Maxima and Minima Questions in English

(A) Let the two nonzero numbers be $x$ and $y$. Given that $x + y = 4$,which implies $y = 4 - x$.
We want to minimize the sum of their reciprocals,$S = \frac{1}{x} + \frac{1}{y}$.
Substituting $y = 4 - x$ into the expression for $S$,we get $S(x) = \frac{1}{x} + \frac{1}{4 - x}$.
To find the minimum,we differentiate $S$ with respect to $x$: $S'(x) = -\frac{1}{x^2} + \frac{1}{(4 - x)^2}$.
Setting $S'(x) = 0$,we get $\frac{1}{x^2} = \frac{1}{(4 - x)^2}$,which implies $x^2 = (4 - x)^2$.
Solving for $x$,$x^2 = 16 - 8x + x^2$,so $8x = 16$,which gives $x = 2$.
If $x = 2$,then $y = 4 - 2 = 2$.
The sum of the reciprocals is $S = \frac{1}{2} + \frac{1}{2} = 1$.
Since $S''(x) = \frac{2}{x^3} + \frac{2}{(4 - x)^3}$,at $x = 2$,$S''(2) = \frac{2}{8} + \frac{2}{8} = \frac{1}{2} > 0$,confirming a local minimum.
Thus,the minimum value is $1$.

(A) Given the function $y = \alpha \log x + \beta x^3 - x$.
For the function to have extreme values at $x = -1$ and $x = 1$,the derivative $\frac{dy}{dx}$ must be zero at these points.
First,find the derivative: $\frac{dy}{dx} = \frac{\alpha}{x} + 3\beta x^2 - 1$.
At $x = 1$: $\frac{dy}{dx} = \alpha + 3\beta - 1 = 0 \implies \alpha + 3\beta = 1$.
At $x = -1$: $\frac{dy}{dx} = \frac{\alpha}{-1} + 3\beta(-1)^2 - 1 = 0 \implies -\alpha + 3\beta = 1$.
Adding the two equations: $(\alpha + 3\beta) + (-\alpha + 3\beta) = 1 + 1 \implies 6\beta = 2 \implies \beta = \frac{1}{3}$.
Substituting $\beta = \frac{1}{3}$ into $\alpha + 3\beta = 1$: $\alpha + 3(\frac{1}{3}) = 1 \implies \alpha + 1 = 1 \implies \alpha = 0$.
Thus,$\alpha = 0$ and $\beta = \frac{1}{3}$.

(C) Let $R(x)$ be the revenue function and $C(x)$ be the cost function.
Revenue $R(x) = x \times \left(6 - \frac{x}{40}\right) = 6x - \frac{x^2}{40}$.
Cost $C(x) = \frac{x}{5} + 193$.
Profit $P(x) = R(x) - C(x) = 6x - \frac{x^2}{40} - \frac{x}{5} - 193$.
$P(x) = -\frac{x^2}{40} + \frac{29x}{5} - 193$.
To find the maximum profit,we find the derivative $P'(x)$ and set it to $0$.
$P'(x) = -\frac{2x}{40} + \frac{29}{5} = -\frac{x}{20} + 5.8$.
Setting $P'(x) = 0$,we get $\frac{x}{20} = 5.8$,so $x = 116$.
Now,find the second derivative $P''(x) = -\frac{1}{20}$. Since $P''(x) < 0$,the profit is maximum at $x = 116$.
Maximum profit $P(116) = -\frac{116^2}{40} + \frac{29(116)}{5} - 193$.
$P(116) = -\frac{13456}{40} + 672.8 - 193 = -336.4 + 672.8 - 193 = 143.4$.

(A) Let $f(x, y) = ax + by$ subject to the constraint $xy = c^2$,where $x, y > 0$.
Substituting $y = \frac{c^2}{x}$ into the expression,we get $f(x) = ax + \frac{bc^2}{x}$.
To find the minimum,we differentiate with respect to $x$:
$f'(x) = a - \frac{bc^2}{x^2}$.
Setting $f'(x) = 0$,we get $a = \frac{bc^2}{x^2}$,which implies $x^2 = \frac{bc^2}{a}$,so $x = c\sqrt{\frac{b}{a}}$.
Since $x, y > 0$,we take the positive root.
Then $y = \frac{c^2}{x} = \frac{c^2}{c\sqrt{b/a}} = c\sqrt{\frac{a}{b}}$.
The minimum value is $f = a(c\sqrt{\frac{b}{a}}) + b(c\sqrt{\frac{a}{b}}) = c\sqrt{ab} + c\sqrt{ab} = 2c\sqrt{ab}$.

(A) Let $f(x) = (\frac{1}{x})^x = x^{-x}$.
Taking the natural logarithm on both sides,we get $\ln(f(x)) = -x \ln(x)$.
To find the critical points,we differentiate with respect to $x$:
$\frac{d}{dx}(\ln(f(x))) = \frac{d}{dx}(-x \ln(x))$
$\frac{f'(x)}{f(x)} = -[\ln(x) + x \cdot \frac{1}{x}] = -(\ln(x) + 1)$.
Setting $f'(x) = 0$,we get $\ln(x) + 1 = 0$,which implies $\ln(x) = -1$,so $x = e^{-1} = \frac{1}{e}$.
Now,we check the second derivative or the sign of $f'(x)$ around $x = \frac{1}{e}$.
For $x < \frac{1}{e}$,$f'(x) > 0$ and for $x > \frac{1}{e}$,$f'(x) < 0$,so $x = \frac{1}{e}$ is a point of local maximum.
The maximum value is $f(\frac{1}{e}) = (\frac{1}{1/e})^{1/e} = e^{1/e}$.

(B) Let $f(x, y) = x^2 y$. Given $x+y=6$,we have $y = 6-x$.
Substituting $y$ in $f$,we get $f(x) = x^2(6-x) = 6x^2 - x^3$.
To find the maximum,we find the derivative $f'(x) = 12x - 3x^2$.
Setting $f'(x) = 0$,we get $3x(4-x) = 0$,which gives $x=0$ or $x=4$.
Since $x=0$ gives $f(0)=0$,we check $x=4$.
For $x=4$,$y = 6-4 = 2$.
The value is $f(4) = 4^2 \times 2 = 16 \times 2 = 32$.
Thus,the maximum value is $32$.

