Chapter 6: Applications of Derivatives

Ans: $f(x) = \dfrac{{\log x}}{x}$

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

${f^\prime }(x) = 0$

$\Rightarrow 1 - \log x = 0$

$\Rightarrow \log x = 1$

$\Rightarrow \log x = \log e$

$\Rightarrow x = e$

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

$= \dfrac{{ - x - 2x(1 - \log x)}}{{{x^4}}}$

$= \dfrac{{ - 3 + 2\log x}}{{{x^3}}}$

${f^{\prime \prime }}(e) =  = \dfrac{{ - 3 + 2\log e}}{{{e^3}}} = \dfrac{{ - 3 + 2}}{{{e^3}}} = \dfrac{{ - 1}}{{{e^3}}} < 0$

$f$ is the maximum at $x = e$

Ans: Let $\vartriangle ABC$ be isosceles where \[BC\]is the base of fixed length $b$. 

Let the length of the two equal sides of $\vartriangle ABC$ be $a$. 

Draw $AD \bot BC$.

$AD = \sqrt {{a^2} - \dfrac{{{b^2}}}{4}} $

Area of triangle $= \dfrac{1}{2}b\sqrt {{a^2} - \dfrac{{{b^2}}}{4}} $

$\dfrac{{dA}}{{dt}} = \dfrac{1}{2}b \cdot \dfrac{{2a}}{{2\sqrt {{a^2} - \dfrac{{{b^2}}}{4}} }}\dfrac{{da}}{{dt}} = \dfrac{{ab}}{{\sqrt {4{a^2} - {b^2}} }}\dfrac{{da}}{{dt}}$

$\dfrac{{da}}{{dt}} =  - 3\;{\text{cm}}/{\text{s}}$

$\therefore \dfrac{{da}}{{dt}} = \dfrac{{ - 3ab}}{{\sqrt {4{a^2} - {b^2}} }}$

when $a = b$

$\dfrac{{dA}}{{dt}} = \dfrac{{ - 3{b^2}}}{{\sqrt {4{b^2} - {b^2}} }} = \dfrac{{ - 3{b^2}}}{{\sqrt {3{b^2}} }} =  - \sqrt {3b}$

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

$\therefore {f^\prime }(x) = \dfrac{{(2 + \cos x)(4\cos x - 2 - \cos x + x\sin x) - (4\sin x - 2x - x\cos x)( - \sin x)}}{{{{(2 + \cos x)}^2}}}$

$= \dfrac{{(2 + \cos x)(3\cos x - 2 + x\sin x) + \sin x(4\sin x - 2x - x\cos x)}}{{{{(2 + \cos x)}^2}}}$

$= \dfrac{{6\cos x - 4 + 2x\sin x + 3{{\cos }^2}x - 2\cos x + x\sin x\cos x + 4{{\sin }^2}x - 2{{\sin }^2}x - 2x\sin x - x\sin x\cos x}}{{{{(2 + \cos x)}^2}}}$

$= \dfrac{{4\cos x - 4 + 3{{\cos }^2}x + 4{{\sin }^2}x}}{{{{(2 + \cos x)}^2}}}$

$= \dfrac{{4\cos x - {{\cos }^2}x}}{{{{(2 + \cos x)}^2}}} = \dfrac{{\cos x(4 - \cos x)}}{{{{(2 + \cos x)}^2}}}$

${f^\prime }(x) = 0$

$\Rightarrow \cos x = 0\cos x = 4$

$\cos x \ne 4$

$\cos x = 0$

$\Rightarrow x = \dfrac{\pi }{2},\dfrac{{3\pi }}{2}$

$\operatorname{In} \left( {0,\dfrac{\pi }{2}} \right)$ and $\left( {\dfrac{{3\pi }}{2},2\pi } \right),{f^\prime }(x) > 0$

$f(x)$ is increasing for $0 < x < \dfrac{x}{2}$ and $\dfrac{{3\pi }}{2} < x < 2\pi $.

$\operatorname{In} \left( {\dfrac{\pi }{2},\dfrac{{3\pi }}{2}} \right),{f^\prime }(x) < 0$

$f(x)$ is decreasing for $\dfrac{\pi }{2} < x < \dfrac{{3\pi }}{2}$.

