Chapter 6: Application of Derivatives (Calculus)

$A=\pi r^2\Rightarrow \frac{dA}{dr}=2\pi r$.

(a) $2\pi (3)=6\pi$ (b) $2\pi (4)=8\pi$

(a) 6π cm²/cm (b) 8π cm²/cm

$V=x^3\Rightarrow \frac{dV}{dt}=3x^2\frac{dx}{dt}=8\Rightarrow \frac{dx}{dt}=\frac{8}{3(144)}=\frac{1}{54}$ cm/s.

$S=6x^2\Rightarrow \frac{dS}{dt}=12x\frac{dx}{dt}=12(12)\cdot \frac{1}{54}=\frac{8}{3}$.

$\frac{8}{3}$ cm²/s

$\frac{dA}{dt}=2\pi r\frac{dr}{dt}=2\pi (10)(3)=60\pi$.

60π cm²/s

$\frac{dV}{dt}=3x^2\frac{dx}{dt}=3(100)(3)=900$.

900 cm³/s

$\frac{dA}{dt}=2\pi r\frac{dr}{dt}=2\pi (8)(5)=80\pi$.

80π cm²/s

$C=2\pi r\Rightarrow \frac{dC}{dt}=2\pi (0.7)=1.4\pi$.

1.4π cm/s

$\frac{dx}{dt}=-5$, $\frac{dy}{dt}=4$.

(a) $P=2(x+y)$: $\frac{dP}{dt}=2(-5+4)=-2$ cm/min (decreasing).

(b) $A=xy$: $\frac{dA}{dt}=x\frac{dy}{dt}+y\frac{dx}{dt}=8(4)+6(-5)=2$ cm²/min.

(a) −2 cm/min (b) 2 cm²/min

$V=\frac{4}{3}\pi r^3\Rightarrow \frac{dV}{dt}=4\pi r^2\frac{dr}{dt}$.

$900=4\pi (225)\frac{dr}{dt}\Rightarrow \frac{dr}{dt}=\frac{900}{900\pi }=\frac{1}{\pi }$.

$\frac{1}{\pi }$ cm/s

$\frac{dV}{dr}=4\pi r^2=4\pi (100)=400\pi$.

400π cm³/cm

Let the foot be x m from the wall and the top y m high: $x^2+y^2=25$. When $x=4$, $y=3$.

Differentiate: $2x\frac{dx}{dt}+2y\frac{dy}{dt}=0\Rightarrow \frac{dy}{dt}=-\frac{x}{y}\frac{dx}{dt}=-\frac{4}{3}(2)=-\frac{8}{3}$ cm/s.

The height decreases at $\frac{8}{3}$ cm/s.

$6\frac{dy}{dt}=3x^2\frac{dx}{dt}$. With $\frac{dy}{dt}=8\frac{dx}{dt}$: $48=3x^2\Rightarrow x=\pm 4$.

$x=4$: $y=\frac{66}{6}=11$. $x=-4$: $y=\frac{-62}{6}=-\frac{31}{3}$.

(4, 11) and $\left(-4,-\frac{31}{3}\right)$

$\frac{dV}{dt}=4\pi r^2\frac{dr}{dt}=4\pi (1)\left(\frac{1}{2}\right)=2\pi$.

2π cm³/s

Radius $r=\frac{3}{4}(2x+1)$, so $V=\frac{4}{3}\pi \cdot \frac{27}{64}(2x+1{)}^3=\frac{9\pi }{16}(2x+1{)}^3$.

$\frac{dV}{dx}=\frac{9\pi }{16}\cdot 3(2x+1{)}^2\cdot 2=\frac{27\pi }{8}(2x+1{)}^2$.

$\frac{27\pi }{8}(2x+1{)}^2$

$h=\frac{r}{6}\Rightarrow r=6h$. $V=\frac{1}{3}\pi r^{2}h=\frac{1}{3}\pi (36h^2)h=12\pi h^3$.

$\frac{dV}{dt}=36\pi h^2\frac{dh}{dt}\Rightarrow 12=36\pi (16)\frac{dh}{dt}\Rightarrow \frac{dh}{dt}=\frac{1}{48\pi }$.

