (2) and (8),

$\dfrac{V_c}{V_s} = \dfrac{4\pi R^3 / 3\sqrt{3}}{4\pi R^3 / 3} = \dfrac{1}{\sqrt{3}}$

$\therefore$ Fraction of the volume of a sphere is taken up by the largest cylinder that can be fit inside the sphere $= \dfrac{1}{\sqrt{3}} \times 100 = 57.5 \%$

Application of Derivatives: Maxima and Minima

A wire of length 36 cm is cut into two pieces. One of the pieces is turned in the form of a square and the other in the form of an equilateral triangle. Find the length of each piece so that the sum of the areas of the two be minimum.

Let the length of one piece of wire $= x$ cm

Then, length of remaining wire $= 36 - x$ cm

Let the wire of length x be turned in the form of a square.

$\therefore$ Length of side of square $= \dfrac{x}{4}$ cm

$\therefore$ Area of square $= \left(\dfrac{x}{4}\right)^2 = \dfrac{x^2}{16} \; cm^2$

$\left(36-x\right)$ cm wire is shaped as an equilateral triangle.

$\therefore$ Length of each side of triangle $=\dfrac{36-x}{3}$ cm

$\therefore$ Area of equilateral triangle $= \dfrac{\sqrt{3}}{4} \times \left(\dfrac{36-x}{3}\right)^2 = \dfrac{\sqrt{3}}{36} \left(36-x\right)^2 \; cm^2$

Sum of areas $= A = \dfrac{x^2}{16} + \dfrac{\sqrt{3}}{36} \left(36-x\right)^2$

For minimum area, $\dfrac{dA}{dx}=0$

Now, $\dfrac{dA}{dx} = \dfrac{2x}{16} + \dfrac{\sqrt{3}}{36} \times 2 \times \left(36-x\right) \times \left(-1\right)$

i.e. $\dfrac{dA}{dx} = \dfrac{x}{8} - \dfrac{\sqrt{3}}{18} \left(36-x\right)$ $\;\; \cdots$ (1)

$\therefore$ $\dfrac{dA}{dx} = 0$ $\implies$ $\dfrac{x}{8}- \dfrac{\sqrt{3}}{18} \left(36-x\right) = 0$

i.e. $9x = 4\sqrt{3}\left(36-x\right)$ $\implies$ $x = \dfrac{144 \sqrt{3}}{9+4\sqrt{3}}$

From equation (1), $\dfrac{d^2A}{dx^2} = \dfrac{1}{8} + \dfrac{\sqrt{3}}{18} > 0$

$\implies$ $x = \dfrac{144\sqrt{3}}{9+4\sqrt{3}}$ gives minimum area.

Length of second piece of wire $= 36 - \dfrac{144\sqrt{3}}{9+4\sqrt{3}} = \dfrac{324}{9+4\sqrt{3}}$

$\therefore$ Length of pieces of wire are $\dfrac{144\sqrt{3}}{9+4\sqrt{3}}$ cm and $\dfrac{324}{9+4\sqrt{3}}$ cm

Application of Derivatives: Maxima and Minima

Find the area of the largest rectangle that fits inside a semicircle of radius 10 units. One side of the rectangle is along the diameter of the semicircle.

Let radius of semicircle $= r$

Let length of rectangle $= 2\ell$

Let breadth of rectangle $= b$

From the figure, $r^2 = \ell^2 + b^2$

$\implies$ $\ell = \sqrt{r^2 - b^2}$ $\;\; \cdots$ (1)

Area of rectangle $=A= 2\ell b$

Substituting the value of $\ell$ from equation (1) gives

$A = 2b \sqrt{r^2 -b^2}$

For maximum area, $\dfrac{dA}{db}=0$

Now, $\dfrac{dA}{db} = 2\sqrt{r^2 - b^2} + \dfrac{2b}{2 \sqrt{r^2-b^2}}\times \left(-2b\right)$

i.e. $\dfrac{dA}{db} = 2 \sqrt{r^2 - b^2} - \dfrac{2b^2}{\sqrt{r^2 - b^2}}$

i.e. $\dfrac{dA}{db} = \dfrac{2r^2 - 4b^2}{\sqrt{r^2 - b^2}}$ $\;\; \cdots$ (2)

