Maxima and Minima Questions in English

(B) Let $R$ be a point on $AB$ such that $AR=x \, m.$
Then $RB=(20-x) \, m$ (as $AB=20 \, m$).
From the geometry of the problem,we have right-angled triangles $\triangle PAR$ and $\triangle QBR$.
$RP^2 = AR^2 + AP^2 = x^2 + 16^2 = x^2 + 256$
$RQ^2 = RB^2 + BQ^2 = (20-x)^2 + 22^2 = (400 - 40x + x^2) + 484 = x^2 - 40x + 884$
Let $S(x) = RP^2 + RQ^2 = (x^2 + 256) + (x^2 - 40x + 884) = 2x^2 - 40x + 1140.$
To find the minimum,we differentiate $S(x)$ with respect to $x$:
$S'(x) = 4x - 40.$
Setting $S'(x) = 0$ gives $4x - 40 = 0 \implies x = 10.$
Now,we check the second derivative: $S''(x) = 4.$
Since $S''(10) = 4 > 0,$ the function $S(x)$ has a local minimum at $x = 10.$
Thus,the distance of point $R$ from $A$ is $10 \, m.$

(A) Let the trapezium be $ABCD$ where $AD = DC = CB = 10 \ cm$. Draw perpendiculars $DP$ and $CQ$ on the base $AB$. Let $AP = x \ cm$. Since the trapezium is isosceles,$QB = x \ cm$. The length $PQ = DC = 10 \ cm$. Thus,the base $AB = x + 10 + x = 2x + 10 \ cm$.
By the Pythagorean theorem in $\Delta APD$,the height $h = DP = \sqrt{10^2 - x^2} = \sqrt{100 - x^2}$.
The area $A$ of the trapezium is given by $A = \frac{1}{2} \times (\text{sum of parallel sides}) \times \text{height}$.
$A(x) = \frac{1}{2} (AB + DC) \times h = \frac{1}{2} (2x + 10 + 10) \sqrt{100 - x^2} = (x + 10) \sqrt{100 - x^2}$.
To find the maximum area,we find the derivative $A'(x)$:
$A'(x) = (1) \sqrt{100 - x^2} + (x + 10) \frac{-2x}{2\sqrt{100 - x^2}} = \frac{100 - x^2 - x^2 - 10x}{\sqrt{100 - x^2}} = \frac{100 - 10x - 2x^2}{\sqrt{100 - x^2}}$.
Setting $A'(x) = 0$,we get $2x^2 + 10x - 100 = 0 \implies x^2 + 5x - 50 = 0 \implies (x + 10)(x - 5) = 0$.
Since $x$ is a length,$x = 5 \ cm$.
The second derivative test confirms this is a maximum.
Substituting $x = 5$ into the area formula:
$A(5) = (5 + 10) \sqrt{100 - 5^2} = 15 \sqrt{75} = 15 \times 5\sqrt{3} = 75\sqrt{3} \ cm^2$.

(N/A) Let $OC = r$ be the radius of the cone and $OA = h$ be its height. Let a cylinder with radius $OE = x$ be inscribed in the given cone. The height $QE$ of the cylinder is given by:
$\frac{QE}{OA} = \frac{EC}{OC}$ (since $\Delta QEC \sim \Delta AOC$)
$\frac{QE}{h} = \frac{r - x}{r}$
$QE = \frac{h(r - x)}{r}$
Let $S$ be the curved surface area of the cylinder. Then:
$S(x) = 2 \pi x \cdot QE = 2 \pi x \cdot \frac{h(r - x)}{r} = \frac{2 \pi h}{r}(rx - x^2)$
To find the maximum surface area,we find the derivative $S'(x)$:
$S'(x) = \frac{2 \pi h}{r}(r - 2x)$
Setting $S'(x) = 0$ gives $r - 2x = 0$,so $x = \frac{r}{2}$.
Now,find the second derivative $S''(x)$:
$S''(x) = \frac{2 \pi h}{r}(-2) = -\frac{4 \pi h}{r}$
Since $S''(x) < 0$ for all $x$,the function $S(x)$ has a maximum at $x = \frac{r}{2}$.
Thus,the radius of the cylinder of greatest curved surface area is half of the radius of the cone.

(A) We have $f(x) = 2x^{3} - 15x^{2} + 36x + 1$.
Taking the derivative,we get $f'(x) = 6x^{2} - 30x + 36 = 6(x - 2)(x - 3)$.
Setting $f'(x) = 0$ gives the critical points $x = 2$ and $x = 3$.
We evaluate the function at the critical points and the endpoints of the interval $[1, 5]$:
$f(1) = 2(1)^{3} - 15(1)^{2} + 36(1) + 1 = 2 - 15 + 36 + 1 = 24$.
$f(2) = 2(2)^{3} - 15(2)^{2} + 36(2) + 1 = 16 - 60 + 72 + 1 = 29$.
$f(3) = 2(3)^{3} - 15(3)^{2} + 36(3) + 1 = 54 - 135 + 108 + 1 = 28$.
$f(5) = 2(5)^{3} - 15(5)^{2} + 36(5) + 1 = 250 - 375 + 180 + 1 = 56$.
Comparing these values,the absolute maximum value is $56$ at $x = 5$ and the absolute minimum value is $24$ at $x = 1$.

(N/A) We have $f(x) = 12x^{\frac{4}{3}} - 6x^{\frac{1}{3}}$.
First,we find the derivative $f'(x)$:
$f'(x) = 12 \cdot \frac{4}{3}x^{\frac{1}{3}} - 6 \cdot \frac{1}{3}x^{-\frac{2}{3}} = 16x^{\frac{1}{3}} - \frac{2}{x^{\frac{2}{3}}} = \frac{16x - 2}{x^{\frac{2}{3}}} = \frac{2(8x - 1)}{x^{\frac{2}{3}}}$.
Setting $f'(x) = 0$ gives $8x - 1 = 0$,so $x = \frac{1}{8}$.
Also,$f'(x)$ is undefined at $x = 0$. Thus,the critical points are $x = 0$ and $x = \frac{1}{8}$.
Now,we evaluate $f(x)$ at the critical points and the endpoints $x = -1$ and $x = 1$:
$f(-1) = 12(-1)^{\frac{4}{3}} - 6(-1)^{\frac{1}{3}} = 12(1) - 6(-1) = 12 + 6 = 18$.
$f(0) = 12(0)^{\frac{4}{3}} - 6(0)^{\frac{1}{3}} = 0$.
$f(\frac{1}{8}) = 12(\frac{1}{8})^{\frac{4}{3}} - 6(\frac{1}{8})^{\frac{1}{3}} = 12(\frac{1}{16}) - 6(\frac{1}{2}) = \frac{3}{4} - 3 = \frac{3 - 12}{4} = -\frac{9}{4}$.
$f(1) = 12(1)^{\frac{4}{3}} - 6(1)^{\frac{1}{3}} = 12 - 6 = 6$.
Comparing these values,the absolute maximum value is $18$ at $x = -1$ and the absolute minimum value is $-\frac{9}{4}$ at $x = \frac{1}{8}$.

(A) For each value of $x$,the helicopter's position is at point $(x, x^{2} + 7)$.
Therefore,the distance $d$ between the helicopter and the soldier at $(3, 7)$ is given by $d = \sqrt{(x - 3)^{2} + (x^{2} + 7 - 7)^{2}} = \sqrt{(x - 3)^{2} + x^{4}}$.
Let $f(x) = (x - 3)^{2} + x^{4}$. To minimize the distance,we minimize $f(x)$.
Taking the derivative,$f'(x) = 2(x - 3) + 4x^{3} = 2(x - 1)(2x^{2} + 2x + 3)$.
Setting $f'(x) = 0$,we get $x = 1$ (since $2x^{2} + 2x + 3 = 0$ has no real roots as its discriminant $D = 4 - 24 = -20 < 0$).
The value of $f(x)$ at $x = 1$ is $f(1) = (1 - 3)^{2} + (1)^{4} = 4 + 1 = 5$.
Thus,the minimum distance is $\sqrt{f(1)} = \sqrt{5}$.

