NCERT Solutions: Class 12 Maths Chapter 6 - Application of Derivatives

The area is $A=\pi r^2$. Hence $\dfrac{dA}{dr}=2\pi r$. At $r=3$, this is $6\pi$; at $r=4$, it is $8\pi$.

Let the edge be $x$ cm. Then $V=x^3$, so $\dfrac{dV}{dt}=3x^2\dfrac{dx}{dt}$. At $x=12$, $8=432\dfrac{dx}{dt}$, giving $\dfrac{dx}{dt}=\dfrac1{54}$. Since $S=6x^2$, $\dfrac{dS}{dt}=12x\dfrac{dx}{dt}=144\cdot\dfrac1{54}=\dfrac83$.

For $A=\pi r^2$, $\dfrac{dA}{dt}=2\pi r\dfrac{dr}{dt}$. With $r=10$ and $\dfrac{dr}{dt}=3$, $\dfrac{dA}{dt}=2\pi(10)(3)=60\pi$.

Let the edge be $x$. Then $V=x^3$ and $\dfrac{dV}{dt}=3x^2\dfrac{dx}{dt}$. At $x=10$ and $\dfrac{dx}{dt}=3$, $\dfrac{dV}{dt}=3(100)(3)=900$.

For the circular wave, $A=\pi r^2$. Thus $\dfrac{dA}{dt}=2\pi r\dfrac{dr}{dt}=2\pi(8)(5)=80\pi$.

The circumference is $C=2\pi r$. Therefore $\dfrac{dC}{dt}=2\pi\dfrac{dr}{dt}=2\pi(0.7)=1.4\pi$.

Here $\dfrac{dx}{dt}=-5$ and $\dfrac{dy}{dt}=4$. Since $P=2(x+y)$, $\dfrac{dP}{dt}=2(-5+4)=-2$. Also $A=xy$, so $\dfrac{dA}{dt}=x\dfrac{dy}{dt}+y\dfrac{dx}{dt}=8(4)+6(-5)=2$.

For a sphere, $V=\dfrac43\pi r^3$, so $\dfrac{dV}{dt}=4\pi r^2\dfrac{dr}{dt}$. Substituting $\dfrac{dV}{dt}=900$ and $r=15$ gives $900=4\pi(225)\dfrac{dr}{dt}$, hence $\dfrac{dr}{dt}=\dfrac1\pi$.

Since $V=\dfrac43\pi r^3$, the rate of change with respect to the radius is $\dfrac{dV}{dr}=4\pi r^2$. At $r=10$, $\dfrac{dV}{dr}=4\pi(100)=400\pi$.

Use centimetres. Let $x$ be the foot distance and $y$ the height. Then $x^2+y^2=500^2$. At $x=400$, $y=300$. Differentiating gives $2x\dfrac{dx}{dt}+2y\dfrac{dy}{dt}=0$, so $\dfrac{dy}{dt}=-\dfrac{x}{y}\dfrac{dx}{dt}=-\dfrac{400}{300}(2)=-\dfrac83$.

Differentiate $6y=x^3+2$ with respect to $t$: $6\dfrac{dy}{dt}=3x^2\dfrac{dx}{dt}$, so $\dfrac{dy}{dt}=\dfrac{x^2}{2}\dfrac{dx}{dt}$. The condition is $\dfrac{dy}{dt}=8\dfrac{dx}{dt}$, hence $x^2/2=8$ and $x=\pm4$. From the curve, for $x=4$, $y=11$; for $x=-4$, $y=-31/3$.

For $V=\dfrac43\pi r^3$, $\dfrac{dV}{dt}=4\pi r^2\dfrac{dr}{dt}$. At $r=1$ and $\dfrac{dr}{dt}=\dfrac12$, the rate is $4\pi(1)^2\cdot\dfrac12=2\pi$.

The radius is $r=\dfrac12\cdot\dfrac32(2x+1)=\dfrac34(2x+1)$, so $\dfrac{dr}{dx}=\dfrac32$. Since $V=\dfrac43\pi r^3$, $\dfrac{dV}{dx}=4\pi r^2\dfrac{dr}{dx}=6\pi\left(\dfrac34(2x+1)\right)^2=\dfrac{27\pi}{8}(2x+1)^2$.