$\therefore$ $\dfrac{dA}{db} = 0$ $\implies$ $\dfrac{2r^2 -4b^2}{\sqrt{r^2 - b^2}} = 0$

i.e. $2b^2 = r^2$

$\implies$ $b = \dfrac{r}{\sqrt{2}}$

Substituting the value of b in equation (1) gives

$\ell = \sqrt{r^2 - \dfrac{r^2}{2}} = \dfrac{r}{\sqrt{2}}$

Now from equation (2) we have

$\dfrac{d^2 A}{db^2} = \dfrac{\sqrt{r^2 - b^2}\left(-2b\right)-\left(2r^2-4b^2\right) \times \dfrac{\left(-2b\right)}{2\sqrt{r^2-b^2}}}{r^2 - b^2}$

i.e. $\dfrac{d^2A}{db^2} = \dfrac{2b\left(r^2-2b^2-r^2+b^2\right)}{\left(r^2-b^2\right)\sqrt{r^2-b^2}}$

i.e. $\dfrac{d^2 A}{db^2} = \dfrac{-2b^3}{\left(r^2 - b^2\right) \sqrt{r^2 - b^2}}$

$\therefore$ $\dfrac{d^2A}{db^2} \bigg |_{b=\frac{r}{\sqrt{2}}} = \dfrac{\dfrac{-2r^3}{2\sqrt{2}}}{\left(r^2 - \dfrac{r^2}{2}\right)\sqrt{r^2 - \dfrac{r^2}{2}}} $ $= \dfrac{\dfrac{-2r^3}{2\sqrt{2}}}{\dfrac{r^3}{2\sqrt{2}}} = -2 < 0$

$\implies$ Area A is maximum when $\ell = \dfrac{r}{\sqrt{2}}$ and $b = \dfrac{r}{\sqrt{2}}$

$\therefore$ Maximum area of the rectangle $=2 \times \dfrac{r}{\sqrt{2}} \times \dfrac{r}{\sqrt{2}} = r^2$

Given: radius of semicircle $= r = 10$ units

$\therefore$ Maximum area of the rectangle $= 100$ sq units

Application of Derivatives: Maxima and Minima

Find the points at which the function f given by $f\left(x\right) = \left(x-2\right)^4 \left(x+1\right)^3$ has local maxima, local minima and point of inflection.

$f\left(x\right) = \left(x-2\right)^4 \left(x+1\right)^3$

$\begin{aligned} \therefore f'\left(x\right) & = 4 \left(x-2\right)^3 \left(x+1\right)^3 + 3 \left(x+1\right)^2 \left(x-2\right)^4 \\ & = \left(x-2\right)^3 \left(x+1\right)^2 \left[4\left(x+1\right)+3\left(x-2\right)\right] \\ & = \left(x-2\right)^3 \left(x+1\right)^2 \left(7x-2\right) \end{aligned}$

For critical points, $f'\left(x\right) = 0$

i.e. $\left(x-2\right)^3 \left(x+1\right)^2 \left(7x-2\right) = 0$

$\implies$ $x-2=0$ or $x+1=0$ or $7x-2=0$

i.e. $x=2$ or $x=-1$ or $x = \dfrac{2}{7}$

Consider $x=2$

When $x<2$, let $x=1.9$

$\begin{aligned} \text{Then, } f'\left(x\right) = f'\left(1.9\right) & = \left(1.9-2\right)^3 \left(1.9+1\right)^2 \left[\left(7\times 1.9\right) -2\right] \\ & = \left(-ve\right) \left(+ve\right) \left(+ve\right) \\ & = -ve \end{aligned}$

When $x>2$, let $x=2.1$

$\begin{aligned} \text{Then, } f'\left(x\right) = f'\left(2.1\right) & = \left(2.1-2\right)^3 \left(2.1+1\right)^2 \left[\left(7\times 2.1\right) -2\right] \\ & = \left(+ve\right) \left(+ve\right) \left(+ve\right) \\ & = +ve \end{aligned}$

i.e. $f'\left(x\right)$ changes sign from negative to positive as x increases through 2.

$\implies$ $x=2$ is a point of local minimum.

Consider $x=-1$

When $x<-1$, let $x=-1.1$

$\begin{aligned} \text{Then, } f'\left(x\right) = f'\left(-1.1\right) & = \left(-1.1-2\right)^3 \left(-1.1+1\right)^2 \left[7 \times\left(-1.1\right) -2\right] \\ & = \left(-ve\right) \left(+ve\right) \left(-ve\right) \\ & = +ve \end{aligned}$

When $x>-1$, let $x=-0.9$

$\begin{aligned} \text{Then, } f'\left(x\right) = f'\left(-0.9\right) & = \left(-0.9-2\right)^3 \left(-0.9+1\right)^2 \left[7 \times \left(-0.9\right) -2\right] \\ & = \left(-ve\right) \left(+ve\right) \left(-ve\right) \\ & = +ve \end{aligned}$

i.e. $f'\left(x\right)$ does not change sign as x increases through $-1$.

$\implies$ $x=-1$ is a point of inflexion.