(B) The given function is $f(x)=(2x-1)^{2}+3$.
It is known that for any real number $x$,the square of a real number is always non-negative,i.e.,$(2x-1)^{2} \geq 0$.
Adding $3$ to both sides,we get $(2x-1)^{2}+3 \geq 0+3$,which means $f(x) \geq 3$ for all $x \in \mathbb{R}$.
The minimum value of $f(x)$ is attained when $(2x-1)^{2} = 0$.
Setting $2x-1 = 0$,we get $x = \frac{1}{2}$.
Thus,the minimum value is $f\left(\frac{1}{2}\right) = (2 \cdot \frac{1}{2} - 1)^{2} + 3 = 0 + 3 = 3$.
Since the function $(2x-1)^{2}$ increases indefinitely as $x \to \infty$ or $x \to -\infty$,the function $f(x)$ does not have a maximum value.

(A) The given function is $f(x) = -(x-1)^{2} + 10$.
We know that for any real number $x$,$(x-1)^{2} \geq 0$.
Multiplying by $-1$,we get $-(x-1)^{2} \leq 0$.
Adding $10$ to both sides,we get $-(x-1)^{2} + 10 \leq 10$.
Thus,$f(x) \leq 10$ for all $x \in R$.
The maximum value is attained when $(x-1)^{2} = 0$,which implies $x = 1$.
Therefore,the maximum value is $f(1) = -(1-1)^{2} + 10 = 10$.
Since the function $f(x) = -(x-1)^{2} + 10$ is a downward-opening parabola,as $x \to \infty$ or $x \to -\infty$,$f(x) \to -\infty$.
Hence,the function does not have a minimum value.

(NONE) The given function is $g(x) = x^{3} + 1$.
To find the critical points,we find the first derivative: $g'(x) = 3x^{2}$.
Setting $g'(x) = 0$,we get $3x^{2} = 0$,which implies $x = 0$.
For $x < 0$,$g'(x) > 0$ and for $x > 0$,$g'(x) > 0$.
Since the sign of $g'(x)$ does not change as $x$ passes through $0$,$x = 0$ is a point of inflection.
As $x \to \infty$,$g(x) \to \infty$ and as $x \to -\infty$,$g(x) \to -\infty$.
Therefore,the function $g(x) = x^{3} + 1$ has neither a local maximum nor a local minimum value.

(N/A) Given the function $f(x) = |x + 2| - 1$.
We know that the absolute value function $|x + 2| \geq 0$ for every $x \in \mathbb{R}$.
Therefore,$f(x) = |x + 2| - 1 \geq -1$ for every $x \in \mathbb{R}$.
The minimum value of $f$ is attained when $|x + 2| = 0$.
Setting $|x + 2| = 0$,we get $x = -2$.
The minimum value of $f$ is $f(-2) = |-2 + 2| - 1 = 0 - 1 = -1$.
Since $|x + 2|$ can take arbitrarily large positive values as $x \to \infty$ or $x \to -\infty$,the function $f(x)$ does not have a maximum value.

(A) Given the function $g(x)=-|x+1|+3$.
We know that the absolute value function $|x+1| \geq 0$ for all $x \in \mathbb{R}$.
Multiplying by $-1$,we get $-|x+1| \leq 0$ for all $x \in \mathbb{R}$.
Adding $3$ to both sides,we get $g(x) = -|x+1|+3 \leq 3$ for all $x \in \mathbb{R}$.
The maximum value of $g(x)$ is $3$,which occurs when $|x+1|=0$,i.e.,at $x=-1$.
Since the range of the absolute value function $|x+1|$ is $[0, \infty)$,the range of $-|x+1|$ is $(-\infty, 0]$.
Therefore,the range of $g(x) = -|x+1|+3$ is $(-\infty, 3]$.
Since the function decreases without bound as $x \to \infty$ or $x \to -\infty$,it does not have a minimum value.
Thus,the maximum value is $3$ and the minimum value does not exist.

(NONE) The given function is $h(x) = x + 1$ defined on the open interval $(-1, 1)$.
For any $x \in (-1, 1)$,we have $-1 < x < 1$.
Adding $1$ to all parts of the inequality,we get $-1 + 1 < x + 1 < 1 + 1$,which simplifies to $0 < h(x) < 2$.
As $x$ approaches $-1$ from the right,$h(x)$ approaches $0$,but $h(x)$ is never equal to $0$ because $-1$ is not included in the domain.
As $x$ approaches $1$ from the left,$h(x)$ approaches $2$,but $h(x)$ is never equal to $2$ because $1$ is not included in the domain.
Since the function is strictly increasing and the interval is open,there is no point $c \in (-1, 1)$ such that $h(c)$ is the maximum or minimum value.
Therefore,the function $h(x)$ has neither a maximum nor a minimum value in the interval $(-1, 1)$.

(N/A) $f(x) = x^{2}$
$\therefore f'(x) = 2x$
Now,set $f'(x) = 0$ to find the critical points:
$2x = 0 \Rightarrow x = 0$
Thus,$x = 0$ is the only critical point.
We calculate the second derivative: $f''(x) = 2$.
Since $f''(0) = 2 > 0$,by the second derivative test,$x = 0$ is a point of local minima.
The local minimum value of $f$ at $x = 0$ is $f(0) = (0)^{2} = 0$.
There is no local maximum value for this function as $f(x) \to \infty$ as $x \to \pm \infty$.

(A) Given function: $g(x) = x^{3} - 3x$.
First,find the derivative: $g'(x) = 3x^{2} - 3$.
To find critical points,set $g'(x) = 0$:
$3x^{2} - 3 = 0 \Rightarrow 3(x^{2} - 1) = 0 \Rightarrow x^{2} = 1 \Rightarrow x = \pm 1$.
Now,find the second derivative: $g''(x) = 6x$.
Apply the second derivative test:
For $x = 1$: $g''(1) = 6(1) = 6 > 0$. Since $g''(1) > 0$,$x = 1$ is a point of local minima.
The local minimum value is $g(1) = (1)^{3} - 3(1) = 1 - 3 = -2$.
For $x = -1$: $g''(-1) = 6(-1) = -6 < 0$. Since $g''(-1) < 0$,$x = -1$ is a point of local maxima.
The local maximum value is $g(-1) = (-1)^{3} - 3(-1) = -1 + 3 = 2$.

(A) Given function: $h(x) = \sin x + \cos x$ for $x \in (0, \frac{\pi}{2})$.
First,find the derivative: $h'(x) = \cos x - \sin x$.
Set $h'(x) = 0$ to find critical points:
$\cos x - \sin x = 0 \Rightarrow \sin x = \cos x \Rightarrow \tan x = 1 \Rightarrow x = \frac{\pi}{4}$.
Find the second derivative: $h''(x) = -\sin x - \cos x$.
Evaluate $h''(x)$ at $x = \frac{\pi}{4}$:
$h''(\frac{\pi}{4}) = -\sin(\frac{\pi}{4}) - \cos(\frac{\pi}{4}) = -\frac{1}{\sqrt{2}} - \frac{1}{\sqrt{2}} = -\frac{2}{\sqrt{2}} = -\sqrt{2}$.
Since $h''(\frac{\pi}{4}) < 0$,$x = \frac{\pi}{4}$ is a point of local maxima.
The local maximum value is $h(\frac{\pi}{4}) = \sin(\frac{\pi}{4}) + \cos(\frac{\pi}{4}) = \frac{1}{\sqrt{2}} + \frac{1}{\sqrt{2}} = \sqrt{2}$.
Since the function is strictly concave down on the interval $(0, \frac{\pi}{2})$,there is no local minimum value in the given open interval.