(B) The slope of the tangent to the curve $y=f(x)$ is given by $m = \frac{dy}{dx}$.
Given $y = x^3 - 3x^2 + 2x + 93$,we differentiate with respect to $x$:
$m = \frac{dy}{dx} = 3x^2 - 6x + 2$.
To find the minimum value of the slope $m$,we differentiate $m$ with respect to $x$ and set it to zero:
$\frac{dm}{dx} = 6x - 6$.
Setting $\frac{dm}{dx} = 0$,we get $6x - 6 = 0$,which implies $x = 1$.
Now,we check the second derivative to confirm the minimum:
$\frac{d^2m}{dx^2} = 6 > 0$,which confirms that the slope has a minimum at $x = 1$.
The minimum value of the slope is $m(1) = 3(1)^2 - 6(1) + 2 = 3 - 6 + 2 = -1$.

(D) Given the set $S = \{x \in R : x^2+30 \leq 11x\}$.
Solving the inequality: $x^2-11x+30 \leq 0$.
$(x-5)(x-6) \leq 0$,which implies $x \in [5, 6]$.
Now,consider the function $f(x) = 3x^3-18x^2+27x-40$.
Find the derivative: $f'(x) = 9x^2-36x+27 = 9(x^2-4x+3) = 9(x-1)(x-3)$.
For $x \in [5, 6]$,both $(x-1)$ and $(x-3)$ are positive,so $f'(x) > 0$.
Since $f'(x) > 0$ for all $x \in [5, 6]$,the function $f(x)$ is strictly increasing on the interval $[5, 6]$.
The maximum value occurs at the right endpoint $x = 6$.
$f(6) = 3(6)^3 - 18(6)^2 + 27(6) - 40 = 3(216) - 18(36) + 162 - 40 = 648 - 648 + 162 - 40 = 122$.

(C) Let $\triangle ABC$ be an isosceles triangle such that $AB = AC = x$.
Let $\angle ABC = \angle ACB = \theta$.
Draw segment $AD \perp$ side $BC$ at point $D$.
In $\triangle ABD$,$AD = x \sin \theta$ and $BD = x \cos \theta$.
Similarly,in $\triangle ACD$,$DC = x \cos \theta$.
Therefore,in $\triangle ABC$,the height is $AD = x \sin \theta$ and the base is $BC = BD + DC = x \cos \theta + x \cos \theta = 2x \cos \theta$.
The area $A$ of $\triangle ABC$ is given by:
$A = \frac{1}{2} \times \text{base} \times \text{height}$
$A = \frac{1}{2} \times (2x \cos \theta) \times (x \sin \theta)$
$A = x^2 \sin \theta \cos \theta = \frac{x^2}{2} \sin(2\theta)$.
Since the maximum value of $\sin(2\theta)$ is $1$ (when $2\theta = 90^\circ$ or $\theta = 45^\circ$),the maximum area is $\frac{x^2}{2}$ sq. units.

(B) Let the two numbers be $a$ and $b$.
Given that $a + b = 3$,so $b = 3 - a$.
Let the product be $P = a \cdot b^2$.
Substituting the value of $b$,we get $P = a(3 - a)^2 = a(9 - 6a + a^2) = a^3 - 6a^2 + 9a$.
To find the maximum value,we differentiate $P$ with respect to $a$: $\frac{dP}{da} = 3a^2 - 12a + 9$.
Setting $\frac{dP}{da} = 0$,we get $3(a^2 - 4a + 3) = 0$,which implies $3(a - 1)(a - 3) = 0$.
So,$a = 1$ or $a = 3$.
Now,find the second derivative: $\frac{d^2P}{da^2} = 6a - 12$.
For $a = 1$,$\frac{d^2P}{da^2} = 6(1) - 12 = -6 < 0$,so $P$ is maximum at $a = 1$.
For $a = 3$,$\frac{d^2P}{da^2} = 6(3) - 12 = 6 > 0$,so $P$ is minimum at $a = 3$.
Substituting $a = 1$ into the product formula: $P = 1(3 - 1)^2 = 1(2)^2 = 4$.
Thus,the maximum value is $4$.

(C) Perimeter of the square $= 4x$.
Perimeter of the circle $= 2\pi r$.
Given total length of wire is $2$ units,so $4x + 2\pi r = 2$.
Dividing by $2$,we get $2x + \pi r = 1$,which implies $r = \frac{1 - 2x}{\pi}$.
Sum of the areas $A = x^2 + \pi r^2$.
Substituting $r$ in terms of $x$: $A = x^2 + \pi \left( \frac{1 - 2x}{\pi} \right)^2 = x^2 + \frac{(1 - 2x)^2}{\pi}$.
Differentiating $A$ with respect to $x$: $\frac{dA}{dx} = 2x + \frac{2(1 - 2x)(-2)}{\pi} = 2x - \frac{4(1 - 2x)}{\pi}$.
For minimum area,set $\frac{dA}{dx} = 0$: $2x - \frac{4}{\pi} + \frac{8x}{\pi} = 0$.
Multiply by $\pi$: $2\pi x - 4 + 8x = 0 \Rightarrow (2\pi + 8)x = 4 \Rightarrow (\pi + 4)x = 2$.
Thus,$x = \frac{2}{\pi + 4}$.
Substituting $x$ back into the perimeter equation: $2(\frac{2}{\pi + 4}) + \pi r = 1 \Rightarrow \pi r = 1 - \frac{4}{\pi + 4} = \frac{\pi + 4 - 4}{\pi + 4} = \frac{\pi}{\pi + 4}$.
Therefore,$r = \frac{1}{\pi + 4}$.
Comparing $x$ and $r$,we see that $x = 2r$.