Ans: $f(x) = {x^3} + \dfrac{1}{{{x^3}}}$

$\therefore {f^\prime }(x) = 3{x^2} - \dfrac{3}{{{x^4}}} = \dfrac{{3{x^6} - 3}}{{{x^4}}}$

${f^\prime }(x) = 0 \Rightarrow 3{x^6} - 3 = 0 \Rightarrow {x^6} = x =  \pm 1$

In $( - \infty ,1)$ and $(1,\infty )$ i.e.., when $x <  - 1$ and $x > 1,{f^\prime }(x) > 0.$ when $x <  - 1$ and $x > 1,f$ is increasing. In $( - 1,1)$ i.e., when $ - 1 < x < 1,{f^\prime }(x) < 0$.

Thus, when $ - 1 < x < 1,f$ is decreasing.

Ans:

Ellipse $\dfrac{{{x^2}}}{{{a^2}}} + \dfrac{{{y^2}}}{{{b^2}}} = 1$

Let $A B C$, be the triangle inscribed in the ellipse where vertex $C$ is at $(a,0)$. Since the ellipse is symmetrical with $x$-axis and $y - axis$ ${y_1} =  \pm \dfrac{b}{a}\sqrt {{a^2} - x_1^2} .$

Coordinates of $A$ are $\left( { - {x_{1,}}\dfrac{b}{a}\sqrt {{a^2} - {x^2}1} } \right)$ and coordinates of $B$ are $\left( {{x_1}, - \dfrac{b}{a}\sqrt {{a^2} - x_1^2} .} \right)$ As the point $\left( { - {x_1},{y_1}} \right)$ lies on the ellipse, the area of triangle $A B C$ is $A = \dfrac{1}{2}\mid a\left( {\dfrac{{2b}}{a}\sqrt {{a^2} - x_1^2} } \right) + \left( { - {x_1}} \right)\left( { - \dfrac{b}{a}\sqrt {{a^2} - x_1^2} } \right) + \left( { - {x_1}} \right)\left( { - \dfrac{b}{a}\sqrt {{a^2} - {x^2}1} } \right)$

$\Rightarrow A = ba\sqrt {{a^2} - {x^2}1}  + {x_1}\dfrac{b}{a}\sqrt {{a^2} - x_1^2} $

$\therefore \dfrac{{dA}}{{d{x_1}}} = \dfrac{{ - 2xb}}{{2\sqrt {{a^2} - x_1^2} }} + \dfrac{b}{a}\sqrt {{a^2} - x_1^2}  - \dfrac{{2bx_1^2}}{{{a^2}\sqrt {{a^2} - x_1^2} }}$

$= \dfrac{b}{{2\sqrt {{a^2} - x_1^2} }}\left[ { - {x_1}a + \left( {{a^2} - x_1^2} \right) - x_1^2} \right]$

$= \dfrac{{b\left( { - 2x_1^2 - x_1^2 + {a^2}} \right)}}{{a\sqrt {{a^2} - x_1^2} }}$

$\dfrac{{dA}}{{d{x_1}}} = 0$

$\Rightarrow  - 2x_1^2 - {x_1}a + {a^2} = 0$

$\Rightarrow {x_1} = \dfrac{{a \pm \sqrt {{a^2} - 4( - 2)\left( {{a^2}} \right)} }}{{2( - 2)}}$

$= \dfrac{{a \pm \sqrt {9{a^2}} }}{{ - 4}}$

$= \dfrac{{a \pm 3a}}{{ - 4}}$

$\Rightarrow {x_1} =  - a,\dfrac{a}{2}$

${x_1}$ cannot be equal to $a$. $\therefore {x_1} = \dfrac{a}{2} \Rightarrow {y_1} = \dfrac{b}{a}\sqrt {{a^2} - \dfrac{{{a^2}}}{4}}  = \dfrac{{ba}}{{2a}}\sqrt 3  = \dfrac{{\sqrt {3b} }}{2}$

Now,$\dfrac{{{d^2}A}}{{d{x^2}1}} = \dfrac{b}{a}\left\{ {\dfrac{{\sqrt {{a^2} - x_1^2} \left( { - 4{x_1} - a} \right) - \left( { - 2x_1^2 - {x_1}a + {a^2}} \right)\dfrac{{\left( { - 2{x_1}} \right)}}{{2\sqrt {{a^2} - x_1^2} }}}}{{{a^2} - x_1^2}}} \right\}$