Consider $x = \dfrac{2}{7}$

When $x < \dfrac{2}{7}$, let $x = \dfrac{1.9}{7}$

$\begin{aligned} \text{Then, } f'\left(x\right) = f'\left(\dfrac{1.9}{7}\right) & = \left(\dfrac{1.9}{7}-2\right)^3 \left(\dfrac{1.9}{7}+1\right)^2 \left[\left(7\times \dfrac{1.9}{7}\right) -2\right] \\ & = \left(-ve\right) \left(+ve\right) \left(-ve\right) \\ & = +ve \end{aligned}$

When $x>\dfrac{2}{7}$, let $x=\dfrac{2.1}{7}$

$\begin{aligned} \text{Then, } f'\left(x\right) = f'\left(\dfrac{2.1}{7}\right) & = \left(\dfrac{2.1}{7}-2\right)^3 \left(\dfrac{2.1}{7}+1\right)^2 \left[\left(7\times \dfrac{2.1}{7}\right) -2\right] \\ & = \left(-ve\right) \left(+ve\right) \left(+ve\right) \\ & = -ve \end{aligned}$

i.e. $f'\left(x\right)$ changes sign from positive to negative as x increases through $\dfrac{2}{7}$.

$\implies$ $x=\dfrac{2}{7}$ is a point of local maximum.

Application of Derivatives: Maxima and Minima

A window is in the form of a rectangle surmounted by semicircular opening. The total perimeter of the window is 10 m. Find the dimensions of the window for maximum air flow through the window.

Let width of window $= \ell$

Then diameter of semicircle $= \ell$

Let height of window $= b$

Perimeter of the window $= P = \ell + 2b + \dfrac{\pi \ell}{2}$

Given: Perimeter of window $= 10 \; m$

$\implies$ $\ell + 2b + \dfrac{\pi \ell}{2} = 10$

i.e. $\ell \left(2+\pi\right) + 4b = 20$

i.e. $b = 5 - \dfrac{\ell \left(2+\pi\right)}{4}$ $\;\; \cdots$ (1)

Area of the window $=A= b \; \ell + \dfrac{\pi}{2} \times \dfrac{\ell^2}{4}$ $\;\; \cdots$ (2)

Substituting the value of b from equation (1) in equation (2) gives

$A = \ell \left[5- \dfrac{\ell \left(2+\pi\right)}{4}\right] + \dfrac{\pi \ell^2}{8}$

i.e. $A= 5 \ell - \dfrac{\pi}{4} \ell^2 -\dfrac{1}{2}\ell^2 + \dfrac{\pi}{8} \ell^2$

i.e. $A = 5 \ell - \dfrac{\pi}{8}\ell^2 - \dfrac{1}{2}\ell^2$

i.e. $A = 5 \ell - \dfrac{\ell^2}{2}\left(\dfrac{\pi}{4}+1\right)$

For maximum air flow, $\dfrac{dA}{d\ell} = 0$

i.e. $5-\ell\left(\dfrac{\pi+4}{4}\right) = 0$

$\implies$ $\ell = \dfrac{20}{\pi+4}$

Substituting the value of $\ell$ in equation (1) gives

$b = 5 - \left(\dfrac{\pi + 2}{4}\right) \left(\dfrac{20}{\pi + 4}\right)$

i.e. $b = 5 - 5 \left(\dfrac{\pi + 2}{\pi + 4}\right)$

i.e. $b = \dfrac{5\pi + 20 - 5\pi -10}{\pi + 4}$

i.e. $b = \dfrac{10}{\pi + 4}$

$\therefore$ Length of the window $\ell = \dfrac{20}{\pi + 4} \; m$; breadth of the window $=b=\dfrac{10}{\pi + 4} \; m$

Application of Derivatives: Maxima and Minima

The total cost of manufacturing x pocket radios per day is ₹ $\left(\dfrac{x^2}{4}+35x+25\right)$ and rate at which they may be sold to a distributor is ₹ $\left(\dfrac{100-x}{2}\right)$ each. What should be the daily output to attain a maximum total profit.

Daily output $=x$ radios

Cost price of x radios $=CP= ₹ \left(\dfrac{x^2}{4}+35x+25\right) $

Sale price of x radios $= SP= ₹ \; x \left(50 - \dfrac{x}{2}\right) = ₹ \left(50x - \dfrac{x^2}{2}\right)$

$\therefore$ Profit function $P\left(x\right) = SP - CP= \left(50x - \dfrac{x^2}{2} - \dfrac{x^2}{4} - 35x - 25\right)$

i.e. $P\left(x\right) = 15x - \dfrac{3x^2}{4}-25$

$\therefore$ For maximum profit, $\dfrac{dP}{dx} = 0$

Now, $\dfrac{dP}{dx} = 15 - \dfrac{6x}{4}$

$\therefore$ $\dfrac{dP}{dx}=0$ $\implies$ $15 - \dfrac{3}{2}x = 0$

i.e. $\dfrac{3}{2}x = 15$ $\implies$ $x = 10$

$\therefore$ Daily output $=10$ radios.