(B) Given $f(x) = \sin^4 x + \cos^4 x$.
We can rewrite this as:
$f(x) = (\sin^2 x + \cos^2 x)^2 - 2 \sin^2 x \cos^2 x$
Since $\sin^2 x + \cos^2 x = 1$,we have:
$f(x) = 1 - 2 \sin^2 x \cos^2 x$
Using the identity $\sin 2x = 2 \sin x \cos x$,we get $\sin^2 x \cos^2 x = \frac{\sin^2 2x}{4}$.
Substituting this into the expression:
$f(x) = 1 - 2 \left(\frac{\sin^2 2x}{4}\right) = 1 - \frac{1}{2} \sin^2 2x$.
For $0 < x < \frac{\pi}{2}$,the range of $2x$ is $0 < 2x < \pi$. Thus,the maximum value of $\sin 2x$ is $1$ at $2x = \frac{\pi}{2}$,which means $x = \frac{\pi}{4}$.
When $\sin 2x = 1$,$f(x) = 1 - \frac{1}{2}(1)^2 = 1 - \frac{1}{2} = \frac{1}{2}$.
Thus,the minimum value is $\frac{1}{2}$ at $x = \frac{\pi}{4}$.

(C) Let $f(x) = \frac{\log x}{x}$.
To find the maximum value,we calculate the derivative $f'(x)$ using the quotient rule:
$f'(x) = \frac{x \cdot \frac{d}{dx}(\log x) - \log x \cdot \frac{d}{dx}(x)}{x^2} = \frac{x \cdot \frac{1}{x} - \log x}{x^2} = \frac{1 - \log x}{x^2}$.
For critical points,set $f'(x) = 0$:
$\frac{1 - \log x}{x^2} = 0 \Rightarrow 1 - \log x = 0 \Rightarrow \log x = 1 \Rightarrow x = e$.
To verify the maximum,we check the second derivative $f''(x)$ at $x = e$:
$f''(x) = \frac{x^2(-\frac{1}{x}) - (1 - \log x)(2x)}{x^4} = \frac{-x - 2x(1 - \log x)}{x^4}$.
At $x = e$,$f''(e) = \frac{-e - 2e(1 - 1)}{e^4} = \frac{-e}{e^4} = -\frac{1}{e^3} < 0$.
Since $f''(e) < 0$,the function has a local maximum at $x = e$.
The maximum value is $f(e) = \frac{\log e}{e} = \frac{1}{e}$.

(D) Given the function $f(x)=2x^3-15x^2+36x-48$.
First,we determine the set $A$ by solving the inequality $x^2-9x+20 \leq 0$.
Factoring the quadratic,we get $(x-4)(x-5) \leq 0$,which implies $x \in [4, 5]$.
Next,we find the critical points of $f(x)$ by setting $f'(x)=0$:
$f'(x)=6x^2-30x+36=6(x^2-5x+6)=6(x-2)(x-3)=0$.
The critical points are $x=2$ and $x=3$.
Since both $x=2$ and $x=3$ are outside the interval $[4, 5]$,the function $f(x)$ is monotonic on the interval $[4, 5]$.
We evaluate $f(x)$ at the endpoints of the interval $A=[4, 5]$:
For $x=4$: $f(4)=2(4)^3-15(4)^2+36(4)-48 = 128-240+144-48 = -16$.
For $x=5$: $f(5)=2(5)^3-15(5)^2+36(5)-48 = 250-375+180-48 = 7$.
Comparing the values,the maximum value of $f(x)$ on the set $A$ is $7$.

(C) Given the function $y = a \log x + b x^2 + x$.
First,find the derivative with respect to $x$: $\frac{dy}{dx} = \frac{a}{x} + 2bx + 1$.
Since the extreme values occur at $x = -1$ and $x = 2$,the derivative must be zero at these points.
For $x = -1$: $\frac{a}{-1} + 2b(-1) + 1 = 0 \Rightarrow -a - 2b + 1 = 0 \Rightarrow a = 1 - 2b$.
For $x = 2$: $\frac{a}{2} + 2b(2) + 1 = 0 \Rightarrow \frac{a}{2} + 4b + 1 = 0 \Rightarrow a + 8b + 2 = 0$.
Substitute $a = 1 - 2b$ into the second equation: $(1 - 2b) + 8b + 2 = 0 \Rightarrow 6b + 3 = 0 \Rightarrow 6b = -3 \Rightarrow b = -\frac{1}{2}$.
Now,find $a$: $a = 1 - 2(-\frac{1}{2}) = 1 + 1 = 2$.
Finally,calculate $\left(\frac{a}{b} + \frac{b}{a}\right) = \left(\frac{2}{-1/2} + \frac{-1/2}{2}\right) = (-4 - \frac{1}{4}) = -\frac{17}{4}$.

(B) Let $x$ be the length of the side of the square bottom and $h$ be the height of the tank.
Given volume $V = x^2 h = 4000 \ cm^3 \dots (i)$.
The surface area $A$ of an open tank is $A = x^2 + 4xh \dots (ii)$.
From $(i)$,$h = \frac{4000}{x^2}$.
Substituting $h$ in $(ii)$,we get $A = x^2 + 4x \left( \frac{4000}{x^2} \right) = x^2 + \frac{16000}{x}$.
Differentiating with respect to $x$,$\frac{dA}{dx} = 2x - \frac{16000}{x^2}$.
For minimum surface area,set $\frac{dA}{dx} = 0$,so $2x = \frac{16000}{x^2} \Rightarrow x^3 = 8000 \Rightarrow x = 20 \ cm$.
Checking the second derivative,$\frac{d^2A}{dx^2} = 2 + \frac{32000}{x^3}$.
At $x = 20$,$\frac{d^2A}{dx^2} = 2 + \frac{32000}{8000} = 2 + 4 = 6 > 0$,hence the area is minimum.
Calculating height,$h = \frac{4000}{20^2} = \frac{4000}{400} = 10 \ cm$.

(B) Given the function $y=a \log x+b x^2+x$.
First,find the derivative with respect to $x$: $\frac{dy}{dx} = \frac{a}{x} + 2bx + 1$.
Since the function has extremum values at $x=-1$ and $x=2$,the derivative must be zero at these points.
For $x=-1$: $\frac{a}{-1} + 2b(-1) + 1 = 0 \Rightarrow -a - 2b + 1 = 0 \Rightarrow a + 2b = 1$ (Equation $i$).
For $x=2$: $\frac{a}{2} + 2b(2) + 1 = 0 \Rightarrow \frac{a}{2} + 4b + 1 = 0 \Rightarrow a + 8b = -2$ (Equation $ii$).
Subtracting Equation $i$ from Equation $ii$: $(a + 8b) - (a + 2b) = -2 - 1 \Rightarrow 6b = -3 \Rightarrow b = -\frac{1}{2}$.
Substituting $b = -\frac{1}{2}$ into Equation $i$: $a + 2(-\frac{1}{2}) = 1 \Rightarrow a - 1 = 1 \Rightarrow a = 2$.
Thus,the values are $a=2$ and $b=-\frac{1}{2}$.