Given $f(x) = \sin x - \cos x$ for $0 < x < 2\pi$.
First,find the derivative $f'(x) = \cos x + \sin x$.
Set $f'(x) = 0$ to find critical points: $\cos x + \sin x = 0 \Rightarrow \tan x = -1$.
In the interval $(0, 2\pi)$,$\tan x = -1$ at $x = \frac{3\pi}{4}$ and $x = \frac{7\pi}{4}$.
Now,find the second derivative $f''(x) = -\sin x + \cos x$.
Evaluate at $x = \frac{3\pi}{4}$: $f''(\frac{3\pi}{4}) = -\sin(\frac{3\pi}{4}) + \cos(\frac{3\pi}{4}) = -\frac{1}{\sqrt{2}} - \frac{1}{\sqrt{2}} = -\sqrt{2} < 0$. Thus,$x = \frac{3\pi}{4}$ is a point of local maxima.
The local maximum value is $f(\frac{3\pi}{4}) = \sin(\frac{3\pi}{4}) - \cos(\frac{3\pi}{4}) = \frac{1}{\sqrt{2}} - (-\frac{1}{\sqrt{2}}) = \sqrt{2}$.
Evaluate at $x = \frac{7\pi}{4}$: $f''(\frac{7\pi}{4}) = -\sin(\frac{7\pi}{4}) + \cos(\frac{7\pi}{4}) = -(-\frac{1}{\sqrt{2}}) + \frac{1}{\sqrt{2}} = \sqrt{2} > 0$. Thus,$x = \frac{7\pi}{4}$ is a point of local minima.
The local minimum value is $f(\frac{7\pi}{4}) = \sin(\frac{7\pi}{4}) - \cos(\frac{7\pi}{4}) = -\frac{1}{\sqrt{2}} - \frac{1}{\sqrt{2}} = -\sqrt{2}$.

(N/A) Given $f(x) = x^{3} - 6x^{2} + 9x + 15$.
First,find the derivative $f^{\prime}(x) = 3x^{2} - 12x + 9$.
Set $f^{\prime}(x) = 0$ to find critical points: $3(x^{2} - 4x + 3) = 0 \Rightarrow 3(x - 1)(x - 3) = 0$.
Thus,the critical points are $x = 1$ and $x = 3$.
Now,find the second derivative $f^{\prime \prime}(x) = 6x - 12 = 6(x - 2)$.
For $x = 1$,$f^{\prime \prime}(1) = 6(1 - 2) = -6 < 0$. Since the second derivative is negative,$x = 1$ is a point of local maxima.
The local maximum value is $f(1) = (1)^{3} - 6(1)^{2} + 9(1) + 15 = 1 - 6 + 9 + 15 = 19$.
For $x = 3$,$f^{\prime \prime}(3) = 6(3 - 2) = 6 > 0$. Since the second derivative is positive,$x = 3$ is a point of local minima.
The local minimum value is $f(3) = (3)^{3} - 6(3)^{2} + 9(3) + 15 = 27 - 54 + 27 + 15 = 15$.

(A) Given $g(x) = \frac{x}{2} + \frac{2}{x}$ for $x > 0$.
First,find the derivative $g'(x)$:
$g'(x) = \frac{1}{2} - \frac{2}{x^2}$.
To find the critical points,set $g'(x) = 0$:
$\frac{1}{2} - \frac{2}{x^2} = 0 \implies \frac{2}{x^2} = \frac{1}{2} \implies x^2 = 4$.
Since $x > 0$,we have $x = 2$.
Now,find the second derivative $g''(x)$:
$g''(x) = \frac{d}{dx}(\frac{1}{2} - 2x^{-2}) = 0 - 2(-2)x^{-3} = \frac{4}{x^3}$.
Evaluate $g''(x)$ at $x = 2$:
$g''(2) = \frac{4}{2^3} = \frac{4}{8} = \frac{1}{2}$.
Since $g''(2) > 0$,by the second derivative test,$x = 2$ is a point of local minima.
The local minimum value is $g(2) = \frac{2}{2} + \frac{2}{2} = 1 + 1 = 2$.
There is no local maximum value for this function as $x > 0$.

(A) Given function: $g(x) = \frac{1}{x^{2}+2}$.
To find the critical points,we find the derivative $g'(x)$ using the chain rule:
$g'(x) = \frac{d}{dx}(x^{2}+2)^{-1} = -1(x^{2}+2)^{-2} \cdot (2x) = \frac{-2x}{(x^{2}+2)^{2}}$.
Set $g'(x) = 0$:
$\frac{-2x}{(x^{2}+2)^{2}} = 0 \Rightarrow -2x = 0 \Rightarrow x = 0$.
Now,we apply the first derivative test:
For $x < 0$,let $x = -1$: $g'(-1) = \frac{-2(-1)}{((-1)^{2}+2)^{2}} = \frac{2}{9} > 0$.
For $x > 0$,let $x = 1$: $g'(1) = \frac{-2(1)}{(1^{2}+2)^{2}} = \frac{-2}{9} < 0$.
Since $g'(x)$ changes from positive to negative at $x = 0$,$x = 0$ is a point of local maxima.
The local maximum value is $g(0) = \frac{1}{0^{2}+2} = \frac{1}{2}$.
Since the denominator $(x^{2}+2)^{2}$ is always positive and the numerator $-2x$ does not change sign in a way that creates a local minimum,there is no local minimum value for this function.

(N/A) Given $f(x) = x\sqrt{1 - x}$ for $0 < x < 1$.
Using the product rule,$f'(x) = (1)\sqrt{1 - x} + x \cdot \frac{1}{2\sqrt{1 - x}}(-1) = \sqrt{1 - x} - \frac{x}{2\sqrt{1 - x}}$.
Simplifying,$f'(x) = \frac{2(1 - x) - x}{2\sqrt{1 - x}} = \frac{2 - 3x}{2\sqrt{1 - x}}$.
Setting $f'(x) = 0$,we get $2 - 3x = 0$,which implies $x = \frac{2}{3}$.
Now,find the second derivative $f''(x) = \frac{d}{dx} \left( \frac{2 - 3x}{2(1 - x)^{1/2}} \right)$.
Using the quotient rule,$f''(x) = \frac{1}{2} \left[ \frac{-3(1 - x)^{1/2} - (2 - 3x) \cdot \frac{1}{2}(1 - x)^{-1/2}(-1)}{1 - x} \right] = \frac{-6(1 - x) + (2 - 3x)}{4(1 - x)^{3/2}} = \frac{3x - 4}{4(1 - x)^{3/2}}$.
At $x = \frac{2}{3}$,$f''\left(\frac{2}{3}\right) = \frac{3(2/3) - 4}{4(1 - 2/3)^{3/2}} = \frac{2 - 4}{4(1/3)^{3/2}} = \frac{-2}{4(1/3)^{3/2}} < 0$.
Since $f''\left(\frac{2}{3}\right) < 0$,$x = \frac{2}{3}$ is a point of local maxima.
The local maximum value is $f\left(\frac{2}{3}\right) = \frac{2}{3}\sqrt{1 - \frac{2}{3}} = \frac{2}{3}\sqrt{\frac{1}{3}} = \frac{2}{3\sqrt{3}} = \frac{2\sqrt{3}}{9}$.
There is no local minimum value in the interval $(0, 1)$.

(N/A) Given the function $f(x) = e^{x}$.
First,we find the derivative of the function with respect to $x$:
$f^{\prime}(x) = \frac{d}{dx}(e^{x}) = e^{x}$.
For a function to have local maxima or minima,the first derivative $f^{\prime}(x)$ must be equal to $0$ at some point $c$ in the domain.
Setting $f^{\prime}(x) = 0$,we get:
$e^{x} = 0$.
However,the exponential function $e^{x}$ is strictly positive for all real values of $x$ ($e^{x} > 0$ for all $x \in \mathbb{R}$).
Since $e^{x}$ can never be equal to $0$,there exists no value $c \in \mathbb{R}$ such that $f^{\prime}(c) = 0$.
Therefore,the function $f(x) = e^{x}$ does not have any points of local maxima or local minima.