Given $h=\dfrac{r}{6}$, so $r=6h$. Then $V=\dfrac13\pi r^2h=\dfrac13\pi(36h^2)h=12\pi h^3$. Differentiating, $\dfrac{dV}{dt}=36\pi h^2\dfrac{dh}{dt}$. With $\dfrac{dV}{dt}=12$ and $h=4$, $12=36\pi(16)\dfrac{dh}{dt}$, hence $\dfrac{dh}{dt}=\dfrac{1}{48\pi}$.

Marginal cost is $C'(x)$. Here $C'(x)=0.021x^2-0.006x+15$. Thus $C'(17)=0.021(289)-0.006(17)+15=6.069-0.102+15=20.967$.

Marginal revenue is $R'(x)=26x+26$. Therefore $R'(7)=26(7)+26=208$.

1. i. $10\pi$
2. ii. $12\pi$
3. iii. $8\pi$
4. iv. $11\pi$

For $A=\pi r^2$, $\dfrac{dA}{dr}=2\pi r$. At $r=6$, the rate is $12\pi$, which is option (ii).

1. i. $116$
2. ii. $96$
3. iii. $90$
4. iv. $126$

Marginal revenue is $R'(x)=6x+36$. Therefore $R'(15)=90+36=126$, which is option (iv).

Here $f'(x)=3$, which is positive for every real $x$. Therefore $f$ is increasing on $\mathbb{R}$.

Since $f'(x)=2e^{2x}$ and $e^{2x}\gt 0$ for all real $x$, $f'(x)\gt 0$ everywhere. Hence $f$ is increasing on $\mathbb{R}$.

Here $f'(x)=\cos x$. On $\left(0,\dfrac{\pi}{2}\right)$, $\cos x\gt 0$, so $f$ is increasing. On $\left(\dfrac{\pi}{2},\pi\right)$, $\cos x\lt 0$, so $f$ is decreasing. Since the sign changes inside $(0,\pi)$, $f$ is neither increasing nor decreasing throughout that interval.

Since $f'(x)=4x-3$, the derivative is positive when $x\gt \dfrac34$ and negative when $x\lt \dfrac34$. Hence the stated intervals follow.

We have $f'(x)=6x^2-6x-36=6(x-3)(x+2)$. This is positive for $x\lt -2$ or $x\gt 3$, and negative for $-2\lt x\lt 3$.

Differentiate each function. (a) $2x+2$ changes sign at $-1$. (b) $-6-4x$ changes sign at $-3/2$. (c) $-6x^2-18x-12=-6(x+1)(x+2)$ is positive only on $(-2,-1)$. (d) $-9-2x$ changes sign at $-9/2$. (e) The derivative is $6(x-1)(x+1)^2(x-3)^2$; its sign is the sign of $x-1$ except at the stationary zeros $-1$ and $3$, so the function decreases before $1$ and increases after $1$.

Differentiate: $\dfrac{dy}{dx}=\dfrac1{1+x}-\dfrac4{(2+x)^2}=\dfrac{x^2}{(1+x)(2+x)^2}$. For $x\gt -1$, the denominator is positive and $x^2\ge0$, so $\dfrac{dy}{dx}\ge0$. Hence $y$ is increasing throughout its domain.

$y=x^2(x-2)^2$. Differentiating, $\dfrac{dy}{dx}=4x(x-1)(x-2)$. This derivative is positive on $(0,1)$ and $(2,\infty)$, so these are the required intervals.

Using the quotient rule, $\dfrac{dy}{d\theta}=\dfrac{4(2\cos\theta+1)}{(2+\cos\theta)^2}-1=\dfrac{\cos\theta(4-\cos\theta)}{(2+\cos\theta)^2}$. On $\left[0,\dfrac{\pi}{2}\right]$, $\cos\theta\ge0$ and $4-\cos\theta\gt 0$, so $\dfrac{dy}{d\theta}\ge0$.

For $f(x)=\log x$, $f'(x)=\dfrac1x$. Since $x\gt 0$ on $(0,\infty)$, $f'(x)\gt 0$. Hence the logarithmic function is increasing there.

Here $f'(x)=2x-1$. It is negative on $(-1,1/2)$ and positive on $(1/2,1)$. Thus $f$ decreases on one part of $(-1,1)$ and increases on another, so it is neither strictly increasing nor decreasing on the whole interval.