(D) Given the constraint $x+2y=8$.
We can express $y$ in terms of $x$ as $2y = 8-x$,which gives $y = \frac{8-x}{2}$.
Let the function to be maximized be $f(x) = xy$.
Substituting the value of $y$,we get $f(x) = x \cdot \frac{8-x}{2} = 4x - \frac{x^2}{2}$.
To find the maximum value,we differentiate $f(x)$ with respect to $x$:
$f'(x) = \frac{d}{dx}(4x - \frac{x^2}{2}) = 4 - x$.
Setting the derivative to zero for critical points:
$4 - x = 0 \implies x = 4$.
To confirm it is a maximum,we check the second derivative:
$f''(x) = -1$.
Since $f''(4) = -1 < 0$,the function has a local maximum at $x = 4$.
Substituting $x = 4$ back into the expression for $y$:
$y = \frac{8-4}{2} = 2$.
The maximum value of $xy$ is $4 \times 2 = 8$.

(A) First,we determine the set $S$ by solving the inequality $x^2 + 30 \leq 11x$.
$x^2 - 11x + 30 \leq 0$
$(x - 5)(x - 6) \leq 0$
Thus,$x \in [5, 6]$.
Now,consider the function $f(x) = 3x^3 - 18x^2 + 27x - 40$.
Find the derivative $f'(x)$:
$f'(x) = 9x^2 - 36x + 27 = 9(x^2 - 4x + 3) = 9(x - 1)(x - 3)$.
For $x \in [5, 6]$,both $(x - 1)$ and $(x - 3)$ are positive,so $f'(x) > 0$.
Since $f'(x) > 0$ for all $x \in [5, 6]$,the function $f(x)$ is strictly increasing on the interval $[5, 6]$.
The maximum value occurs at the right endpoint $x = 6$.
$f(6) = 3(6)^3 - 18(6)^2 + 27(6) - 40$
$f(6) = 3(216) - 18(36) + 162 - 40$
$f(6) = 648 - 648 + 162 - 40 = 122$.

(B) Let the side of the square base be $x \,m$ and the height of the tank be $y \,m$.
The volume $V = x^2 y = 500$.
Thus,$y = \frac{500}{x^2}$.
The surface area $S$ of the open tank is $S = x^2 + 4xy$.
Substituting $y$,we get $S = x^2 + 4x \left(\frac{500}{x^2}\right) = x^2 + \frac{2000}{x}$.
Differentiating with respect to $x$: $\frac{dS}{dx} = 2x - \frac{2000}{x^2}$.
Setting $\frac{dS}{dx} = 0$,we get $2x = \frac{2000}{x^2} \Rightarrow x^3 = 1000 \Rightarrow x = 10 \,m$.
The second derivative $\frac{d^2S}{dx^2} = 2 + \frac{4000}{x^3}$. At $x = 10$,$\frac{d^2S}{dx^2} = 2 + 4 = 6 > 0$,so the area is minimum.
The height $y = \frac{500}{10^2} = 5 \,m$.
Thus,the dimensions are $10 \,m, 10 \,m, 5 \,m$.

(A) Given $f(x) = \alpha \log x + \beta x^2 + x$.
Since $x=1$ and $x=2$ are extreme points,the derivative $f'(x)$ must be zero at these points.
$f'(x) = \frac{\alpha}{x} + 2\beta x + 1$.
At $x=1$,$f'(1) = \frac{\alpha}{1} + 2\beta(1) + 1 = 0 \implies \alpha + 2\beta = -1$ (Equation $1$).
At $x=2$,$f'(2) = \frac{\alpha}{2} + 2\beta(2) + 1 = 0 \implies \frac{\alpha}{2} + 4\beta = -1 \implies \alpha + 8\beta = -2$ (Equation $2$).
Subtracting Equation $1$ from Equation $2$: $(\alpha + 8\beta) - (\alpha + 2\beta) = -2 - (-1) \implies 6\beta = -1 \implies \beta = -\frac{1}{6}$.
Substituting $\beta = -\frac{1}{6}$ into Equation $1$: $\alpha + 2(-\frac{1}{6}) = -1 \implies \alpha - \frac{1}{3} = -1 \implies \alpha = -\frac{2}{3}$.
Now,calculate $\alpha^2 + 2\beta$: $(-\frac{2}{3})^2 + 2(-\frac{1}{6}) = \frac{4}{9} - \frac{1}{3} = \frac{4-3}{9} = \frac{1}{9}$.

(A) Given function: $f(x) = x^3 - 6x^2 + 9x + 2$
First,find the derivative $f'(x)$:
$f'(x) = 3x^2 - 12x + 9$
To find the critical points,set $f'(x) = 0$:
$3(x^2 - 4x + 3) = 0$
$3(x - 1)(x - 3) = 0$
So,the critical points are $x = 1$ and $x = 3$.
Now,find the second derivative $f''(x)$:
$f''(x) = 6x - 12$
Check the nature of the function at critical points using the second derivative test:
At $x = 1$: $f''(1) = 6(1) - 12 = -6$. Since $f''(1) < 0$,the function has a local maximum at $x = 1$.
At $x = 3$: $f''(3) = 6(3) - 12 = 6$. Since $f''(3) > 0$,the function has a local minimum at $x = 3$.
Therefore,the function has a maximum value when $x = 1$.