$= \dfrac{b}{a}\left\{ {\dfrac{{\left( {{a^2} - x_1^2} \right)\left( { - 4{x_1} - a} \right) + {x_1}\left( { - 2x_1^2 - {x_1}a + {a^2}} \right)}}{{{{\left( {{a^2} - x_1^2} \right)}^{\dfrac{2}{3}}}}}} \right\}$

$= \dfrac{b}{a}\left\{ {\dfrac{{2{x^3} - 3{a^2}x - {a^3}}}{{{{\left( {{a^2} - x_1^2} \right)}^{\dfrac{3}{2}}}}}} \right\}$

when ${x_1} = \dfrac{a}{2}$,

$\dfrac{{{d^2}A}}{{dx_1^2}} = \dfrac{b}{a}\left\{ {\dfrac{{2\dfrac{{{a^3}}}{8} - 3\dfrac{{{a^3}}}{2} - {a^3}}}{{{{\left( {\dfrac{{3{a^2}}}{4}} \right)}^{\dfrac{3}{2}}}}}} \right\} = \dfrac{b}{a}\left\{ {\dfrac{{\dfrac{{{a^3}}}{4} - \dfrac{3}{2}{a^3} - {a^3}}}{{{{\left( {\dfrac{{3{a^2}}}{4}} \right)}^{\dfrac{3}{2}}}}}} \right\}$

$= \dfrac{b}{a}\left\{ {\dfrac{{\dfrac{9}{4}{a^3}}}{{{{\left( {\dfrac{{3{a^2}}}{4}} \right)}^{\dfrac{3}{2}}}}}} \right\} < 0$

Area is the maximum when ${x_1} = \dfrac{a}{2}$. Maximum area of the triangle is $A = b\sqrt {{a^2} - \dfrac{{{a^2}}}{4}}  + \left( {\dfrac{a}{2}} \right)\dfrac{b}{a}\sqrt {{a^2} - \dfrac{{{a^2}}}{4}} $

$= ab\dfrac{{\sqrt 3 }}{2} + \left( {\dfrac{a}{2}} \right)\dfrac{b}{a} \times \dfrac{{a\sqrt 3 }}{2}$

$= \dfrac{{ab\sqrt 3 }}{2} + \dfrac{{ab\sqrt 3 }}{4} = \dfrac{{3\sqrt 3 }}{4}ab$

Ans: Let \[l,b\]and $h$ represent the length, breadth, and height of the tank respectively.

height $(h) = 2m$

Volume of the tank $= 8\;{{\text{m}}^3}$ Volume of the tank $= l \times b \times {\text{h}}$ $8 = l \times b \times 2$

$\Rightarrow lb = 4 \Rightarrow b = \dfrac{4}{l}$

Area of the base $= lb = 4$ 

Area of 4 walls $(A) = 2h(l + b)$

$\therefore A = 4\left( {l + \dfrac{4}{l}} \right)$

$\Rightarrow \dfrac{{dA}}{{dl}} = 4\left( {1 - \dfrac{4}{{{l^2}}}} \right)$

$Now,\dfrac{{dA}}{{dl}} = 0$

$\Rightarrow l - \dfrac{4}{{{l^2}}} = 0$

$\Rightarrow {l^2} = 4$

$\Rightarrow l =  \pm 2$

Therefore, we have $l = 4$. 

$\therefore b = \dfrac{4}{l} = \dfrac{4}{2} = 2$

$\dfrac{{{d^2}A}}{{d{l^2}}} = \dfrac{{32}}{{{l^3}}}$

$l = 2,\dfrac{{{d^2}A}}{{d{l^2}}} = \dfrac{{32}}{8} = 4 > 0.$

Area is the minimum when $l = 2$. 

We have $l = b = h = 2$.

Cost of building base \[ = \] Rs $70{\text{x}}(lb) = $ Rs$70(4) = $ Rs 280

Cost of building walls $= Rs2h(l + h) \times 45 = Rs90(2)(2 + 2) = $ 

Rs $8(90) = $ Rs 720

Required total cost $= Rs(280 + 720) = $ Rs 1000

Ans: $2\pi r + 4a = k$ (where $k$ is constant) $\Rightarrow a = \dfrac{{k - 2\pi r}}{4}$

the sum of the areas of the circle and the square $(A)$ is given by,

$A = \pi {r^2} + {a^2} = \pi {r^2} + \dfrac{{{{(k = 2\pi r)}^2}}}{{16}}$

$\therefore \dfrac{{dA}}{{dr}} = 2\pi r + \dfrac{{2(k - 2\pi r)(2\pi )}}{{16}} = 2\pi r$