1. i. $\cos x$
2. ii. $\cos 2x$
3. iii. $\cos 3x$
4. iv. $\tan x$

For $\cos x$, derivative $-\sin x\lt 0$ on the interval. For $\cos2x$, derivative $-2\sin2x\lt 0$ on the interval. For $\cos3x$, the derivative changes sign, and $\tan x$ has positive derivative $\sec^2x$. Therefore only options (i) and (ii) are decreasing throughout.

1. i. $(0,1)$
2. ii. $\left(\dfrac{\pi}{2},\pi\right)$
3. iii. $\left(0,\dfrac{\pi}{2}\right)$
4. iv. None of these

Here $f'(x)=100x^{99}+\cos x$. On $(0,1)$ and $\left(0,\dfrac{\pi}{2}\right)$ this is positive. On $\left(\dfrac{\pi}{2},\pi\right)$, $100x^{99}$ dominates while $\cos x\ge-1$, so the derivative remains positive. Hence none of the listed intervals is an interval of decrease.

We need $f'(x)=2x+a\ge0$ for all $x\in[1,2]$. The minimum of $2x+a$ on this interval is at $x=1$, so $2+a\ge0$. Hence $a\ge-2$.

For $f(x)=x+\dfrac1x$, $f'(x)=1-\dfrac1{x^2}=\dfrac{x^2-1}{x^2}$. If $I$ is disjoint from $[-1,1]$, then $|x|\gt 1$ for every $x\in I$, so $x^2-1\gt 0$ and $f'(x)\gt 0$. Hence $f$ is increasing on $I$.

Here $f'(x)=\dfrac{\cos x}{\sin x}=\cot x$. On $\left(0,\dfrac{\pi}{2}\right)$, $\cot x\gt 0$; on $\left(\dfrac{\pi}{2},\pi\right)$, $\cot x\lt 0$. Therefore the required monotonicity follows.

For intervals where $\cos x\ne0$, $f'(x)=-\tan x$. On $\left(0,\dfrac{\pi}{2}\right)$, $\tan x\gt 0$, so $f'(x)\lt 0$. On $\left(\dfrac{3\pi}{2},2\pi\right)$, $\tan x\lt 0$, so $f'(x)\gt 0$.

We have $f'(x)=3x^2-6x+3=3(x-1)^2\ge0$ for all real $x$. Therefore $f$ is increasing on $\mathbb{R}$.

1. i. $(-\infty,\infty)$
2. ii. $(-2,0)$
3. iii. $(2,\infty)$
4. iv. $(0,2)$

Differentiate: $y'=2xe^{-x}-x^2e^{-x}=xe^{-x}(2-x)$. Since $e^{-x}\gt 0$, $y'\gt 0$ exactly when $0\lt x\lt 2$. Thus the answer is option (iv).

(i) A square is non-negative, so $(2x-1)^2+3\ge3$, with equality at $x=1/2$. (ii) $9x^2+12x+2=9(x+2/3)^2-2$, so the least value is $-2$. (iii) $-(x-1)^2+10\le10$, with equality at $x=1$. (iv) $x^3+1$ is unbounded above and below.

Use the ranges of absolute value and sine. (i) $|x+2|\ge0$. (ii) $-|x+1|\le0$. (iii) $-1\le\sin2x\le1$. (iv) Since $2\le\sin4x+3\le4$, the absolute value has range $[2,4]$. (v) On $(-1,1)$, $x+1$ has range $(0,2)$, so neither endpoint value is attained.

Differentiate and use the first derivative test. (i) $2x$ changes from negative to positive at $0$. (ii) $3x^2-3=3(x-1)(x+1)$. (iii) $\cos x-\sin x=0$ gives $x=\pi/4$. (iv) $\cos x+\sin x=0$ gives $3\pi/4$ and $7\pi/4$. (v) $3(x-1)(x-3)$ changes sign at $1$ and $3$. (vi) $1/2-2/x^2=0$ gives $x=2$. (vii) $-2x/(x^2+2)^2=0$ gives $x=0$. (viii) $f'(x)=\dfrac{1-3x/2}{\sqrt{1-x}}$, so $x=2/3$.