$\frac{1}{48\pi }$ cm/s

$MC=C^{\prime }(x)=0.021x^2-0.006x+15$.

$C^{\prime }(17)=0.021(289)-0.006(17)+15=6.069-0.102+15=20.967$.

₹20.967

$R^{\prime }(x)=26x+26$; $R^{\prime }(7)=182+26=208$.

₹208

$2\pi r=12\pi$.

(B)

$R^{\prime }(x)=6x+36$; $R^{\prime }(15)=126$.

(D)

$f^{\prime }(x)=3>0$ for all x.

f′(x) = 3 > 0, so f is increasing on R.

$f^{\prime }(x)=2e^{2x}>0$ for all x.

f′(x) = 2e²ˣ > 0, so f is increasing on R.

$f^{\prime }(x)=\cos x$.

(a) On $(0,\pi /2)$, $\cos x>0$: increasing.

(b) On $(\pi /2,\pi )$, $\cos x<0$: decreasing.

(c) On $(0,\pi )$, f′ changes sign at $\pi /2$, so f is neither increasing nor decreasing there.

Shown using the sign of cos x.

$f^{\prime }(x)=4x-3=0\Rightarrow x=\frac{3}{4}$. $f^{\prime }>0$ for $x>\frac{3}{4}$, $f^{\prime }<0$ for $x<\frac{3}{4}$.

(a) $\left(\frac{3}{4},\infty \right)$ (b) $\left(-\infty ,\frac{3}{4}\right)$

$f^{\prime }(x)=6x^2-6x-36=6(x-3)(x+2)$.

Sign: $+$ on $(-\infty ,-2)$, $-$ on $(-2,3)$, $+$ on $(3,\infty )$.

(a) (−∞, −2) and (3, ∞) (b) (−2, 3)

(a) $f^{\prime }=2x+2$: decreasing for $x<-1$, increasing for $x>-1$.

(b) $f^{\prime }=-6-4x$: increasing for $x<-\frac{3}{2}$, decreasing for $x>-\frac{3}{2}$.

(c) $f^{\prime }=-6x^2-18x-12=-6(x+1)(x+2)$: positive for $-2<x<-1$; negative for $x<-2$ and $x>-1$.

(d) $f^{\prime }=-9-2x$: increasing for $x<-\frac{9}{2}$, decreasing for $x>-\frac{9}{2}$.

(e) $f^{\prime }=3(x+1{)}^2(x-3{)}^3+3(x+1{)}^3(x-3{)}^2=3(x+1{)}^2(x-3{)}^2(2x-2)=6(x+1{)}^2(x-3{)}^2(x-1)$. The squares are ≥ 0, so the sign is that of $(x-1)$.

(a) dec x < −1, inc x > −1. (b) inc x < −3/2, dec x > −3/2. (c) inc −2 < x < −1; dec x < −2 and x > −1. (d) inc x < −9/2, dec x > −9/2. (e) inc on (1, 3) and (3, ∞); dec on (−∞, −1) and (−1, 1).

$$\frac{dy}{dx}=\frac{1}{1+x}-\frac{2(2+x)-2x}{(2+x{)}^2}=\frac{1}{1+x}-\frac{4}{(2+x{)}^2}=\frac{(2+x{)}^2-4(1+x)}{(1+x)(2+x{)}^2}=\frac{x^2}{(1+x)(2+x{)}^2}$$

For $x>-1$, the denominator is positive and $x^2\ge 0$, so $\frac{dy}{dx}\ge 0$ (zero only at the single point $x=0$).

dy/dx = x²/[(1 + x)(2 + x)²] ≥ 0, so y is increasing for x > −1.

$y=x^2(x-2{)}^2$; $\frac{dy}{dx}=2x(x-2{)}^2+2x^2(x-2)=2x(x-2)(2x-2)=4x(x-1)(x-2)$.

Sign: $-$ for $x<0$, $+$ for $0<x<1$, $-$ for $1<x<2$, $+$ for $x>2$.