$=  - \dfrac{{\pi (k - 2\pi r)}}{4}$

Now, $\dfrac{{dA}}{{dr}} = 0$

$\Rightarrow 2\pi r = \dfrac{{\pi (k - 2\pi r)}}{4}$

$8r = k - 2\pi r$

$\Rightarrow (8 + 2\pi )r = k$

$\Rightarrow r = \dfrac{k}{{8 + 2\pi }} = \dfrac{k}{{2(4 + \pi )}}$

Now, $\dfrac{{{d^2}A}}{{d{r^2}}} = 2\pi  + \dfrac{{{\pi ^2}}}{2} > 0$

$\therefore $ where $r = \dfrac{k}{{2(4 + \pi )}},\dfrac{{{d^2}A}}{{d{r^2}}} > 0.$

area is least when $r = \dfrac{k}{{2(4 + \pi )}}$ where $r = \dfrac{k}{{2(4 + \pi )}},$

$a = \dfrac{{k - 2\pi \left[ {\dfrac{k}{{2(4 + \pi )}}} \right]}}{4} = \dfrac{{8k + 2\pi k - 2\pi k}}{{2(4 + \pi ) \times 4}} = \dfrac{k}{{4 + \pi }} = 2r$

Ans: $x$ and $y$ are the length and breadth of the rectangular window.

Radius of semicircular opening $= \dfrac{x}{2}$

$\therefore x + 2y + \dfrac{{\pi {\text{x}}}}{2} = 10$

$\Rightarrow x\left( {1 + \dfrac{\pi }{2}} \right) + 2y = 10$

$\Rightarrow 2y = 10 - x\left( {1 + \dfrac{\pi }{2}} \right)$

$\Rightarrow y = 5 - x\left( {\dfrac{1}{2} + \dfrac{\pi }{4}} \right)$

$A = xy + \dfrac{\pi }{2}{\left( {\dfrac{x}{2}} \right)^2}$

$= x\left[ {5 - x\left( {\dfrac{1}{2} + \dfrac{\pi }{4}} \right)} \right] + \dfrac{\pi }{8}{\text{x}}{x^2}$

$= 5x - {x^2}\left( {\dfrac{1}{2} + \dfrac{\pi }{4}} \right) + \dfrac{\pi }{8}{\text{x}}{x^2}$

$\therefore \dfrac{{dA}}{{dx}} = 5 - 2x\left( {\dfrac{1}{2} + \dfrac{\pi }{4}} \right) + \dfrac{\pi }{4}x$

$\dfrac{{{d^2}A}}{{d{x^2}}} =  - \left( {1 - \dfrac{\pi }{2}} \right) + \dfrac{\pi }{4} =  - 1 - \dfrac{\pi }{4}$

$\dfrac{{dA}}{{dx}} = 0$

$\Rightarrow 5 - x\left( {1 + \dfrac{\pi }{2}} \right) + \dfrac{\pi }{4}x = 0$

$\Rightarrow 5 - x - \dfrac{\pi }{4}x = 0$

$\Rightarrow x\left( {1 + \dfrac{\pi }{4}} \right) = 5$

$\Rightarrow x = \dfrac{5}{{\left( {1 + \dfrac{\pi }{4}} \right)}} = \dfrac{{20}}{{\pi  + 4}}$

$x = \dfrac{{20}}{{\pi  + 4}},\dfrac{{{d^2}A}}{{d{x^2}}} < 0.$

area is maximum when length $x = \dfrac{{20}}{{\pi  + 4}}m$. 

Now, $y = 5 - \dfrac{{20}}{{\pi  + 4}}\left( {\dfrac{{2 + \pi }}{4}} \right) = 5 - \dfrac{{5(2 + \pi )}}{{\pi  + 4}} = \dfrac{{10}}{{\pi  + 4}}m$

the required dimensions length $= \dfrac{{20}}{{\pi  + 4}}m$ and breadth $= \dfrac{{10}}{{\pi  + 4}}m$.

Ans: $\vartriangle ABC$ right-angled at $B$. $AB = x$ and $BC = y$.

$P$ be a point on hypotenuse such that ${\text{P}}$ is at a distance of a and ${\text{b}}$ from the sides \[AB\]and \[BC\]respectively.