(N/A) Given the function $g(x) = \log x$.
First,we find the derivative of the function with respect to $x$:
$g'(x) = \frac{d}{dx}(\log x) = \frac{1}{x}$.
Since the domain of the logarithmic function $g(x) = \log x$ is $x > 0$,the derivative $g'(x) = \frac{1}{x}$ is always positive for all $x$ in its domain ($g'(x) > 0$ for all $x > 0$).
For a function to have a local maximum or minimum,there must exist a point $c$ in the domain such that $g'(c) = 0$ or $g'(c)$ does not exist.
In this case,$\frac{1}{x}$ is never equal to $0$ for any real value of $x$.
Since $g'(x) \neq 0$ for any $x$ in the domain,the function $g(x) = \log x$ does not have any points of local maxima or local minima.

(N/A) Given the function $h(x) = x^{3} + x^{2} + x + 1$.
First,we find the derivative of the function with respect to $x$:
$h'(x) = \frac{d}{dx}(x^{3} + x^{2} + x + 1) = 3x^{2} + 2x + 1$.
To find the critical points,we set $h'(x) = 0$:
$3x^{2} + 2x + 1 = 0$.
For this quadratic equation $ax^{2} + bx + c = 0$,the discriminant $D$ is given by $D = b^{2} - 4ac$.
Here,$a = 3$,$b = 2$,and $c = 1$.
$D = (2)^{2} - 4(3)(1) = 4 - 12 = -8$.
Since the discriminant $D < 0$,the equation $3x^{2} + 2x + 1 = 0$ has no real roots.
This implies that $h'(x)$ is always positive (since the coefficient of $x^{2}$ is positive) for all $x \in \mathbb{R}$.
Since $h'(x) \neq 0$ for any $x \in \mathbb{R}$,the function $h(x)$ does not have any local maxima or local minima.

(A) The given function is $f(x) = x^{3}$.
$\therefore f^{\prime}(x) = 3x^{2}$.
Now,setting $f^{\prime}(x) = 0$ gives $3x^{2} = 0$,which implies $x = 0$.
We evaluate the value of $f$ at the critical point $x = 0$ and at the endpoints of the interval $[-2, 2]$:
$f(0) = 0^{3} = 0$.
$f(-2) = (-2)^{3} = -8$.
$f(2) = (2)^{3} = 8$.
Comparing these values,the absolute maximum value of $f$ on $[-2, 2]$ is $8$ (at $x = 2$) and the absolute minimum value of $f$ on $[-2, 2]$ is $-8$ (at $x = -2$).

(A) The given function is $f(x) = \sin x + \cos x$.
First,we find the derivative: $f'(x) = \cos x - \sin x$.
To find the critical points,set $f'(x) = 0$:
$\cos x - \sin x = 0 \Rightarrow \sin x = \cos x \Rightarrow \tan x = 1$.
Since $x \in [0, \pi]$,the only critical point is $x = \frac{\pi}{4}$.
Now,we evaluate $f(x)$ at the critical point and the endpoints of the interval $[0, \pi]$:
$f\left(\frac{\pi}{4}\right) = \sin\left(\frac{\pi}{4}\right) + \cos\left(\frac{\pi}{4}\right) = \frac{1}{\sqrt{2}} + \frac{1}{\sqrt{2}} = \sqrt{2}$.
$f(0) = \sin(0) + \cos(0) = 0 + 1 = 1$.
$f(\pi) = \sin(\pi) + \cos(\pi) = 0 - 1 = -1$.
Comparing these values,the absolute maximum value is $\sqrt{2}$ at $x = \frac{\pi}{4}$ and the absolute minimum value is $-1$ at $x = \pi$.

(N/A) The given function is $f(x) = 4x - \frac{1}{2}x^2$.
First,we find the derivative of the function:
$f'(x) = \frac{d}{dx}(4x - \frac{1}{2}x^2) = 4 - x$.
To find the critical points,we set $f'(x) = 0$:
$4 - x = 0 \implies x = 4$.
Since $x = 4$ lies within the interval $\left[-2, \frac{9}{2}\right]$,we evaluate the function at the critical point and the endpoints of the interval:
$f(4) = 4(4) - \frac{1}{2}(4)^2 = 16 - 8 = 8$.
$f(-2) = 4(-2) - \frac{1}{2}(-2)^2 = -8 - 2 = -10$.
$f\left(\frac{9}{2}\right) = 4\left(\frac{9}{2}\right) - \frac{1}{2}\left(\frac{9}{2}\right)^2 = 18 - \frac{81}{8} = 18 - 10.125 = 7.875$.
Comparing these values,the absolute maximum value is $8$ at $x = 4$ and the absolute minimum value is $-10$ at $x = -2$.

(A) The given function is $f(x) = (x - 1)^{2} + 3$ defined on the interval $[-3, 1]$.
First,we find the derivative of the function:
$f'(x) = 2(x - 1)$.
To find the critical points,we set $f'(x) = 0$:
$2(x - 1) = 0 \Rightarrow x = 1$.
Now,we evaluate the function $f(x)$ at the critical point $x = 1$ and at the endpoints of the interval $x = -3$ and $x = 1$:
$f(1) = (1 - 1)^{2} + 3 = 0 + 3 = 3$.
$f(-3) = (-3 - 1)^{2} + 3 = (-4)^{2} + 3 = 16 + 3 = 19$.
Comparing these values,the absolute maximum value is $19$ at $x = -3$ and the absolute minimum value is $3$ at $x = 1$.

(A) The profit function is given as $p(x) = 41 - 72x - 18x^{2}$.
To find the maximum profit,we find the first derivative $p'(x)$:
$p'(x) = \frac{d}{dx}(41 - 72x - 18x^{2}) = -72 - 36x$.
Set $p'(x) = 0$ to find the critical point:
$-72 - 36x = 0 \Rightarrow 36x = -72 \Rightarrow x = -2$.
Now,find the second derivative $p''(x)$ to check for maxima:
$p''(x) = \frac{d}{dx}(-72 - 36x) = -36$.
Since $p''(-2) = -36 < 0$,the function has a local maximum at $x = -2$.
Now,calculate the maximum profit by substituting $x = -2$ into $p(x)$:
$p(-2) = 41 - 72(-2) - 18(-2)^{2}$
$p(-2) = 41 + 144 - 18(4)$
$p(-2) = 41 + 144 - 72$
$p(-2) = 113$.
Hence,the maximum profit is $113$ units.

(A) Let $f(x) = 3x^{4}-8x^{3}+12x^{2}-48x+25$.
First,find the derivative $f'(x) = 12x^{3}-24x^{2}+24x-48$.
Set $f'(x) = 0$ to find critical points: $12(x^{3}-2x^{2}+2x-4) = 0$.
Factoring the cubic: $12[x^{2}(x-2)+2(x-2)] = 0$,which gives $12(x-2)(x^{2}+2) = 0$.
The only real critical point in the interval $[0,3]$ is $x=2$.
Now,evaluate $f(x)$ at the critical point and the endpoints of the interval:
At $x=0$: $f(0) = 3(0)^{4}-8(0)^{3}+12(0)^{2}-48(0)+25 = 25$.
At $x=2$: $f(2) = 3(2)^{4}-8(2)^{3}+12(2)^{2}-48(2)+25 = 3(16)-8(8)+12(4)-96+25 = 48-64+48-96+25 = -39$.
At $x=3$: $f(3) = 3(3)^{4}-8(3)^{3}+12(3)^{2}-48(3)+25 = 3(81)-8(27)+12(9)-144+25 = 243-216+108-144+25 = 16$.
Comparing these values: $25, -39, 16$.
The maximum value is $25$ at $x=0$ and the minimum value is $-39$ at $x=2$.