0 < x < 1 and x > 2

$$\frac{dy}{d\theta }=\frac{4\cos \theta (2+\cos \theta )+4{\sin }^2\theta }{(2+\cos \theta {)}^2}-1=\frac{8\cos \theta +4-(2+\cos \theta {)}^2}{(2+\cos \theta {)}^2}=\frac{\cos \theta (4-\cos \theta )}{(2+\cos \theta {)}^2}$$

On $[0,\pi /2]$, $\cos \theta \ge 0$ and $4-\cos \theta >0$, so $\frac{dy}{d\theta }\ge 0$ (zero only at $\theta =\pi /2$).

dy/dθ = cos θ(4 − cos θ)/(2 + cos θ)² ≥ 0, so y is increasing.

$f(x)=\log x$, $f^{\prime }(x)=\frac{1}{x}>0$ for $x>0$.

f′(x) = 1/x > 0 on (0, ∞).

$f^{\prime }(x)=2x-1$: negative on $\left(-1,\frac{1}{2}\right)$ and positive on $\left(\frac{1}{2},1\right)$. f decreases and then increases, so it is neither on the whole interval.

f′ changes sign at x = 1/2.

(A) $-\sin x<0$ ✔. (B) $-2\sin 2x<0$ as $2x\in (0,\pi )$ ✔. (C) $-3\sin 3x>0$ when $3x\in (\pi ,3\pi /2)$, so not decreasing throughout. (D) ${\sec }^{2}x>0$, increasing.

(A) and (B)

$f^{\prime }(x)=100x^{99}+\cos x$.

On (0, 1) and (0, π/2): both terms ≥ 0 and the sum is positive. On (π/2, π): $x>1$, so $100x^{99}>100>|\cos x|$, hence $f^{\prime }>0$. f is increasing on all three.

(D)

$f^{\prime }(x)=2x+a$. Its least value on [1, 2] is at $x=1$: $2+a$. Need $2+a>0$.

a > −2

$f^{\prime }(x)=1-\frac{1}{x^2}=\frac{x^2-1}{x^2}$. On I, $|x|>1$, so $x^2-1>0$ and $f^{\prime }(x)>0$.

f′(x) > 0 for |x| > 1, so f is increasing on I.

$f^{\prime }(x)=\frac{\cos x}{\sin x}=\cot x$: positive on $(0,\pi /2)$, negative on $(\pi /2,\pi )$.

f′(x) = cot x: > 0 on (0, π/2), < 0 on (π/2, π).

$f^{\prime }(x)=\frac{-\sin x}{\cos x}=-\tan x$. On $(0,\pi /2)$, $\tan x>0\Rightarrow f^{\prime }<0$. On $(3\pi /2,2\pi )$, $\tan x<0\Rightarrow f^{\prime }>0$.

f′(x) = −tan x gives the required signs.

$f^{\prime }(x)=3x^2-6x+3=3(x-1{)}^2\ge 0$, zero only at $x=1$.

f′(x) = 3(x − 1)² ≥ 0, so f is increasing on R.

$y^{\prime }=2x{e}^{-x}-x^2{e}^{-x}=x(2-x)e^{-x}>0$ for $0<x<2$.

(D)

(i) $(2x-1{)}^2\ge 0$, so $f(x)\ge 3$, equality at $x=\frac{1}{2}$. Min = 3; no maximum (f grows without bound).

(ii) $9x^2+12x+2=(3x+2{)}^2-2\ge -2$. Min = −2 (at $x=-\frac{2}{3}$); no maximum.

(iii) $-(x-1{)}^2\le 0$, so $f(x)\le 10$. Max = 10 (at $x=1$); no minimum.

(iv) $x^3+1$ takes all real values, so it has neither.

(i) min 3 (ii) min −2 (iii) max 10 (iv) neither max nor min.

(i) $|x+2|\ge 0\Rightarrow f\ge -1$ (at $x=-2$). Min −1; no max.

(ii) $-|x+1|\le 0\Rightarrow g\le 3$ (at $x=-1$). Max 3; no min.

(iii) $-1\le \sin 2x\le 1\Rightarrow 4\le h\le 6$. Min 4, max 6.

(iv) $2\le \sin 4x+3\le 4$, so $|\sin 4x+3|$ lies in [2, 4]. Min 2, max 4.