(A) Given function: $f(x) = 3x^3 - 18x^2 + 27x - 40$.
First,find the derivative: $f'(x) = 9x^2 - 36x + 27 = 9(x^2 - 4x + 3) = 9(x - 1)(x - 3)$.
The critical points are $x = 1$ and $x = 3$.
Next,determine the set $S$ by solving the inequality $x^2 + 30 \leq 11x$:
$x^2 - 11x + 30 \leq 0$
$(x - 5)(x - 6) \leq 0$
This gives $x \in [5, 6]$.
Now,evaluate the behavior of $f(x)$ on the interval $[5, 6]$. Since $f'(x) = 9(x - 1)(x - 3)$,for any $x \geq 5$,both $(x - 1)$ and $(x - 3)$ are positive,so $f'(x) > 0$.
Thus,$f(x)$ is strictly increasing on the interval $[5, 6]$.
The maximum value on the set $S = [5, 6]$ occurs at the right endpoint $x = 6$.
$f(6) = 3(6)^3 - 18(6)^2 + 27(6) - 40$
$f(6) = 3(216) - 18(36) + 162 - 40$
$f(6) = 648 - 648 + 162 - 40 = 122$.
Therefore,the maximum value is $122$.

(B) Given function: $f(x) = (x-1)(x+2)^2$.
Find the derivative $f'(x)$ using the product rule:
$f'(x) = (1)(x+2)^2 + (x-1)(2)(x+2)$
$f'(x) = (x+2)[(x+2) + 2(x-1)]$
$f'(x) = (x+2)(x+2+2x-2) = 3x(x+2)$.
Set $f'(x) = 0$ to find critical points:
$3x(x+2) = 0 \implies x = 0, x = -2$.
Use the first derivative test:
For $x < -2$,$f'(x) > 0$ (increasing).
For $-2 < x < 0$,$f'(x) < 0$ (decreasing).
For $x > 0$,$f'(x) > 0$ (increasing).
At $x = -2$,the function changes from increasing to decreasing,so $f(-2)$ is a local maximum:
$f(-2) = (-2-1)(-2+2)^2 = (-3)(0) = 0$.
At $x = 0$,the function changes from decreasing to increasing,so $f(0)$ is a local minimum:
$f(0) = (0-1)(0+2)^2 = (-1)(4) = -4$.
Thus,the local maximum and local minimum values are $0$ and $-4$ respectively.

(C) Given function: $f(x) = x \sqrt{1-x}$
To find the local maximum,we first find the derivative $f^{\prime}(x)$:
$f^{\prime}(x) = (1) \cdot \sqrt{1-x} + x \cdot \frac{1}{2\sqrt{1-x}} \cdot (-1)$
$f^{\prime}(x) = \sqrt{1-x} - \frac{x}{2\sqrt{1-x}}$
$f^{\prime}(x) = \frac{2(1-x) - x}{2\sqrt{1-x}} = \frac{2 - 3x}{2\sqrt{1-x}}$
Set $f^{\prime}(x) = 0$ to find critical points:
$\frac{2 - 3x}{2\sqrt{1-x}} = 0 \Rightarrow 2 - 3x = 0 \Rightarrow x = \frac{2}{3}$
Now,check the sign of $f^{\prime}(x)$ around $x = \frac{2}{3}$:
For $x < \frac{2}{3}$,$f^{\prime}(x) > 0$ (function is increasing).
For $x > \frac{2}{3}$,$f^{\prime}(x) < 0$ (function is decreasing).
Since the derivative changes sign from positive to negative at $x = \frac{2}{3}$,the function has a local maximum at $x = \frac{2}{3}$.

(B) Let the distance of point $M$ from $A$ be $x$. Then $AM = x$ and $MB = 10 - x$.
In right-angled triangles $\triangle DAM$ and $\triangle CBM$:
$MD^2 = AM^2 + AD^2 = x^2 + 8^2 = x^2 + 64$
$MC^2 = MB^2 + BC^2 = (10 - x)^2 + 11^2 = (10 - x)^2 + 121$
Let $f(x) = MD^2 + MC^2 = x^2 + 64 + (10 - x)^2 + 121$
$f(x) = x^2 + 64 + 100 - 20x + x^2 + 121$
$f(x) = 2x^2 - 20x + 285$
To find the minimum,differentiate $f(x)$ with respect to $x$:
$f'(x) = 4x - 20$
Set $f'(x) = 0$ for critical points:
$4x - 20 = 0 \Rightarrow x = 5$
Since $f''(x) = 4 > 0$,the function $f(x)$ is minimum at $x = 5$ meters.

(C) Let $f(x) = x^{25}(1-x)^{75}$.
To find the critical points,we differentiate $f(x)$ with respect to $x$ using the product rule:
$f'(x) = 25x^{24}(1-x)^{75} + x^{25} \cdot 75(1-x)^{74} \cdot (-1)$
$f'(x) = 25x^{24}(1-x)^{74} [(1-x) - 3x]$
$f'(x) = 25x^{24}(1-x)^{74} (1-4x)$
Setting $f'(x) = 0$,we get critical points at $x = 0$,$x = 1$,and $x = \frac{1}{4}$.
Since $f(0) = 0$ and $f(1) = 0$,and $f(x) > 0$ for $x \in (0,1)$,the function must attain its maximum at the critical point $x = \frac{1}{4}$.

(B) Let the two parts be $x$ and $y$. Given $x + y = 20$,so $y = 20 - x$.
Let the function be $f(x) = x(20 - x)^3$.
To find the maximum,we differentiate $f(x)$ with respect to $x$:
$f'(x) = (20 - x)^3 + x \cdot 3(20 - x)^2(-1)$
$f'(x) = (20 - x)^2 [20 - x - 3x] = (20 - x)^2 (20 - 4x)$.
Setting $f'(x) = 0$,we get $x = 20$ or $x = 5$.
Since $x=20$ gives $f(x)=0$ (minimum),we check $x = 5$.
For $x = 5$,$y = 20 - 5 = 15$.
The product of the two parts is $xy = 5 \times 15 = 75$.