(A) Let $f(x) = \sin 2x$.
Then,$f'(x) = 2 \cos 2x$.
For critical points,set $f'(x) = 0$,which implies $2 \cos 2x = 0$,so $\cos 2x = 0$.
In the interval $[0, 2\pi]$,$2x$ lies in $[0, 4\pi]$. Thus,$2x = \frac{\pi}{2}, \frac{3\pi}{2}, \frac{5\pi}{2}, \frac{7\pi}{2}$.
This gives $x = \frac{\pi}{4}, \frac{3\pi}{4}, \frac{5\pi}{4}, \frac{7\pi}{4}$.
Evaluating $f(x)$ at these points:
$f(\frac{\pi}{4}) = \sin(\frac{\pi}{2}) = 1$
$f(\frac{3\pi}{4}) = \sin(\frac{3\pi}{2}) = -1$
$f(\frac{5\pi}{4}) = \sin(\frac{5\pi}{2}) = 1$
$f(\frac{7\pi}{4}) = \sin(\frac{7\pi}{2}) = -1$
Also,at the endpoints $f(0) = 0$ and $f(2\pi) = 0$.
The maximum value is $1$,which occurs at $x = \frac{\pi}{4}$ and $x = \frac{5\pi}{4}$.

(N/A) Let $f(x) = 2x^{3}-24x+107$.
$\therefore f'(x) = 6x^{2}-24 = 6(x^{2}-4)$.
Now,$f'(x) = 0 \Rightarrow 6(x^{2}-4) = 0 \Rightarrow x^{2} = 4 \Rightarrow x = \pm 2$.
Case $1$: Interval $[1,3]$.
We evaluate $f(x)$ at the critical point $x = 2 \in [1,3]$ and at the endpoints $x = 1$ and $x = 3$.
$f(2) = 2(8) - 24(2) + 107 = 16 - 48 + 107 = 75$.
$f(1) = 2(1) - 24(1) + 107 = 2 - 24 + 107 = 85$.
$f(3) = 2(27) - 24(3) + 107 = 54 - 72 + 107 = 89$.
Thus,the absolute maximum value of $f(x)$ in $[1,3]$ is $89$ at $x = 3$.
Case $2$: Interval $[-3,-1]$.
We evaluate $f(x)$ at the critical point $x = -2 \in [-3,-1]$ and at the endpoints $x = -3$ and $x = -1$.
$f(-3) = 2(-27) - 24(-3) + 107 = -54 + 72 + 107 = 125$.
$f(-1) = 2(-1) - 24(-1) + 107 = -2 + 24 + 107 = 129$.
$f(-2) = 2(-8) - 24(-2) + 107 = -16 + 48 + 107 = 139$.
Thus,the absolute maximum value of $f(x)$ in $[-3,-1]$ is $139$ at $x = -2$.

(A) Let $f(x) = x^{4}-62x^{2}+ax+9.$
First,we find the derivative of the function $f(x)$ with respect to $x$:
$f'(x) = \frac{d}{dx}(x^{4}-62x^{2}+ax+9) = 4x^{3}-124x+a.$
It is given that the function $f(x)$ attains its maximum value on the interval $[0,2]$ at $x=1.$
Since $x=1$ is an interior point of the interval $[0,2],$ the derivative at this point must be zero,i.e.,$f'(1) = 0.$
Substituting $x=1$ into the derivative expression:
$f'(1) = 4(1)^{3}-124(1)+a = 0.$
$4-124+a = 0.$
$-120+a = 0.$
$a = 120.$
Thus,the value of $a$ is $120.$

Let $f(x) = x + \sin 2x$.
$\therefore f'(x) = 1 + 2 \cos 2x$.
Now,$f'(x) = 0 \Rightarrow \cos 2x = -\frac{1}{2}$.
Since $\cos \frac{2\pi}{3} = -\frac{1}{2}$ and $\cos \frac{4\pi}{3} = -\frac{1}{2}$,we have $2x = \frac{2\pi}{3}, \frac{4\pi}{3}, \frac{8\pi}{3}, \frac{10\pi}{3}$ for $x \in [0, 2\pi]$.
Thus,the critical points are $x = \frac{\pi}{3}, \frac{2\pi}{3}, \frac{4\pi}{3}, \frac{5\pi}{3}$.
Evaluating $f(x)$ at critical points and endpoints:
$f(0) = 0 + \sin(0) = 0$.
$f(\frac{\pi}{3}) = \frac{\pi}{3} + \sin(\frac{2\pi}{3}) = \frac{\pi}{3} + \frac{\sqrt{3}}{2}$.
$f(\frac{2\pi}{3}) = \frac{2\pi}{3} + \sin(\frac{4\pi}{3}) = \frac{2\pi}{3} - \frac{\sqrt{3}}{2}$.
$f(\frac{4\pi}{3}) = \frac{4\pi}{3} + \sin(\frac{8\pi}{3}) = \frac{4\pi}{3} + \frac{\sqrt{3}}{2}$.
$f(\frac{5\pi}{3}) = \frac{5\pi}{3} + \sin(\frac{10\pi}{3}) = \frac{5\pi}{3} - \frac{\sqrt{3}}{2}$.
$f(2\pi) = 2\pi + \sin(4\pi) = 2\pi$.
Comparing these values,the absolute maximum value is $2\pi$ at $x = 2\pi$ and the absolute minimum value is $0$ at $x = 0$.

(A) Let the two numbers be $x$ and $y$. Given that their sum is $x + y = 24$,we can express $y$ as $y = 24 - x$.
Let $P$ be the product of the two numbers,so $P = x \cdot y = x(24 - x) = 24x - x^2$.
To find the maximum product,we find the derivative of $P$ with respect to $x$:
$\frac{dP}{dx} = 24 - 2x$.
Setting the derivative to zero for critical points:
$24 - 2x = 0 \Rightarrow x = 12$.
To confirm this is a maximum,we find the second derivative:
$\frac{d^2P}{dx^2} = -2$.
Since the second derivative is negative $(-2 < 0)$,the function has a local maximum at $x = 12$.
If $x = 12$,then $y = 24 - 12 = 12$.
Thus,the two numbers are $12$ and $12$.

(A) Given $x+y=60$,so $y=60-x$.
Let $f(x) = x y^3 = x(60-x)^3$.
To find the maximum,we find the derivative $f'(x)$:
$f'(x) = (60-x)^3 + x \cdot 3(60-x)^2(-1) = (60-x)^2 [60-x - 3x] = (60-x)^2 (60-4x)$.
Setting $f'(x) = 0$,we get $x=60$ or $x=15$.
Since $x$ and $y$ are positive numbers and $x+y=60$,$x$ cannot be $60$ (as that would make $y=0$).
Thus,$x=15$.
Now,$y = 60 - 15 = 45$.
Checking the second derivative $f''(x) = 2(60-x)(-1)(60-4x) + (60-x)^2(-4) = -2(60-x)(60-4x + 2(60-x)) = -2(60-x)(180-6x)$.
For $x=15$,$f''(15) = -2(45)(180-90) = -2(45)(90) < 0$.
Since the second derivative is negative,$x=15$ is a point of local maxima.
Therefore,the numbers are $x=15$ and $y=45$.