(v) On the open interval (−1, 1), $x+1$ takes values in (0, 2) but never reaches 0 or 2.

(i) min −1 (ii) max 3 (iii) min 4, max 6 (iv) min 2, max 4 (v) neither max nor min.

(i) $f^{\prime }=2x=0\Rightarrow x=0$; $f^{\prime \prime }=2>0$. Local min at 0, value 0.

(ii) $g^{\prime }=3x^2-3=0\Rightarrow x=\pm 1$; $g^{\prime \prime }=6x$. At $x=1$: $g^{\prime \prime }>0$, local min $g(1)=-2$. At $x=-1$: $g^{\prime \prime }<0$, local max $g(-1)=2$.

(iii) $h^{\prime }=\cos x-\sin x=0\Rightarrow x=\frac{\pi }{4}$; $h^{\prime \prime }=-\sin x-\cos x<0$. Local max at $\frac{\pi }{4}$, value $\sqrt{2}$.

(iv) $f^{\prime }=\cos x+\sin x=0\Rightarrow \tan x=-1\Rightarrow x=\frac{3\pi }{4},\frac{7\pi }{4}$. $f^{\prime \prime }=-\sin x+\cos x$: at $\frac{3\pi }{4}$ it is $-\sqrt{2}<0$ (local max $=\sqrt{2}$); at $\frac{7\pi }{4}$ it is $\sqrt{2}>0$ (local min $=-\sqrt{2}$).

(v) $f^{\prime }=3x^2-12x+9=3(x-1)(x-3)$; $f^{\prime \prime }=6x-12$. At 1: $f^{\prime \prime }<0$, local max $f(1)=19$. At 3: $f^{\prime \prime }>0$, local min $f(3)=15$.

(vi) $g^{\prime }=\frac{1}{2}-\frac{2}{x^2}=0\Rightarrow x=2$ ($x>0$); $g^{\prime \prime }=\frac{4}{x^3}>0$. Local min $g(2)=2$.

(vii) $g^{\prime }=\frac{-2x}{(x^2+2{)}^2}=0\Rightarrow x=0$; g′ changes from + to −. Local max $g(0)=\frac{1}{2}$.

(viii) $f^{\prime }=\sqrt{1-x}-\frac{x}{2\sqrt{1-x}}=\frac{2-3x}{2\sqrt{1-x}}=0\Rightarrow x=\frac{2}{3}$; f′ changes from + to −. Local max $f\left(\frac{2}{3}\right)=\frac{2}{3}\sqrt{\frac{1}{3}}=\frac{2\sqrt{3}}{9}$.

(i) min 0 at 0 (ii) max 2 at −1, min −2 at 1 (iii) max √2 at π/4 (iv) max √2 at 3π/4, min −√2 at 7π/4 (v) max 19 at 1, min 15 at 3 (vi) min 2 at 2 (vii) max 1/2 at 0 (viii) max 2√3/9 at 2/3

(i) $f^{\prime }(x)=e^x>0$ never zero: f is strictly increasing.

(ii) $g^{\prime }(x)=\frac{1}{x}>0$ for $x>0$, never zero.

(iii) $h^{\prime }(x)=3x^2+2x+1$; discriminant $4-12<0$ and leading coefficient positive, so $h^{\prime }>0$ always.

In each case f′ never vanishes, so there is no local maximum or minimum.

The derivative never becomes zero, so no maxima or minima.

(i) $f^{\prime }=3x^2=0\Rightarrow x=0$. $f(-2)=-8$, $f(0)=0$, $f(2)=8$.

(ii) $f^{\prime }=\cos x-\sin x=0\Rightarrow x=\frac{\pi }{4}$. $f(0)=1$, $f(\frac{\pi }{4})=\sqrt{2}$, $f(\pi )=-1$.

(iii) $f^{\prime }=4-x=0\Rightarrow x=4$. $f(-2)=-8-2=-10$, $f(4)=16-8=8$, $f(\frac{9}{2})=18-\frac{81}{8}=\frac{63}{8}$.