(D) The volume of a cone is given by $V = \frac{1}{3} \pi r^2 h$.
Given the slant height $\ell = 3 \text{ cm}$,we have the relation $\ell^2 = r^2 + h^2$,which implies $r^2 = 9 - h^2$.
Substituting $r^2$ into the volume formula: $V = \frac{1}{3} \pi (9 - h^2) h = 3 \pi h - \frac{\pi}{3} h^3$.
To find the maximum volume,we differentiate $V$ with respect to $h$: $\frac{dV}{dh} = 3 \pi - \pi h^2$.
Setting $\frac{dV}{dh} = 0$,we get $3 \pi = \pi h^2$,which gives $h^2 = 3$,so $h = \sqrt{3} \text{ cm}$ (since height must be positive).
Checking the second derivative: $\frac{d^2V}{dh^2} = -2 \pi h$.
At $h = \sqrt{3}$,$\frac{d^2V}{dh^2} = -2 \sqrt{3} \pi < 0$.
Since the second derivative is negative,the volume is maximum when $h = \sqrt{3} \text{ cm}$.

(A) Given the function $f(x) = x^2 + ax + b$.
To find the critical points,we calculate the first derivative: $f'(x) = 2x + a$.
Setting $f'(x) = 0$,we get $x = -a/2$.
Since the second derivative $f''(x) = 2 > 0$,the function has a local minimum at $x = -a/2$.
According to the problem,the minimum occurs at $x = 3$.
Therefore,$-a/2 = 3$,which implies $a = -6$.
Given that the minimum value of the function is $5$ at $x = 3$,we substitute these values into the original function:
$f(3) = (3)^2 + a(3) + b = 5$.
Substituting $a = -6$:
$9 + (-6)(3) + b = 5$.
$9 - 18 + b = 5$.
$-9 + b = 5$.
$b = 14$.
Thus,the values are $a = -6$ and $b = 14$.

(A) Let the rectangle $ABCD$ be inscribed in a circle of radius $r$ centered at the origin. Let the coordinates of vertex $B$ be $(r \cos \theta, r \sin \theta)$.
Then,the length $AB = 2r \cos \theta$ and the width $BC = 2r \sin \theta$.
The area of the rectangle $A(\theta) = AB \times BC = (2r \cos \theta)(2r \sin \theta) = 2r^2 \sin 2\theta$.
To find the maximum area,we differentiate $A(\theta)$ with respect to $\theta$:
$A'(\theta) = 4r^2 \cos 2\theta$.
Setting $A'(\theta) = 0$,we get $\cos 2\theta = 0$,which implies $2\theta = \frac{\pi}{2}$,so $\theta = \frac{\pi}{4}$.
Checking the second derivative: $A''(\theta) = -8r^2 \sin 2\theta$.
At $\theta = \frac{\pi}{4}$,$A''(\frac{\pi}{4}) = -8r^2 \sin(\frac{\pi}{2}) = -8r^2 < 0$.
Since the second derivative is negative,the area is maximum at $\theta = \frac{\pi}{4}$.
Substituting $\theta = \frac{\pi}{4}$ into the area formula:
Maximum Area $= 2r^2 \sin(2 \times \frac{\pi}{4}) = 2r^2 \sin(\frac{\pi}{2}) = 2r^2(1) = 2r^2$ sq. units.

(A) Given the function $y=x^3-\alpha x^2-\beta x+5$.
Find the derivative with respect to $x$:
$\frac{dy}{dx} = 3x^2 - 2\alpha x - \beta$.
Since $x=-2$ and $x=4$ are extreme points,the derivative must be zero at these values.
For $x=-2$:
$3(-2)^2 - 2\alpha(-2) - \beta = 0 \implies 12 + 4\alpha - \beta = 0 \implies 4\alpha - \beta = -12$ (Equation $1$).
For $x=4$:
$3(4)^2 - 2\alpha(4) - \beta = 0 \implies 48 - 8\alpha - \beta = 0 \implies 8\alpha + \beta = 48$ (Equation $2$).
Adding Equation $1$ and Equation $2$:
$(4\alpha - \beta) + (8\alpha + \beta) = -12 + 48 \implies 12\alpha = 36 \implies \alpha = 3$.
Substitute $\alpha = 3$ into Equation $1$:
$4(3) - \beta = -12 \implies 12 - \beta = -12 \implies \beta = 24$.
Thus,$\alpha=3$ and $\beta=24$.

(A) Let the two parts of $10$ be $x$ and $(10-x)$.
Define the function $f(x) = 2x + (10-x)^2$.
Expanding the function: $f(x) = 2x + 100 - 20x + x^2 = x^2 - 18x + 100$.
To find the minimum,find the first derivative: $f'(x) = 2x - 18$.
Set $f'(x) = 0$ to find the critical point: $2x - 18 = 0 \implies x = 9$.
Find the second derivative: $f''(x) = 2$.
Since $f''(9) = 2 > 0$,the function has a minimum at $x = 9$.
The first part is $x = 9$ and the second part is $10 - 9 = 1$.
Therefore,the numbers are $9$ and $1$.

(A) We have $f(x)=\frac{1-x+x^2}{1+x+x^2}$.
To find the critical points,we calculate the derivative $f^{\prime}(x)$ using the quotient rule:
$f^{\prime}(x)=\frac{(1+x+x^2)(2x-1)-(1-x+x^2)(2x+1)}{(1+x+x^2)^2}$.
Expanding the numerator:
$f^{\prime}(x)=\frac{(2x-1+2x^2-x+2x^3-x^2)-(2x+1-2x^2-x+2x^3+x^2)}{(1+x+x^2)^2}$.
$f^{\prime}(x)=\frac{(2x^3+x^2+x-1)-(2x^3-x^2+x+1)}{(1+x+x^2)^2}$.
$f^{\prime}(x)=\frac{2x^2-2}{(1+x+x^2)^2} = \frac{2(x^2-1)}{(1+x+x^2)^2}$.
Setting $f^{\prime}(x)=0$,we get $x^2-1=0$,which implies $x=1$ or $x=-1$.
Evaluating the function at these points:
For $x=1$,$f(1)=\frac{1-1+1}{1+1+1} = \frac{1}{3}$.
For $x=-1$,$f(-1)=\frac{1-(-1)+(-1)^2}{1+(-1)+(-1)^2} = \frac{1+1+1}{1-1+1} = 3$.
Comparing the values,the minimum value of $f(x)$ is $\frac{1}{3}$.