(A) Let one number be $x$. Then,the other number is $y = 35 - x$.
Let $P(x) = x^{2} y^{5}$. Substituting $y$,we get:
$P(x) = x^{2} (35 - x)^{5}$.
To find the critical points,we differentiate $P(x)$ with respect to $x$:
$P'(x) = 2x(35 - x)^{5} + x^{2} \cdot 5(35 - x)^{4} \cdot (-1)$
$P'(x) = x(35 - x)^{4} [2(35 - x) - 5x]$
$P'(x) = x(35 - x)^{4} (70 - 2x - 5x)$
$P'(x) = x(35 - x)^{4} (70 - 7x) = 7x(35 - x)^{4} (10 - x)$.
Setting $P'(x) = 0$,we get $x = 0$,$x = 35$,or $x = 10$.
Since $x$ and $y$ are positive numbers,$x = 0$ and $x = 35$ are rejected as they lead to a product of $0$.
Now,we check the second derivative $P''(x)$ at $x = 10$:
$P''(x) = \frac{d}{dx} [7x(35 - x)^{4} (10 - x)]$.
Using the product rule,at $x = 10$,the term $(10 - x)$ becomes $0$,so:
$P''(10) = 7(10)(35 - 10)^{4} (-1) = -70(25)^{4} < 0$.
Since $P''(10) < 0$,the function $P(x)$ is maximum at $x = 10$.
Thus,$x = 10$ and $y = 35 - 10 = 25$.

(A) Let one number be $x$. Then,the other number is $(16-x)$.
Let the sum of the cubes of these numbers be denoted by $S(x)$. Then,
$S(x) = x^3 + (16-x)^3$
To find the minimum,we find the first derivative:
$S'(x) = 3x^2 - 3(16-x)^2$
Setting $S'(x) = 0$:
$3x^2 - 3(16-x)^2 = 0$
$x^2 - (256 - 32x + x^2) = 0$
$32x - 256 = 0$
$x = 8$
Now,we find the second derivative to check for minima:
$S''(x) = 6x + 6(16-x)$
$S''(8) = 6(8) + 6(16-8) = 48 + 48 = 96$
Since $S''(8) > 0$,the function has a local minimum at $x = 8$.
Thus,the two numbers are $8$ and $16-8 = 8$.

(A) Let the side of the square to be cut off be $x \, cm$.
Then,the length and the breadth of the box will be $(18-2x) \, cm$ each,and the height of the box will be $x \, cm$.
Therefore,the volume $V(x)$ of the box is given by:
$V(x) = x(18-2x)^2 = x(324 - 72x + 4x^2) = 4x^3 - 72x^2 + 324x$.
To find the maximum volume,we find the derivative $V'(x)$:
$V'(x) = 12x^2 - 144x + 324$.
Setting $V'(x) = 0$:
$12(x^2 - 12x + 27) = 0
\Rightarrow 12(x-9)(x-3) = 0$.
So,$x = 9$ or $x = 3$.
If $x = 9$,the side of the box $(18-2x)$ becomes $0$,which is not possible.
Thus,$x = 3$.
Now,we check the second derivative $V''(x) = 24x - 144$.
$V''(3) = 24(3) - 144 = 72 - 144 = -72 < 0$.
Since $V''(3) < 0$,the volume is maximum at $x = 3 \, cm$.

(A) Let the side of the square to be cut be $x \, cm$. Then,the height of the box is $x$,the length is $45-2x$,and the breadth is $24-2x$.
Therefore,the volume $V(x)$ of the box is given by:
$V(x) = x(45-2x)(24-2x) = x(1080 - 90x - 48x + 4x^2) = 4x^3 - 138x^2 + 1080x$.
Now,find the first derivative:
$V'(x) = 12x^2 - 276x + 1080 = 12(x^2 - 23x + 90) = 12(x-18)(x-5)$.
Setting $V'(x) = 0$,we get $x = 18$ or $x = 5$.
Since the breadth is $24 \, cm$,$x$ cannot be $18$ (as $2x$ would be $36 > 24$). Thus,$x = 5$.
Now,find the second derivative:
$V''(x) = 24x - 276$.
At $x = 5$,$V''(5) = 24(5) - 276 = 120 - 276 = -156 < 0$.
Since $V''(5) < 0$,the volume is maximum at $x = 5 \, cm$.

(N/A) Let a rectangle of length $l$ and breadth $b$ be inscribed in the given circle of radius $R$.
The diagonal of the rectangle passes through the center of the circle and is equal to the diameter,$2R$.
By applying the Pythagoras theorem,we have:
$(2R)^{2} = l^{2} + b^{2}$
$\Rightarrow b^{2} = 4R^{2} - l^{2}$
$\Rightarrow b = \sqrt{4R^{2} - l^{2}}$
Area of the rectangle,$A = l \times b = l \sqrt{4R^{2} - l^{2}}$
To maximize the area,we maximize $A^{2}$. Let $S = A^{2} = l^{2}(4R^{2} - l^{2}) = 4R^{2}l^{2} - l^{4}$.
Differentiating with respect to $l$:
$\frac{dS}{dl} = 8R^{2}l - 4l^{3}$
Setting $\frac{dS}{dl} = 0$:
$4l(2R^{2} - l^{2}) = 0$
Since $l \neq 0$,we have $l^{2} = 2R^{2} \Rightarrow l = R\sqrt{2}$.
Now,$b = \sqrt{4R^{2} - (R\sqrt{2})^{2}} = \sqrt{4R^{2} - 2R^{2}} = \sqrt{2R^{2}} = R\sqrt{2}$.
Since $l = b = R\sqrt{2}$,the rectangle is a square.
Checking the second derivative:
$\frac{d^{2}S}{dl^{2}} = 8R^{2} - 12l^{2}$
At $l^{2} = 2R^{2}$,$\frac{d^{2}S}{dl^{2}} = 8R^{2} - 12(2R^{2}) = 8R^{2} - 24R^{2} = -16R^{2} < 0$.
Since the second derivative is negative,the area is maximum when $l = b = R\sqrt{2}$.
Thus,it is proved that of all rectangles inscribed in a fixed circle,the square has the maximum area.

(N/A) Let $r$ and $h$ be the radius and height of the cylinder respectively.
The surface area $S$ of the cylinder is given by $S = 2\pi r^2 + 2\pi rh$.
From this,we can express $h$ in terms of $r$ and $S$:
$h = \frac{S - 2\pi r^2}{2\pi r} = \frac{S}{2\pi r} - r$.
The volume $V$ of the cylinder is $V = \pi r^2 h$.
Substituting the expression for $h$:
$V = \pi r^2 \left( \frac{S}{2\pi r} - r \right) = \frac{Sr}{2} - \pi r^3$.
To find the maximum volume,we differentiate $V$ with respect to $r$:
$\frac{dV}{dr} = \frac{S}{2} - 3\pi r^2$.
Setting $\frac{dV}{dr} = 0$ gives $\frac{S}{2} = 3\pi r^2$,so $S = 6\pi r^2$.
Now,find the second derivative to check for maximum:
$\frac{d^2V}{dr^2} = -6\pi r$.
Since $r > 0$,$\frac{d^2V}{dr^2} < 0$,which confirms that the volume is maximum at $S = 6\pi r^2$.
Substituting $S = 6\pi r^2$ into the expression for $h$:
$h = \frac{6\pi r^2 - 2\pi r^2}{2\pi r} = \frac{4\pi r^2}{2\pi r} = 2r$.
Since $2r$ is the diameter of the base,the height of the cylinder is equal to its diameter.

(A) Let $r$ be the radius and $h$ be the height of the cylinder.
Given volume $V = \pi r^2 h = 100 \text{ cm}^3$.
Thus,$h = \frac{100}{\pi r^2}$.
The surface area $S$ is given by $S = 2\pi r^2 + 2\pi rh$.
Substituting $h$,we get $S = 2\pi r^2 + 2\pi r \left(\frac{100}{\pi r^2}\right) = 2\pi r^2 + \frac{200}{r}$.
Differentiating with respect to $r$,we get $\frac{dS}{dr} = 4\pi r - \frac{200}{r^2}$.
Setting $\frac{dS}{dr} = 0$,we have $4\pi r = \frac{200}{r^2} \Rightarrow r^3 = \frac{50}{\pi} \Rightarrow r = \left(\frac{50}{\pi}\right)^{1/3}$.
Checking the second derivative,$\frac{d^2S}{dr^2} = 4\pi + \frac{400}{r^3}$. Since $r > 0$,$\frac{d^2S}{dr^2} > 0$,so $S$ is minimum at $r = \left(\frac{50}{\pi}\right)^{1/3}$.
Then $h = \frac{100}{\pi (50/\pi)^{2/3}} = \frac{100}{\pi} \cdot \frac{\pi^{2/3}}{50^{2/3}} = 2 \cdot \frac{50}{\pi^{1/3} \cdot 50^{2/3}} = 2 \left(\frac{50}{\pi}\right)^{1/3}$.
Thus,the dimensions are $r = \left(\frac{50}{\pi}\right)^{1/3} \text{ cm}$ and $h = 2\left(\frac{50}{\pi}\right)^{1/3} \text{ cm}$.