(iv) $f^{\prime }=2(x-1)=0\Rightarrow x=1$. $f(-3)=16+3=19$, $f(1)=3$.

(i) max 8, min −8 (ii) max √2, min −1 (iii) max 8, min −10 (iv) max 19, min 3

$p^{\prime }(x)=-72-36x=0\Rightarrow x=-2$; $p^{\prime \prime }(x)=-36<0$, a maximum.

$p(-2)=41+144-72=113$.

Maximum profit = 113 units

$f^{\prime }(x)=12x^3-24x^2+24x-48=12(x-2)(x^2+2)$, zero only at $x=2$.

$f(0)=25$, $f(2)=48-64+48-96+25=-39$, $f(3)=243-216+108-144+25=16$.

Maximum 25 (at x = 0), minimum −39 (at x = 2).

$\sin 2x=1\Rightarrow 2x=\frac{\pi }{2},\frac{5\pi }{2}\Rightarrow x=\frac{\pi }{4},\frac{5\pi }{4}$ (both in [0, 2π]).

$x=\frac{\pi }{4}$ and $\frac{5\pi }{4}$

$\sin x+\cos x=\sqrt{2}\sin \left(x+\frac{\pi }{4}\right)\le \sqrt{2}$, with equality at $x=\frac{\pi }{4}$. (Or: $f^{\prime }=0$ at $x=\frac{\pi }{4}$, $f^{\prime \prime }=-\sqrt{2}<0$.)

√2

$f^{\prime }(x)=6x^2-24=0\Rightarrow x=\pm 2$.

On [1, 3]: $f(1)=85$, $f(2)=75$, $f(3)=54-72+107=89$. Max 89 at $x=3$.

On [−3, −1]: $f(-3)=125$, $f(-2)=-16+48+107=139$, $f(-1)=129$. Max 139 at $x=-2$.

89 (at x = 3) on [1, 3]; 139 (at x = −2) on [−3, −1].

$f^{\prime }(x)=4x^3-124x+a$. At an interior maximum, $f^{\prime }(1)=0$: $4-124+a=0$.

a = 120

$f^{\prime }(x)=1+2\cos 2x=0\Rightarrow \cos 2x=-\frac{1}{2}\Rightarrow x=\frac{\pi }{3},\frac{2\pi }{3},\frac{4\pi }{3},\frac{5\pi }{3}$.

$f(0)=0$, $f(\frac{\pi }{3})=\frac{\pi }{3}+\frac{\sqrt{3}}{2}$, $f(\frac{2\pi }{3})=\frac{2\pi }{3}-\frac{\sqrt{3}}{2}$, $f(\frac{4\pi }{3})=\frac{4\pi }{3}+\frac{\sqrt{3}}{2}\approx 5.05$, $f(\frac{5\pi }{3})=\frac{5\pi }{3}-\frac{\sqrt{3}}{2}$, $f(2\pi )=2\pi \approx 6.28$.

Maximum 2π (at x = 2π), minimum 0 (at x = 0).

$P=x(24-x)$; $P^{\prime }=24-2x=0\Rightarrow x=12$; $P^{\prime \prime }=-2<0$.

12 and 12

$P=x(60-x{)}^3$; $P^{\prime }=(60-x{)}^3-3x(60-x{)}^2=(60-x{)}^2(60-4x)=0\Rightarrow x=15$ (x = 60 gives y = 0). P′ changes from + to − at 15.

x = 15, y = 45

$P=x^2(35-x{)}^5$; $P^{\prime }=2x(35-x{)}^5-5x^2(35-x{)}^4=x(35-x{)}^4[70-2x-5x]=x(35-x{)}^4(70-7x)$.

$P^{\prime }=0\Rightarrow x=10$ (P′ changes from + to −).

x = 10, y = 25

$S=x^3+(16-x{)}^3$; $S^{\prime }=3x^2-3(16-x{)}^2=0\Rightarrow x=16-x\Rightarrow x=8$; $S^{\prime \prime }=6x+6(16-x)=96>0$.

8 and 8

Cut squares of side x: $V=x(18-2x{)}^2$, $0<x<9$.