For $e^x$, $f'(x)=e^x\gt 0$, so it is strictly increasing and unbounded. For $\log x$, $g'(x)=1/x\gt 0$ on $(0,\infty)$ and the function is unbounded below near $0$ and unbounded above as $x\to\infty$. For $h$, $h'(x)=3x^2+2x+1$, whose discriminant is $4-12\lt 0$ and leading coefficient is positive; hence $h'(x)\gt 0$ for all $x$, and $h$ is strictly increasing and unbounded.

Check critical points and endpoints. (i) $x^3$ is increasing, so extrema occur at endpoints. (ii) $f'(x)=\cos x-\sin x$, giving $x=\pi/4$; compare with endpoints. (iii) $f'(x)=4-x$, so $x=4$ gives $8$; endpoints give $-10$ and $63/8$. (iv) The parabola has vertex at $x=1$, giving $3$; the far endpoint $x=-3$ gives $19$.

$p'(x)=-72-36x$. Setting $p'(x)=0$ gives $x=-2$. The second derivative is $-36$, which is negative, so this point gives a maximum. The maximum profit is $p(-2)=41+144-72=113$.

Let $f(x)=3x^4-8x^3+12x^2-48x+25$. Then $f'(x)=12x^3-24x^2+24x-48=12(x-2)(x^2+2)$. The only critical point in $[0,3]$ is $x=2$. Compare $f(0)=25$, $f(2)=-39$, and $f(3)=16$.

The maximum value of $\sin2x$ is $1$. Thus $2x=\dfrac{\pi}{2}+2n\pi$, so $x=\dfrac{\pi}{4}+n\pi$. In $[0,2\pi]$, this gives $x=\dfrac{\pi}{4}$ and $x=\dfrac{5\pi}{4}$.

$\sin x+\cos x=\sqrt2\sin\left(x+\dfrac{\pi}{4}\right)$. Since the maximum value of sine is $1$, the maximum value is $\sqrt2$.

For $f(x)=2x^3-24x+107$, $f'(x)=6x^2-24=6(x-2)(x+2)$. On $[1,3]$, compare $f(1)=85$, $f(2)=75$, $f(3)=161$. On $[-3,-1]$, compare $f(-3)=125$, $f(-2)=139$, $f(-1)=129$.

Since the maximum occurs at the interior point $x=1$, $f'(1)=0$. Now $f'(x)=4x^3-124x+a$, so $0=f'(1)=4-124+a=a-120$. Hence $a=120$.

Let $f(x)=x+\sin2x$. Then $f'(x)=1+2\cos2x$. Critical points satisfy $\cos2x=-1/2$. Comparing those values with endpoints gives $f(0)=0$ and $f(2\pi)=2\pi$; all interior critical values lie between these. Therefore the absolute minimum is $0$ and the absolute maximum is $2\pi$.

Let the numbers be $x$ and $24-x$. Their product is $P=x(24-x)=24x-x^2$. Then $P'=24-2x$, so $P'=0$ gives $x=12$. Since the second derivative is $-2$, the product is maximum when both numbers are $12$.

Put $x=60-y$. Then $P=xy^3=(60-y)y^3=60y^3-y^4$. Differentiating, $P'=180y^2-4y^3=4y^2(45-y)$. Since $y\gt 0$, the critical point is $y=45$, giving $x=15$. The second derivative is negative there, so this is a maximum.

Let $y=35-x$ and maximize $P=x^2(35-x)^5$. For positive $x,y$, maximize $\log P=2\log x+5\log(35-x)$. Differentiating gives $\dfrac2x-\dfrac5{35-x}=0$, so $2(35-x)=5x$, hence $x=10$ and $y=25$.

Let the numbers be $x$ and $16-x$. Then $S=x^3+(16-x)^3$. Differentiating, $S'=3x^2-3(16-x)^2=3(2x-16)(16)$. Thus $S'=0$ gives $x=8$, and the two numbers are $8$ and $8$.

Let $x$ cm be cut from each corner. The box has base $(18-2x)\times(18-2x)$ and height $x$, so $V=x(18-2x)^2$. Then $V'=(18-2x)(18-6x)$. The feasible critical point is $x=3$; endpoints give zero volume, so the maximum occurs at $x=3$ cm.