(A) Let the length of the wire used for the square be $l \, m$. Then the length of the wire used for the circle is $(28-l) \, m$.
The side of the square is $s = \frac{l}{4}$. The area of the square is $A_s = s^2 = \frac{l^2}{16}$.
For the circle,the circumference is $2\pi r = 28-l$,so the radius is $r = \frac{28-l}{2\pi}$. The area of the circle is $A_c = \pi r^2 = \pi \left( \frac{28-l}{2\pi} \right)^2 = \frac{(28-l)^2}{4\pi}$.
The total area $A = A_s + A_c = \frac{l^2}{16} + \frac{(28-l)^2}{4\pi}$.
To find the minimum,we differentiate $A$ with respect to $l$: $\frac{dA}{dl} = \frac{2l}{16} + \frac{2(28-l)(-1)}{4\pi} = \frac{l}{8} - \frac{28-l}{2\pi}$.
Setting $\frac{dA}{dl} = 0$,we get $\frac{l}{8} = \frac{28-l}{2\pi} \Rightarrow \pi l = 4(28-l) \Rightarrow \pi l = 112 - 4l \Rightarrow l(\pi+4) = 112 \Rightarrow l = \frac{112}{\pi+4}$.
The second derivative $\frac{d^2A}{dl^2} = \frac{1}{8} + \frac{1}{2\pi} > 0$,confirming a minimum.
The length for the square is $\frac{112}{\pi+4} \, m$ and the length for the circle is $28 - \frac{112}{\pi+4} = \frac{28\pi+112-112}{\pi+4} = \frac{28\pi}{\pi+4} \, m$.

(N/A) 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 $h = R + AB = R + \sqrt{R^{2} - r^{2}}$.
$V = \frac{1}{3} \pi r^{2} (R + \sqrt{R^{2} - r^{2}})$.
$V = \frac{1}{3} \pi r^{2} R + \frac{1}{3} \pi r^{2} \sqrt{R^{2} - r^{2}}$.
Differentiating with respect to $r$:
$\frac{dV}{dr} = \frac{2}{3} \pi r R + \frac{1}{3} \pi \left( 2r \sqrt{R^{2} - r^{2}} + r^{2} \cdot \frac{-2r}{2\sqrt{R^{2} - r^{2}}} \right)$.
$\frac{dV}{dr} = \frac{2}{3} \pi r R + \frac{1}{3} \pi \left( \frac{2r(R^{2} - r^{2}) - r^{3}}{\sqrt{R^{2} - r^{2}}} \right) = \frac{2}{3} \pi r R + \frac{\pi r (2R^{2} - 3r^{2})}{3\sqrt{R^{2} - r^{2}}}$.
Setting $\frac{dV}{dr} = 0$:
$\frac{2}{3} \pi r R = - \frac{\pi r (2R^{2} - 3r^{2})}{3\sqrt{R^{2} - r^{2}}} \Rightarrow 2R\sqrt{R^{2} - r^{2}} = 3r^{2} - 2R^{2}$.
Squaring both sides: $4R^{2}(R^{2} - r^{2}) = (3r^{2} - 2R^{2})^{2}$.
$4R^{4} - 4R^{2}r^{2} = 9r^{4} + 4R^{4} - 12r^{2}R^{2}$.
$9r^{4} = 8R^{2}r^{2} \Rightarrow r^{2} = \frac{8}{9}R^{2}$.
Then $h = R + \sqrt{R^{2} - \frac{8}{9}R^{2}} = R + \frac{R}{3} = \frac{4}{3}R$.
Maximum volume $V = \frac{1}{3} \pi (\frac{8}{9}R^{2}) (\frac{4}{3}R) = \frac{32}{81} \pi R^{3}$.
Volume of sphere $V_{s} = \frac{4}{3} \pi R^{3}$.
Ratio $\frac{V}{V_{s}} = \frac{32/81 \pi R^{3}}{4/3 \pi R^{3}} = \frac{32}{81} \cdot \frac{3}{4} = \frac{8}{27}$.
Hence,the volume of the largest cone is $\frac{8}{27}$ of the volume of the sphere.

(N/A) Let $r$ and $h$ be the radius and the height (altitude) of the cone respectively.
The volume $(V)$ of the cone is given by $V = \frac{1}{3} \pi r^{2} h$,which implies $h = \frac{3V}{\pi r^{2}}$.
The curved surface area $(S)$ of the cone is given by $S = \pi r l = \pi r \sqrt{r^{2} + h^{2}}$.
Substituting $h$,we get $S = \pi r \sqrt{r^{2} + \frac{9V^{2}}{\pi^{2} r^{4}}} = \frac{1}{r} \sqrt{\pi^{2} r^{6} + 9V^{2}}$.
To minimize $S$,we minimize $S^{2} = \frac{\pi^{2} r^{6} + 9V^{2}}{r^{2}} = \pi^{2} r^{4} + 9V^{2} r^{-2}$.
Let $f(r) = S^{2}$. Then $f'(r) = 4 \pi^{2} r^{3} - 18V^{2} r^{-3}$.
Setting $f'(r) = 0$,we get $4 \pi^{2} r^{3} = \frac{18V^{2}}{r^{3}}$,so $r^{6} = \frac{18V^{2}}{4 \pi^{2}} = \frac{9V^{2}}{2 \pi^{2}}$.
Substituting $V = \frac{1}{3} \pi r^{2} h$ into $r^{6} = \frac{9V^{2}}{2 \pi^{2}}$,we get $r^{6} = \frac{9}{2 \pi^{2}} \cdot (\frac{1}{9} \pi^{2} r^{4} h^{2}) = \frac{1}{2} r^{4} h^{2}$.
Thus,$r^{2} = \frac{1}{2} h^{2}$,which implies $h^{2} = 2r^{2}$,or $h = \sqrt{2} r$.
Since $f''(r) = 12 \pi^{2} r^{2} + 54V^{2} r^{-4} > 0$ for $r > 0$,the surface area is indeed minimized at $h = \sqrt{2} r$.