$V^{\prime }=(18-2x{)}^2-4x(18-2x)=(18-2x)(18-6x)=0\Rightarrow x=3$ (x = 9 gives no box).

$V^{\prime \prime }=24x-144$; at $x=3$: $-72<0$, maximum.

3 cm

$V=x(45-2x)(24-2x)=4x^3-138x^2+1080x$, $0<x<12$.

$V^{\prime }=12x^2-276x+1080=12(x-5)(x-18)=0\Rightarrow x=5$ (18 is not allowed).

$V^{\prime \prime }=24x-276$; at 5: $-156<0$, maximum.

5 cm

Let the circle have radius r; the diagonal of the rectangle is 2r. Sides x and $\sqrt{4r^2-x^2}$.

$A^2=x^2(4r^2-x^2)=4r^2{x}^2-x^4$. $\frac{d}{dx}(A^2)=8r^{2}x-4x^3=0\Rightarrow x^2=2r^2$; second derivative $8r^2-12x^2=-16r^2<0$, maximum.

Then the other side $=\sqrt{4r^2-2r^2}=\sqrt{2}r=x$: both sides equal.

Maximum when x = √2 r = other side, i.e. a square.

$S=2\pi r^2+2\pi rh$ (fixed) $\Rightarrow h=\frac{S-2\pi r^2}{2\pi r}$.

$V=\pi r^{2}h=\frac{r(S-2\pi r^2)}{2}=\frac{Sr-2\pi r^3}{2}$.

$V^{\prime }=\frac{S-6\pi r^2}{2}=0\Rightarrow S=6\pi r^2$; $V^{\prime \prime }=-6\pi r<0$.

Then $2\pi r^2+2\pi rh=6\pi r^2\Rightarrow h=2r$.

At maximum volume, h = 2r = diameter.

$\pi r^{2}h=100\Rightarrow h=\frac{100}{\pi r^2}$. $S=2\pi r^2+2\pi rh=2\pi r^2+\frac{200}{r}$.

$S^{\prime }=4\pi r-\frac{200}{r^2}=0\Rightarrow r^3=\frac{50}{\pi }$; $S^{\prime \prime }=4\pi +\frac{400}{r^3}>0$, minimum.

$h=\frac{100}{\pi r^2}=\frac{2\cdot 50}{\pi r^3}\cdot r=2r$.

$r={\left(\frac{50}{\pi }\right)}^{1/3}$ cm, $h=2{\left(\frac{50}{\pi }\right)}^{1/3}$ cm

Square piece x (side $\frac{x}{4}$), circle piece $28-x$ (radius $\frac{28-x}{2\pi }$).

$A=\frac{x^2}{16}+\frac{(28-x{)}^2}{4\pi }$; $A^{\prime }=\frac{x}{8}-\frac{28-x}{2\pi }=0\Rightarrow \pi x=4(28-x)\Rightarrow x=\frac{112}{\pi +4}$.

$A^{\prime \prime }=\frac{1}{8}+\frac{1}{2\pi }>0$, minimum. Circle piece $=28-\frac{112}{\pi +4}=\frac{28\pi }{\pi +4}$.

Square: $\frac{112}{\pi +4}$ m, circle: $\frac{28\pi }{\pi +4}$ m

Let the base of the cone be at distance x from the centre (on the far side from the vertex). Height $h=R+x$, base radius² $=R^2-x^2$.

$V=\frac{\pi }{3}(R^2-x^2)(R+x)$. $\frac{dV}{dx}=\frac{\pi }{3}[(R^2-x^2)-2x(R+x)]=\frac{\pi }{3}(R+x)(R-3x)=0\Rightarrow x=\frac{R}{3}$ (V′ changes + to −).

$V_{\max }=\frac{\pi }{3}\cdot \frac{8R^2}{9}\cdot \frac{4R}{3}=\frac{32\pi R^3}{81}=\frac{8}{27}\cdot \frac{4}{3}\pi R^3$.

Largest cone = (8/27) × volume of sphere.

$V=\frac{1}{3}\pi r^{2}h$ (fixed) $\Rightarrow h=\frac{3V}{\pi r^2}$. Curved surface $S=\pi r\sqrt{r^2+h^2}$.