$\angle c = \theta $

$AC = \sqrt {{x^2} + {y^2}} $

$PC = b\operatorname{cosec} \theta $

$AP = a\sec \theta $

$AC = AP + PC$

$AC = b\operatorname{cosec} \theta  + a\sec \theta  \ldots ..(1)$

$\therefore \dfrac{{d(AC)}}{{d\theta }} =  - b\operatorname{cosec} \theta \cot \theta  + a\sec \theta \tan \theta $

$\therefore \dfrac{{d(AC)}}{{d\theta }} = 0$

$\Rightarrow a\sec \theta \tan \theta  = b\cos \operatorname{ec} \theta \cot \theta $

$\Rightarrow \dfrac{a}{{\cos \theta }} \cdot \dfrac{{\sin \theta }}{{\cos \theta }} = \dfrac{b}{{\sin \theta }}\dfrac{{\cos \theta }}{{\sin \theta }}$

$\Rightarrow a{\sin ^3}\theta  = b{\cos ^3}\theta $

$\Rightarrow {(a)^{\dfrac{1}{3}}}\sin \theta  = {(b)^{\dfrac{1}{3}}}\cos \theta $

$\Rightarrow \tan \theta  = {\left( {\dfrac{b}{a}} \right)^{\dfrac{1}{3}}}$

$\therefore \sin \theta  = \dfrac{{{{(b)}^{\dfrac{1}{3}}}}}{{\sqrt {{a^{\dfrac{2}{3}}} + {b^{\dfrac{2}{3}}}} }}$ and $\cos \theta  = \dfrac{{{{(a)}^{\dfrac{1}{3}}}}}{{\sqrt {{a^{\dfrac{2}{3}}} + {b^{\dfrac{2}{3}}}} }}\quad  \ldots ..(2)$

clearly $\dfrac{{{d^2}(AC)}}{{d{\theta ^2}}} < 0$ when $\tan \theta  = {\left( {\dfrac{b}{a}} \right)^{\dfrac{1}{3}}}$.

the length of the hypotenuse is the maximum when $\tan \theta  = {\left( {\dfrac{b}{a}} \right)^{\dfrac{1}{3}}}$. 

Now, when $\tan \theta  = {\left( {\dfrac{b}{a}} \right)^{\dfrac{1}{3}}}$

$\tan \theta  = {\left( {\dfrac{b}{a}} \right)^{\dfrac{1}{3}}}$

$AC = \dfrac{{b\sqrt {{a^{\dfrac{2}{3}}} + {b^{\dfrac{2}{3}}}} }}{{{b^{\dfrac{1}{3}}}}} + \dfrac{{a\sqrt {{a^{\dfrac{2}{3}}} + {b^{\dfrac{2}{3}}}} }}{{{a^{\dfrac{1}{3}}}}}$

$= a\sqrt {{a^{\dfrac{2}{3}}} + {b^{\dfrac{2}{3}}}} \left( {{b^{\dfrac{2}{3}}} + {a^{\dfrac{2}{3}}}} \right)$

$= {\left( {{a^{\dfrac{2}{3}}} + {b^{\dfrac{2}{3}}}} \right)^{\dfrac{3}{2}}}$

maximum length of the hypotenuse is $= {\left( {{a^{\dfrac{2}{3}}} + {b^{\dfrac{2}{3}}}} \right)^{\dfrac{3}{2}}}$.

Ans: $f(x) = {(x - 2)^4}{(x + 1)^3}$

$\therefore {f^\prime }(x) = 4{(x - 2)^3}{(x + 1)^3} + 3{(x + 1)^2}{(x - 2)^4}$

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

$= {(x - 2)^3}{(x + 1)^2}(7x - 2)$

${f^\prime }(x) = 0 \Rightarrow x =  - 1$ and $x = \dfrac{2}{7}$ or $x = 2$

for $x$ close to $\dfrac{2}{7}$ and to left of $\dfrac{2}{7},{f^\prime }(x) > 0$.

for $x$ close to $\dfrac{2}{7}$ and to right of $\dfrac{2}{7},{f^\prime }(x) > 0$. $x = \dfrac{2}{7}$ is point of local minima.

as the value of $x$ varies ${f^\prime }(x)$ does not changes its sign.

$x =  - 1$ is point of inflexion.