(A) Let $\theta$ be the semi-vertical angle of the cone.
It is clear that $\theta \in [0, \frac{\pi}{2}]$.
Let $r$,$h$,and $l$ be the radius,height,and the slant height of the cone respectively.
The slant height $l$ of the cone is given as constant.
Now,$r = l \sin \theta$ and $h = l \cos \theta$.
The volume $(V)$ of the cone is given by $V = \frac{1}{3} \pi r^2 h$.
Substituting the values of $r$ and $h$,we get:
$V = \frac{1}{3} \pi (l^2 \sin^2 \theta)(l \cos \theta) = \frac{1}{3} \pi l^3 \sin^2 \theta \cos \theta$.
Differentiating $V$ with respect to $\theta$:
$\frac{dV}{d\theta} = \frac{\pi l^3}{3} [\sin^2 \theta(-\sin \theta) + \cos \theta(2 \sin \theta \cos \theta)]$
$= \frac{\pi l^3}{3} [-\sin^3 \theta + 2 \sin \theta \cos^2 \theta]$.
For maximum or minimum volume,set $\frac{dV}{d\theta} = 0$:
$\sin^3 \theta = 2 \sin \theta \cos^2 \theta \Rightarrow \tan^2 \theta = 2 \Rightarrow \tan \theta = \sqrt{2} \Rightarrow \theta = \tan^{-1} \sqrt{2}$.
Now,check the second derivative:
$\frac{d^2V}{d\theta^2} = \frac{\pi l^3}{3} [-3 \sin^2 \theta \cos \theta + 2 \cos^3 \theta - 4 \sin^2 \theta \cos \theta] = \frac{\pi l^3}{3} [2 \cos^3 \theta - 7 \sin^2 \theta \cos \theta]$.
Substituting $\sin^2 \theta = 2 \cos^2 \theta$ at $\tan \theta = \sqrt{2}$:
$\frac{d^2V}{d\theta^2} = \frac{\pi l^3}{3} [2 \cos^3 \theta - 7(2 \cos^2 \theta) \cos \theta] = \frac{\pi l^3}{3} [2 \cos^3 \theta - 14 \cos^3 \theta] = -4 \pi l^3 \cos^3 \theta$.
Since $\theta \in [0, \frac{\pi}{2}]$,$\cos \theta > 0$,so $\frac{d^2V}{d\theta^2} < 0$.
Thus,the volume is maximum when $\theta = \tan^{-1} \sqrt{2}$.

(N/A) Let $r, h, l$ be the radius,height,and slant height of the right circular cone respectively. Let $S$ be the given surface area of the cone.
We have,$l^{2}=r^{2}+h^{2} \quad \dots (1)$
$S=\pi r l+\pi r^{2}$
$S-\pi r^{2}=\pi r l$
$\Rightarrow l=\frac{S-\pi r^{2}}{\pi r}$
$V=\frac{1}{3} \pi r^{2} h$
$\Rightarrow V=\frac{1}{3} \pi r^{2} \sqrt{l^{2}-r^{2}} \quad (\text{by } (1))$
$\Rightarrow V^{2}=\frac{1}{9} \pi^{2} r^{4}\left(l^{2}-r^{2}\right)$
$\Rightarrow V^{2}=\frac{1}{9} \pi^{2} r^{4}\left[\left(\frac{S-\pi r^{2}}{\pi r}\right)^{2}-r^{2}\right]$
$\Rightarrow V^{2}=\frac{1}{9} \pi^{2} r^{4}\left[\frac{\left(S-\pi r^{2}\right)^{2}-\pi^{2} r^{4}}{\pi^{2} r^{2}}\right]$
$\Rightarrow V^{2}=\frac{1}{9} r^{2}\left[\left(S-\pi r^{2}\right)^{2}-\pi^{2} r^{4}\right]$
$\Rightarrow V^{2}=\frac{1}{9} r^{2}\left[S^{2}-2 \pi S r^{2}+\pi^{2} r^{4}-\pi^{2} r^{4}\right]$
$\Rightarrow V^{2}=\frac{1}{9}\left(r^{2} S^{2}-2 \pi S r^{4}\right)$
$2 V \frac{d V}{d r}=\frac{S^{2}}{9} (2 r)-\frac{2 \pi S}{9} (4 r^{3})$
$2 V \frac{d V}{d r}=\frac{2 r S}{9}\left(S-4 \pi r^{2}\right)$
For maximum volume,$\frac{d V}{d r}=0$
$\Rightarrow \frac{2 r S}{9}\left(S-4 \pi r^{2}\right)=0$
Since $r \neq 0$,we have $S=4 \pi r^{2}$
$\Rightarrow r^{2}=\frac{S}{4 \pi}$
Substituting $S=\pi r l+\pi r^{2}$,we get $4 \pi r^{2}=\pi r l+\pi r^{2}$
$\Rightarrow 3 \pi r^{2}=\pi r l$
$\Rightarrow l=3 r$
Let $\alpha$ be the semi-vertical angle.
$\sin \alpha=\frac{r}{l}=\frac{r}{3 r}=\frac{1}{3}$
$\Rightarrow \alpha=\sin ^{-1}\left(\frac{1}{3}\right)$

(D) Let $f(x) = \frac{1-x+x^{2}}{1+x+x^{2}}$.
Using the quotient rule,$f^{\prime}(x) = \frac{(1+x+x^{2})(-1+2x) - (1-x+x^{2})(1+2x)}{(1+x+x^{2})^{2}}$.
Simplifying the numerator: $(2x^{2}+x-1) - (2x^{3}+x^{2}-x+1) = 2x^{2}-2 = 2(x^{2}-1)$.
So,$f^{\prime}(x) = \frac{2(x^{2}-1)}{(1+x+x^{2})^{2}}$.
Setting $f^{\prime}(x) = 0$,we get $x^{2}-1 = 0$,which implies $x = 1$ or $x = -1$.
Evaluating $f(x)$ at these critical points:
$f(1) = \frac{1-1+1}{1+1+1} = \frac{1}{3}$.
$f(-1) = \frac{1-(-1)+(-1)^{2}}{1+(-1)+(-1)^{2}} = \frac{1+1+1}{1-1+1} = 3$.
Since $f(x) \to 1$ as $x \to \pm \infty$,the minimum value is $\frac{1}{3}$.

(A) Let $f(x) = [x(x-1)+1]^{\frac{1}{3}} = [x^2-x+1]^{\frac{1}{3}}$.
To find the critical points,we calculate the derivative $f'(x)$:
$f'(x) = \frac{1}{3}[x^2-x+1]^{-\frac{2}{3}} \cdot (2x-1) = \frac{2x-1}{3(x^2-x+1)^{\frac{2}{3}}}$.
Setting $f'(x) = 0$,we get $2x-1 = 0$,which implies $x = \frac{1}{2}$.
Since $x = \frac{1}{2}$ lies within the interval $[0, 1]$,we evaluate $f(x)$ at the critical point and the endpoints:
$f(0) = [0(0-1)+1]^{\frac{1}{3}} = 1^{\frac{1}{3}} = 1$.
$f(1) = [1(1-1)+1]^{\frac{1}{3}} = 1^{\frac{1}{3}} = 1$.
$f(\frac{1}{2}) = [\frac{1}{2}(\frac{1}{2}-1)+1]^{\frac{1}{3}} = [\frac{1}{2}(-\frac{1}{2})+1]^{\frac{1}{3}} = [-\frac{1}{4}+1]^{\frac{1}{3}} = (\frac{3}{4})^{\frac{1}{3}}$.
Comparing the values $1$,$1$,and $(\frac{3}{4})^{\frac{1}{3}}$,the maximum value is $1$.
Thus,the correct answer is $A$.

$ (A) $ Let $x$ metres be the side length of the removed squares. The height of the box is $x$, the length is $8-2x$, and the breadth is $3-2x$.
Since the dimensions must be positive, $x > 0$, $8-2x > 0$, and $3-2x > 0$, which implies $0 < x < 1.5$.
The volume $V(x)$ is given by:
$V(x) = x(3-2x)(8-2x) = x(24 - 6x - 16x + 4x^2) = 4x^3 - 22x^2 + 24x$.
To find the maximum volume, we find the derivative $V'(x)$:
$V'(x) = 12x^2 - 44x + 24$.
Setting $V'(x) = 0$:
$12x^2 - 44x + 24 = 0 \implies 4(3x^2 - 11x + 6) = 0 \implies 4(3x-2)(x-3) = 0$.
The critical points are $x = 2/3$ and $x = 3$.
Since $x < 1.5$, we discard $x = 3$.
Using the second derivative test: $V''(x) = 24x - 44$.
$V''(2/3) = 24(2/3) - 44 = 16 - 44 = -28 < 0$.
Thus, $x = 2/3$ is the point of local maxima.
The maximum volume is:
$V(2/3) = (2/3)(3 - 2(2/3))(8 - 2(2/3)) = (2/3)(3 - 4/3)(8 - 4/3) = (2/3)(5/3)(20/3) = 200/27 \text{ m}^3$.