Minimise $S^2={\pi }^2{r}^2(r^2+h^2)={\pi }^2{r}^4+\frac{9V^2}{r^2}$.

$\frac{d(S^2)}{dr}=4{\pi }^2{r}^3-\frac{18V^2}{r^3}=0\Rightarrow 2{\pi }^2{r}^6=9V^2={\pi }^2{r}^4{h}^2\Rightarrow h^2=2r^2$.

Second derivative $12{\pi }^2{r}^2+\frac{54V^2}{r^4}>0$, a minimum.

h = √2 r

Slant height l fixed, semi-vertical angle θ: $r=l\sin \theta$, $h=l\cos \theta$.

$V=\frac{\pi }{3}{l}^3{\sin }^2\theta \cos \theta$. $\frac{dV}{d\theta }=\frac{\pi }{3}{l}^3\sin \theta (2{\cos }^2\theta -{\sin }^2\theta )=0\Rightarrow {\tan }^2\theta =2$.

V′ changes from + to − there, so it is a maximum: $\theta ={\tan }^{-1}\sqrt{2}$.

θ = tan⁻¹ √2

$S=\pi r^2+\pi rl$ (fixed) $\Rightarrow l=\frac{S-\pi r^2}{\pi r}$.

$V^2=\frac{{\pi }^2{r}^4}{9}(l^2-r^2)=\frac{r^2}{9}[(S-\pi r^2{)}^2-{\pi }^2{r}^4]=\frac{S}{9}(S{r}^2-2\pi r^4)$.

$\frac{d(V^2)}{dr}=\frac{S}{9}(2Sr-8\pi r^3)=0\Rightarrow S=4\pi r^2$ (second derivative $\frac{S}{9}(2S-24\pi r^2)=\frac{S}{9}(-16\pi r^2)<0$, maximum).

Then $\pi rl=3\pi r^2\Rightarrow l=3r\Rightarrow \sin \theta =\frac{r}{l}=\frac{1}{3}$.

θ = sin⁻¹ (1/3)

$D^2=x^2+(y-5{)}^2=2y+(y-5{)}^2$. $\frac{d}{dy}=2+2(y-5)=0\Rightarrow y=4$, $x=\pm 2\sqrt{2}$.

(A)

$f^{\prime }(x)=\frac{(2x-1)(x^2+x+1)-(x^2-x+1)(2x+1)}{(1+x+x^2{)}^2}=\frac{2(x^2-1)}{(1+x+x^2{)}^2}=0\Rightarrow x=\pm 1$.

$f(1)=\frac{1}{3}$, $f(-1)=3$, and $f\to 1$ as $x\to \pm \infty$. Min $=\frac{1}{3}$.

(D)

$x^2-x+1$ on [0, 1] has least value $\frac{3}{4}$ at $x=\frac{1}{2}$ and greatest value 1 at $x=0$ and $x=1$. So the maximum is $1^{1/3}=1$.

(C)

$f^{\prime }(x)=\frac{1-\log x}{x^2}=0\Rightarrow x=e$.

$f^{\prime \prime }(x)=\frac{-x-2x(1-\log x)}{x^4}$; at $x=e$: $\frac{-e}{e^4}<0$.

f′(e) = 0 and f″(e) < 0, so f has a maximum at x = e (value 1/e).

Equal sides x: height $=\sqrt{x^2-\frac{b^2}{4}}$, so $A=\frac{b}{4}\sqrt{4x^2-b^2}$.

$\frac{dA}{dt}=\frac{b}{4}\cdot \frac{4x}{\sqrt{4x^2-b^2}}\frac{dx}{dt}=\frac{bx}{\sqrt{4x^2-b^2}}\frac{dx}{dt}$.

At $x=b$, $\frac{dx}{dt}=-3$: $\frac{dA}{dt}=\frac{b^2}{\sqrt{3}b}(-3)=-\sqrt{3}b$.

The area decreases at √3 b cm²/s.

$f(x)=\frac{4\sin x}{2+\cos x}-x$, so (as in Ex 6.2 Q9) $f^{\prime }(x)=\frac{\cos x(4-\cos x)}{(2+\cos x{)}^2}$.