(D) Given the function $f(x) = x \log x$.
To find the minimum value,we first find the derivative $f'(x)$.
Using the product rule: $f'(x) = x \cdot \frac{d}{dx}(\log x) + \log x \cdot \frac{d}{dx}(x) = x \cdot \frac{1}{x} + \log x = 1 + \log x$.
For critical points,set $f'(x) = 0$:
$1 + \log x = 0 \implies \log x = -1 \implies x = e^{-1} = \frac{1}{e}$.
Now,find the second derivative $f''(x)$:
$f''(x) = \frac{d}{dx}(1 + \log x) = \frac{1}{x}$.
Evaluate $f''(x)$ at $x = \frac{1}{e}$:
$f''(\frac{1}{e}) = \frac{1}{1/e} = e$.
Since $e > 0$,the function has a local minimum at $x = \frac{1}{e}$.
The minimum value is $f(\frac{1}{e}) = \frac{1}{e} \log(\frac{1}{e}) = \frac{1}{e} \log(e^{-1}) = \frac{1}{e} (-1) = -\frac{1}{e}$.

(C) Let $a, b, c$ be the sides of the triangle. The perimeter of the triangle is $a+b+c = 10 \text{ cm}$.
Given $a = 4 \text{ cm}$,so $b+c = 6 \text{ cm}$,which implies $c = 6-b$.
The semi-perimeter $s = \frac{10}{2} = 5 \text{ cm}$.
Using Heron's formula,the area $\Delta = \sqrt{s(s-a)(s-b)(s-c)} = \sqrt{5(5-4)(5-b)(5-c)} = \sqrt{5(1)(5-b)(5-(6-b))} = \sqrt{5(5-b)(b-1)}$.
To maximize the area,we maximize $f(b) = 5(5-b)(b-1) = 5(-b^2 + 6b - 5)$.
Differentiating with respect to $b$,$f'(b) = 5(-2b + 6)$.
Setting $f'(b) = 0$ gives $b = 3$.
Since $f''(b) = -10 < 0$,the area is maximum at $b = 3 \text{ cm}$.
Then $c = 6 - 3 = 3 \text{ cm}$.
Thus,the remaining sides are $3 \text{ cm}$ and $3 \text{ cm}$.

(B) Let $f(x) = \frac{\log x}{x}$.
To find the maximum value,we find the derivative $f'(x)$ using the quotient rule:
$f'(x) = \frac{x \cdot \frac{d}{dx}(\log x) - \log x \cdot \frac{d}{dx}(x)}{x^2} = \frac{x \cdot \frac{1}{x} - \log x \cdot 1}{x^2} = \frac{1 - \log x}{x^2}$.
Set $f'(x) = 0$ to find critical points:
$1 - \log x = 0 \Rightarrow \log x = 1 \Rightarrow x = e$.
Now,we use the second derivative test:
$f''(x) = \frac{x^2 \cdot (-\frac{1}{x}) - (1 - \log x) \cdot 2x}{x^4} = \frac{-x - 2x + 2x \log x}{x^4} = \frac{2 \log x - 3}{x^3}$.
At $x = e$,$f''(e) = \frac{2 \log e - 3}{e^3} = \frac{2(1) - 3}{e^3} = -\frac{1}{e^3} < 0$.
Since $f''(e) < 0$,the function has a local maximum at $x = e$.
The maximum value is $f(e) = \frac{\log e}{e} = \frac{1}{e}$.

(B) Let the sides of the rectangle be $x$ and $y$.
Given the perimeter is $108 \ m$,we have $2x + 2y = 108$,which simplifies to $x + y = 54$,or $y = 54 - x$.
The area of the rectangle is $A = x \times y$.
Substituting $y$,we get $A = x(54 - x) = 54x - x^2$.
To find the maximum area,we differentiate $A$ with respect to $x$: $\frac{dA}{dx} = 54 - 2x$.
Setting $\frac{dA}{dx} = 0$,we get $54 - 2x = 0$,which implies $x = 27$.
Since $\frac{d^2A}{dx^2} = -2 < 0$,the area is maximum at $x = 27$.
Then $y = 54 - 27 = 27$.
Thus,the dimensions are $27 \ m$ and $27 \ m$.

(C) Let $r$ be the radius and $\ell$ be the arc length of the circular sector as shown in the figure. The perimeter of the sector is given by $P = 2r + \ell = 20 \ m$.
Thus,$\ell = 20 - 2r$.
The area $A$ of the circular sector is given by $A = \frac{1}{2} \ell r$.
Substituting the value of $\ell$,we get $A = \frac{1}{2}(20 - 2r)r = 10r - r^2$.
To find the maximum area,we differentiate $A$ with respect to $r$:
$\frac{dA}{dr} = 10 - 2r$.
Setting $\frac{dA}{dr} = 0$,we get $10 - 2r = 0$,which implies $r = 5 \ m$.
To verify the maximum,we find the second derivative: $\frac{d^2A}{dr^2} = -2$.
Since $\frac{d^2A}{dr^2} < 0$,the area is maximum at $r = 5 \ m$.

(D) Let $ABCD$ be the rectangle inscribed in a circle of radius $r$.
Since the rectangle is inscribed in the circle,its diagonal is the diameter of the circle.
$\Rightarrow AC = BD = 2r = \text{diameter}$.
Let $x$ and $y$ be the length and breadth of the rectangle.
By Pythagoras theorem,$x^{2} + y^{2} = (2r)^{2} = 4r^{2}$.
$\Rightarrow y = \sqrt{4r^{2} - x^{2}}$.
Now,the area of the rectangle $A = xy = x\sqrt{4r^{2} - x^{2}}$.
To find the maximum area,differentiate $A$ with respect to $x$:
$\frac{dA}{dx} = \sqrt{4r^{2} - x^{2}} + x \cdot \frac{1}{2\sqrt{4r^{2} - x^{2}}} \cdot (-2x) = \sqrt{4r^{2} - x^{2}} - \frac{x^{2}}{\sqrt{4r^{2} - x^{2}}} = \frac{4r^{2} - 2x^{2}}{\sqrt{4r^{2} - x^{2}}}$.
For maximum area,set $\frac{dA}{dx} = 0$:
$4r^{2} - 2x^{2} = 0 \Rightarrow x^{2} = 2r^{2} \Rightarrow x = \sqrt{2}r$.
Substituting $x = \sqrt{2}r$ in the expression for $y$:
$y = \sqrt{4r^{2} - (\sqrt{2}r)^{2}} = \sqrt{4r^{2} - 2r^{2}} = \sqrt{2r^{2}} = \sqrt{2}r$.
Thus,the dimensions are $\sqrt{2}r$ units and $\sqrt{2}r$ units.