Let the cut-off square have side $x$. Then $V=x(45-2x)(24-2x)$. Expanding and differentiating, $V'=12x^2-276x+1080=12(x^2-23x+90)$. Thus $x=5$ or $x=18$; only $x=5$ is feasible for the $24$ cm side. Hence the required side is $5$ cm.

Let the rectangle have sides $x$ and $y$, and let the fixed diagonal be the circle diameter $d$. Then $x^2+y^2=d^2$. Its area is $A=xy$. For a fixed sum $x^2+y^2$, $2xy\le x^2+y^2=d^2$, so $A\le d^2/2$. Equality holds when $x=y$, i.e. when the rectangle is a square.

Let radius be $r$, height $h$, and total surface area be fixed: $S=2\pi r^2+2\pi rh$. Then $h=\dfrac{S-2\pi r^2}{2\pi r}$. Volume $V=\pi r^2h=\dfrac{Sr}{2}-\pi r^3$. Thus $V'=\dfrac S2-3\pi r^2=0$, so $S=6\pi r^2$. Substituting in $2\pi r^2+2\pi rh=S$ gives $2\pi rh=4\pi r^2$, hence $h=2r$.

For a closed cylinder, $V=\pi r^2h=100$, so $h=100/(\pi r^2)$. Surface area $S=2\pi r^2+2\pi rh=2\pi r^2+200/r$. Then $S'=4\pi r-200/r^2$. Setting $S'=0$ gives $4\pi r^3=200$, so $r^3=50/\pi$. Also $h=100/(\pi r^2)=2r$.

Let $x$ m be used for the square, so its side is $x/4$ and area is $x^2/16$. The circle uses $28-x$ m, so its radius is $(28-x)/(2\pi)$ and area is $(28-x)^2/(4\pi)$. Thus $A=x^2/16+(28-x)^2/(4\pi)$. Setting $A'=x/8-(28-x)/(2\pi)=0$ gives $x=112/(\pi+4)$. The remaining length is $28\pi/(\pi+4)$.

Let the cone height be $h$. In the axial section, the base radius satisfies $r^2=h(2R-h)$. Hence $V=\dfrac13\pi r^2h=\dfrac13\pi h^2(2R-h)$. Then $V'=\dfrac13\pi(4Rh-3h^2)=\dfrac13\pi h(4R-3h)$. The maximum occurs at $h=4R/3$. Therefore $V=\dfrac13\pi\cdot\dfrac{16R^2}{9}\cdot\dfrac{2R}{3}=\dfrac{32}{81}\pi R^3$. Since sphere volume is $\dfrac43\pi R^3$, the ratio is $\dfrac{32}{81}\div\dfrac43=\dfrac{8}{27}$.

Let the volume $V=\dfrac13\pi r^2h$ be fixed, so $h=3V/(\pi r^2)$. Curved surface area $S=\pi r\sqrt{r^2+h^2}$. Minimizing $S^2=\pi^2r^2(r^2+h^2)=\pi^2r^4+9V^2/r^2$ is equivalent. Differentiate: $\dfrac{d}{dr}(S^2)=4\pi^2r^3-18V^2/r^3=0$, so $2\pi^2r^6=9V^2$. Using $h=3V/(\pi r^2)$ gives $h^2=2r^2$, hence $h=\sqrt2r$.

Let the slant height be $l$, base radius $r$, height $h$, and semi-vertical angle $\alpha$. Then $h^2+r^2=l^2$ and $V=\dfrac13\pi r^2h=\dfrac13\pi r^2\sqrt{l^2-r^2}$. Maximizing gives $2(l^2-r^2)-r^2=0$, so $2l^2=3r^2$ and $h^2=l^2-r^2=l^2/3$. Thus $r/h=\sqrt2$, so $\tan\alpha=\sqrt2$ and $\alpha=\tan^{-1}\sqrt2$.

Let total surface area $S=\pi r^2+\pi rl$ be fixed, where $l$ is slant height. Since $l=(S-\pi r^2)/(\pi r)$ and $h^2=l^2-r^2$, maximize $V^2=\dfrac{\pi^2r^4}{9}(l^2-r^2)$. Substitution and differentiation give the maximum condition $l=3r$. Therefore $\sin\alpha=r/l=1/3$, so $\alpha=\sin^{-1}(1/3)$.