Since $4-\cos x>0$, the sign of f′ is the sign of cos x. Taking $0\le x\le 2\pi$:

(i) Increasing on $\left[0,\frac{\pi }{2}\right)$ and $\left(\frac{3\pi }{2},2\pi \right]$ (ii) decreasing on $\left(\frac{\pi }{2},\frac{3\pi }{2}\right)$

$f^{\prime }(x)=3x^2-\frac{3}{x^4}=\frac{3(x^6-1)}{x^4}$. Positive when $x^6>1$, i.e. $|x|>1$; negative when $0<|x|<1$.

(i) x < −1 and x > 1 (ii) −1 < x < 1, x ≠ 0

Vertex at $(a,0)$; the other two vertices $(a\cos \theta ,\pm b\sin \theta )$.

$A=\frac{1}{2}(2b\sin \theta )(a-a\cos \theta )=ab\sin \theta (1-\cos \theta )$.

$\frac{dA}{d\theta }=ab(\cos \theta -\cos 2\theta )=0\Rightarrow \cos 2\theta =\cos \theta \Rightarrow \theta =\frac{2\pi }{3}$ (a maximum).

$A=ab\cdot \frac{\sqrt{3}}{2}\cdot \frac{3}{2}=\frac{3\sqrt{3}}{4}ab$.

$\frac{3\sqrt{3}}{4}ab$

Base $l\times w$ with $2lw=8\Rightarrow lw=4$, $w=\frac{4}{l}$.

Cost $C=70(4)+45\cdot 2\cdot 2(l+w)=280+180\left(l+\frac{4}{l}\right)$.

$C^{\prime }=180\left(1-\frac{4}{l^2}\right)=0\Rightarrow l=2$, $w=2$; $C^{\prime \prime }>0$.

$C=280+180(4)=1000$.

₹1000

$2\pi r+4a=k\Rightarrow a=\frac{k-2\pi r}{4}$. $A=\pi r^2+a^2$.

$\frac{dA}{dr}=2\pi r+2a\cdot \left(-\frac{\pi }{2}\right)=\pi (2r-a)=0\Rightarrow a=2r$.

$\frac{d^{2}A}{d{r}^2}=2\pi +\frac{{\pi }^2}{2}>0$, a minimum.

Least total area when a = 2r.

Let the rectangle be 2r wide and y high (semicircle radius r). Perimeter: $2r+2y+\pi r=10\Rightarrow y=\frac{10-(2+\pi )r}{2}$.

$A=2ry+\frac{1}{2}\pi r^2=10r-2r^2-\frac{\pi r^2}{2}$.

$A^{\prime }=10-(4+\pi )r=0\Rightarrow r=\frac{10}{\pi +4}$; $A^{\prime \prime }<0$, maximum.

Width $2r=\frac{20}{\pi +4}$; $y=\frac{10-(2+\pi )\frac{10}{\pi +4}}{2}=\frac{10}{\pi +4}$.

Rectangle: length $\frac{20}{\pi +4}$ m, width (height) $\frac{10}{\pi +4}$ m

Let the hypotenuse make angle θ with one leg. The point splits it into $a\sec \theta$ and $b\mathrm{cosec}\theta$, so $L=a\sec \theta +b\mathrm{cosec}\theta$, $0<\theta <\frac{\pi }{2}$.

$L^{\prime }=a\sec \theta \tan \theta -b\mathrm{cosec}\theta \cot \theta =0\Rightarrow {\tan }^3\theta =\frac{b}{a}$ (L′ changes − to +, a minimum).

Then $\sec \theta =\frac{\sqrt{a^{2/3}+b^{2/3}}}{a^{1/3}}$, $\mathrm{cosec}\theta =\frac{\sqrt{a^{2/3}+b^{2/3}}}{b^{1/3}}$.

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

$L_{\min }={\left(a^{2/3}+b^{2/3}\right)}^{3/2}$

$f^{\prime }(x)=4(x-2{)}^3(x+1{)}^3+3(x-2{)}^4(x+1{)}^2=(x-2{)}^3(x+1{)}^2(7x-2)$.