(B) Let the radius of the cylinder be $r$ and the height be $h$. Given $r + h = 6$, so $h = 6 - r$.
The volume $V$ of the cylinder is given by $V = \pi r^2 h$.
Substituting $h$, we get $V(r) = \pi r^2 (6 - r) = \pi (6r^2 - r^3)$.
To find the maximum volume, we differentiate $V$ with respect to $r$ and set it to zero:
$\frac{dV}{dr} = \pi (12r - 3r^2) = 0$.
$3r(4 - r) = 0$, which gives $r = 0$ (not possible) or $r = 4$.
For $r = 4$, $h = 6 - 4 = 2$.
Checking the second derivative: $\frac{d^2V}{dr^2} = \pi (12 - 6r)$. At $r = 4$, $\frac{d^2V}{dr^2} = \pi (12 - 24) = -12\pi < 0$, so the volume is maximum at $r = 4$.
The maximum volume is $V = \pi (4)^2 (2) = 32\pi \text{ m}^3$.

(C) Let $d(AP) = x$. Then $d(BP) = 12 - x$.
Define the function $f(x) = AP^{2} + BP^{2} = x^{2} + (12 - x)^{2}$.
Expanding this,we get $f(x) = x^{2} + 144 - 24x + x^{2} = 2x^{2} - 24x + 144$.
To find the minimum,we calculate the first derivative: $f'(x) = 4x - 24$.
Setting $f'(x) = 0$,we get $4x = 24$,which implies $x = 6$.
Checking the second derivative,$f''(x) = 4$. Since $f''(x) > 0$,the function $f(x)$ attains a minimum at $x = 6$.
Since $x = 6$ is exactly half of the total length $12 \text{ cm}$,$P$ is the midpoint of segment $AB$.

(B) Given $f(x) = x + \frac{1}{x}$,$x \neq 0$.
Differentiating with respect to $x$,we get $f'(x) = 1 - \frac{1}{x^2}$.
For local maxima or minima,we set $f'(x) = 0$.
$1 - \frac{1}{x^2} = 0 \Rightarrow x^2 = 1 \Rightarrow x = 1, -1$.
Now,we check the sign of $f'(x)$ around these points:
For $x < -1$,$f'(x) > 0$. For $-1 < x < 0$,$f'(x) < 0$. Since $f'(x)$ changes from positive to negative at $x = -1$,$x = -1$ is a point of local maxima.
The local maximum value is $f(-1) = -1 + \frac{1}{-1} = -2$.
For $0 < x < 1$,$f'(x) < 0$. For $x > 1$,$f'(x) > 0$. Since $f'(x)$ changes from negative to positive at $x = 1$,$x = 1$ is a point of local minima.
The local minimum value is $f(1) = 1 + \frac{1}{1} = 2$.
Thus,the local maximum value is $-2$ and the local minimum value is $2$.

(B) Given the function $f(x) = 3x^3 - 9x^2 - 27x + 15$.
To find the critical points,we calculate the first derivative: $f'(x) = 9x^2 - 18x - 27$.
Setting $f'(x) = 0$,we get $9(x^2 - 2x - 3) = 0$,which factors to $9(x - 3)(x + 1) = 0$.
Thus,the critical points are $x = 3$ and $x = -1$.
To determine the nature of these points,we find the second derivative: $f''(x) = 18x - 18$.
At $x = 3$,$f''(3) = 18(3) - 18 = 36 > 0$,so $f(x)$ has a local minimum at $x = 3$.
At $x = -1$,$f''(-1) = 18(-1) - 18 = -36 < 0$,so $f(x)$ has a local maximum at $x = -1$.
The maximum value is $f(-1) = 3(-1)^3 - 9(-1)^2 - 27(-1) + 15 = -3 - 9 + 27 + 15 = 30$.

(B) Given function is $f(x) = \frac{\log x}{x}$.
To find the maximum value,we find the derivative $f'(x)$ using the quotient rule:
$f'(x) = \frac{x \cdot \frac{d}{dx}(\log x) - \log x \cdot \frac{d}{dx}(x)}{x^2} = \frac{x \cdot \frac{1}{x} - \log x \cdot 1}{x^2} = \frac{1 - \log x}{x^2}$.
For critical points,set $f'(x) = 0$:
$\frac{1 - \log x}{x^2} = 0 \Rightarrow 1 - \log x = 0 \Rightarrow \log x = 1 \Rightarrow x = e$.
To confirm this is a maximum,we check the second derivative or the sign change of $f'(x)$. Since $f'(x) > 0$ for $x < e$ and $f'(x) < 0$ for $x > e$,$x = e$ is a point of local maximum.
The maximum value is $f(e) = \frac{\log e}{e} = \frac{1}{e}$.

(B) Let $f(x) = x^{3} - 12x^{2} + 36x + 17$.
Find the derivative: $f'(x) = 3x^{2} - 24x + 36$.
Set $f'(x) = 0$ to find critical points: $3(x^{2} - 8x + 12) = 0 \Rightarrow 3(x - 2)(x - 6) = 0$.
The critical points are $x = 2$ and $x = 6$.
Evaluate the function at the critical points and the interval endpoints $[1, 10]$:
$f(1) = (1)^{3} - 12(1)^{2} + 36(1) + 17 = 1 - 12 + 36 + 17 = 42$.
$f(2) = (2)^{3} - 12(2)^{2} + 36(2) + 17 = 8 - 48 + 72 + 17 = 49$.
$f(6) = (6)^{3} - 12(6)^{2} + 36(6) + 17 = 216 - 432 + 216 + 17 = 17$.
$f(10) = (10)^{3} - 12(10)^{2} + 36(10) + 17 = 1000 - 1200 + 360 + 17 = 177$.
Comparing these values,the maximum value is $177$.