1. i. $(2\sqrt2,4)$
2. ii. $(2\sqrt2,0)$
3. iii. $(0,0)$
4. iv. $(2,2)$

A point on the curve is $(x,x^2/2)$. Its squared distance from $(0,5)$ is $D=x^2+(x^2/2-5)^2$. Then $D'=2x+2(x^2/2-5)x=x(x^2-8)$. Thus $x=0$ or $x=\pm2\sqrt2$. The nearest listed point is $(2\sqrt2,4)$.

1. i. $0$
2. ii. $1$
3. iii. $3$
4. iv. $\dfrac13$

Let $y=\dfrac{x^2-x+1}{x^2+x+1}$. Then $y(x^2+x+1)=x^2-x+1$, so $(y-1)x^2+(y+1)x+(y-1)=0$ must have a real solution. Its discriminant is non-negative: $(y+1)^2-4(y-1)^2\ge0$. This gives $-3y^2+10y-3\ge0$, or $1/3\le y\le3$. Hence the minimum is $1/3$.

1. i. $\left(\dfrac13\right)^{1/3}$
2. ii. $\dfrac12$
3. iii. $1$
4. iv. $0$

The inside expression is $x^2-x+1=(x-1/2)^2+3/4$. On $[0,1]$, its maximum is $1$, attained at $x=0$ and $x=1$. The cube-root function is increasing, so the maximum value is $1$.

For $x\gt 0$, $f'(x)=\dfrac{1-\log x}{x^2}$. Thus $f'(x)=0$ gives $\log x=1$, so $x=e$. The derivative is positive for $0\lt x\lt e$ and negative for $x\gt e$, hence $f$ has a maximum at $x=e$.

Let each equal side be $x$ and height be $h=\sqrt{x^2-b^2/4}$. Area $A=\dfrac b2h$. Hence $\dfrac{dA}{dt}=\dfrac b2\cdot\dfrac{x}{h}\dfrac{dx}{dt}$. When $x=b$, $h=\dfrac{\sqrt3b}{2}$ and $\dfrac{dx}{dt}=-3$. Thus $\dfrac{dA}{dt}=\dfrac b2\cdot\dfrac{2}{\sqrt3}(-3)=-\sqrt3b$.

Let $N=4\sin x-2x-x\cos x$ and $D=2+\cos x$. By the quotient rule, $f'(x)=\dfrac{(3\cos x-2+x\sin x)(2+\cos x)+(4\sin x-2x-x\cos x)\sin x}{(2+\cos x)^2}=\dfrac{\cos x(4-\cos x)}{(2+\cos x)^2}$. Since $2+\cos x\gt 0$ and $4-\cos x\gt 0$, the sign is the sign of $\cos x$.

Here $f'(x)=3x^2-3/x^4=\dfrac{3(x^6-1)}{x^4}$. Since $x^4\gt 0$ for $x\ne0$, the derivative is positive when $|x|\gt 1$ and negative when $0\lt |x|\lt 1$.

Take the fixed vertex as $(a,0)$ and the other two vertices as $(x,\pm y)$ on the ellipse. Then $y=\dfrac ba\sqrt{a^2-x^2}$ and the area is $A=y(a-x)=\dfrac ba(a-x)\sqrt{a^2-x^2}$. Maximizing $A^2=\dfrac{b^2}{a^2}(a-x)^2(a^2-x^2)$ gives $x=-a/2$. Then $y=\dfrac{\sqrt3b}{2}$, so $A=(3a/2)(\sqrt3b/2)=\dfrac{3\sqrt3}{4}ab$.

Let base dimensions be $x$ and $y$. Since depth is $2$ and volume is $8$, $xy=4$. Cost of base is $70xy=280$. Area of sides is $2(2x)+2(2y)=4x+4y$, so side cost is $45(4x+4y)=180(x+y)$. The sum $x+y$ is minimum for fixed $xy=4$ when $x=y=2$. Hence minimum cost is $280+180(4)=1000$.

Let the circle radius be $r$ and the square side be $a$. The constraint is $2\pi r+4a=k$, so $a=(k-2\pi r)/4$. Total area $A=\pi r^2+a^2=\pi r^2+\dfrac{(k-2\pi r)^2}{16}$. Differentiating and setting $A'=0$ gives $2\pi r-\dfrac{\pi}{4}(k-2\pi r)=0$, hence $k=2(\pi+4)r$. Then $a=(k-2\pi r)/4=2r$.

Let the semicircle radius be $r$ and the rectangle height be $h$. The width is $2r$ and the perimeter condition is $2h+2r+\pi r=10$. Thus $h=\dfrac{10-(\pi+2)r}{2}$. Area $A=2rh+\dfrac12\pi r^2=10r-\left(2+\dfrac\pi2\right)r^2$. Setting $A'=0$ gives $r=10/(\pi+4)$, and then $h=10/(\pi+4)$.

Let the acute angles made by the hypotenuse with the two sides be represented so that the hypotenuse length is $L=a\csc\theta+b\sec\theta$. Differentiate: $L'=-a\csc\theta\cot\theta+b\sec\theta\tan\theta$. At the minimum, $b\sin^3\theta=a\cos^3\theta$, so $\tan\theta=(a/b)^{1/3}$. Substitution in $L=a/\sin\theta+b/\cos\theta$ gives $L_{\min}=(a^{2/3}+b^{2/3})^{3/2}$.

Differentiating, $f'(x)=(x-2)^3(x+1)^2(7x-2)$. The derivative changes from positive to negative at $x=2/7$, so there is a local maximum there; it changes from negative to positive at $x=2$, so there is a local minimum there. The second derivative is $6(x-2)^2(x+1)(7x^2-4x-2)$, so inflexion occurs at $x=-1$ and at $x=(2\pm3\sqrt2)/7$.

Let $s=\sin x$. On $[0,\pi]$, $0\le s\le1$, and $f=\cos^2x+\sin x=1-s^2+s=\dfrac54-(s-1/2)^2$. The maximum is $5/4$ when $s=1/2$, i.e. at $x=\pi/6,5\pi/6$. The minimum is $1$, attained when $s=0$ or $s=1$, i.e. at $x=0,\pi/2,\pi$.

Let the cone height be $h$ and the sphere radius be $r$. The cone base radius satisfies $a^2=h(2r-h)$. Thus $V=\dfrac13\pi a^2h=\dfrac13\pi h^2(2r-h)$. Differentiating, $V'=\dfrac13\pi h(4r-3h)$. The non-zero critical value is $h=4r/3$, and it gives the maximum volume.

Take any $x_1,x_2\in(a,b)$ with $x_1\lt x_2$. By the Mean Value Theorem, there is $c\in(x_1,x_2)$ such that $f(x_2)-f(x_1)=f'(c)(x_2-x_1)$. Since $f'(c)\gt 0$ and $x_2-x_1\gt 0$, we get $f(x_2)-f(x_1)\gt 0$. Hence $f(x_1)\lt f(x_2)$, so $f$ is increasing.

Let the cylinder have radius $a$ and height $h$. From the axial section, $a^2+(h/2)^2=R^2$, so $a^2=R^2-h^2/4$. Volume $V=\pi a^2h=\pi h(R^2-h^2/4)$. Differentiating, $V'=\pi(R^2-3h^2/4)$. Thus $h=2R/\sqrt3$. Then $a^2=2R^2/3$, giving $V_{\max}=\pi(2R^2/3)(2R/\sqrt3)=\dfrac{4\pi R^3}{3\sqrt3}$.

Let the cylinder height be $x$. At height $x$ from the base of the cone, the cylinder radius is $(h-x)\tan\alpha$. Thus $V=\pi x(h-x)^2\tan^2\alpha$. Differentiating the variable part, $\dfrac{d}{dx}[x(h-x)^2]=(h-x)(h-3x)$. The interior maximum is at $x=h/3$. Hence radius is $(2h/3)\tan\alpha$ and $V_{\max}=\pi(h/3)(4h^2/9)\tan^2\alpha=\dfrac{4}{27}\pi h^3\tan^2\alpha$.

1. i. $1$ m/h
2. ii. $0.1$ m/h
3. iii. $1.1$ m/h
4. iv. $0.5$ m/h

For the cylinder, $V=\pi r^2h=100\pi h$. Hence $\dfrac{dV}{dt}=100\pi\dfrac{dh}{dt}$. Given $\dfrac{dV}{dt}=314$ and using $\pi=3.14$, $314=314\dfrac{dh}{dt}$, so $\dfrac{dh}{dt}=1$ m/h.