Maxima and Minima Questions in English

(D) Given function is $f(x) = \frac{x}{8} + \frac{2}{x}$.
First, find the derivative $f'(x) = \frac{1}{8} - \frac{2}{x^2} = \frac{x^2 - 16}{8x^2}$.
To find critical points, set $f'(x) = 0$, which gives $x^2 - 16 = 0$, so $x = 4$ or $x = -4$.
Since the interval is $[1, 6]$, we only consider $x = 4$.
Now, evaluate $f(x)$ at the critical point and the endpoints of the interval $[1, 6]$:
$f(1) = \frac{1}{8} + \frac{2}{1} = \frac{1}{8} + 2 = \frac{17}{8} = 2.125$.
$f(4) = \frac{4}{8} + \frac{2}{4} = \frac{1}{2} + \frac{1}{2} = 1$.
$f(6) = \frac{6}{8} + \frac{2}{6} = \frac{3}{4} + \frac{1}{3} = \frac{9+4}{12} = \frac{13}{12} \approx 1.083$.
Comparing the values $f(1) = \frac{17}{8}$, $f(4) = 1$, and $f(6) = \frac{13}{12}$, the maximum value is $\frac{17}{8}$.

(A) Given the function $f(x) = x^3 e^{-3x}$ for $x > 0$.
To find the maximum value, we first find the derivative $f'(x)$ using the product rule:
$f'(x) = \frac{d}{dx}(x^3) \cdot e^{-3x} + x^3 \cdot \frac{d}{dx}(e^{-3x})$
$f'(x) = 3x^2 e^{-3x} + x^3 (-3 e^{-3x})$
$f'(x) = 3x^2 e^{-3x} (1 - x)$
Setting $f'(x) = 0$ for critical points:
$3x^2 e^{-3x} (1 - x) = 0$
Since $x > 0$ and $e^{-3x} \neq 0$, we have $1 - x = 0$, which gives $x = 1$.
To confirm it is a maximum, we check the sign of $f'(x)$ around $x = 1$. For $x < 1$, $f'(x) > 0$ and for $x > 1$, $f'(x) < 0$.
Thus, $f(x)$ has a local maximum at $x = 1$.
The maximum value is $f(1) = (1)^3 e^{-3(1)} = e^{-3}$.

(A) The displacement is given by $x = t^4 - k t^3$.
The velocity $v$ is the derivative of displacement with respect to time $t$:
$v = \frac{dx}{dt} = 4t^3 - 3kt^2$.
To find the condition for minimum velocity, we find the acceleration $a = \frac{dv}{dt}$:
$a = \frac{dv}{dt} = 12t^2 - 6kt$.
For the velocity to be minimum at $t = 2$, the derivative of velocity with respect to time must be zero at $t = 2$:
$\frac{dv}{dt} \big|_{t=2} = 12(2)^2 - 6k(2) = 0$.
$12(4) - 12k = 0$.
$48 - 12k = 0$.
$12k = 48$.
$k = 4$.
Thus, the value of $k$ is $4$.

(B) To find the point where the slope is maximum, we need to maximize $f'(x)$. Let $g(x) = f'(x) = e^x(\sin x + \cos x)$.
For $g(x)$ to have a maximum, we set $g'(x) = f''(x) = 0$.
$f'(x) = e^x(\sin x + \cos x)$
$f''(x) = e^x(\sin x + \cos x) + e^x(\cos x - \sin x) = 2e^x \cos x$.
Setting $f''(x) = 0$ gives $2e^x \cos x = 0$. Since $e^x \neq 0$, we have $\cos x = 0$.
In the interval $[0, 2\pi]$, $\cos x = 0$ at $x = \frac{\pi}{2}$ and $x = \frac{3\pi}{2}$.
We check the second derivative of $g(x)$, which is $g'(x) = f''(x) = 2e^x \cos x$.
$g''(x) = f'''(x) = 2e^x \cos x - 2e^x \sin x = 2e^x(\cos x - \sin x)$.
At $x = \frac{\pi}{2}$, $g''(\frac{\pi}{2}) = 2e^{\pi/2}(0 - 1) = -2e^{\pi/2} < 0$ (Local maximum).
At $x = \frac{3\pi}{2}$, $g''(\frac{3\pi}{2}) = 2e^{3\pi/2}(0 - (-1)) = 2e^{3\pi/2} > 0$ (Local minimum).
Thus, the slope is maximum at $x = \frac{\pi}{2}$.

(B) Given, $f(x) = x^{4} - 4x^{3} + 4x^{2} + c$.
Evaluating at the boundaries of the interval $(1, 2)$:
$f(1) = 1^{4} - 4(1)^{3} + 4(1)^{2} + c = 1 - 4 + 4 + c = 1 + c$.
$f(2) = 2^{4} - 4(2)^{3} + 4(2)^{2} + c = 16 - 32 + 16 + c = c$.
According to the Intermediate Value Theorem, if $f(1) \cdot f(2) < 0$, then there exists at least one root in the interval $(1, 2)$.
$f(1) \cdot f(2) = (1 + c)c$.
For $f(1) \cdot f(2) < 0$, we require $c(c + 1) < 0$, which implies $c \in (-1, 0)$.
Since $f'(x) = 4x^{3} - 12x^{2} + 8x = 4x(x^{2} - 3x + 2) = 4x(x - 1)(x - 2)$, we observe that $f'(x) = 0$ at $x = 0, 1, 2$.
In the interval $(1, 2)$, $f'(x) < 0$, meaning the function is strictly decreasing.
Since the function is strictly decreasing on $(1, 2)$ and changes sign, it must have exactly one zero in $(1, 2)$ for $c \in (-1, 0)$.

(A) Given $f(x)=1-2x+\int_{0}^{x}e^{(x-t)}f(t)dt$.
Multiplying by $e^{-x}$, we get $e^{-x}f(x) = (1-2x)e^{-x} + \int_{0}^{x}e^{-t}f(t)dt$.
Differentiating with respect to $x$ using Leibniz rule:
$e^{-x}f'(x) - e^{-x}f(x) = -2e^{-x} - (1-2x)e^{-x} + e^{-x}f(x)$.
$f'(x) - f(x) = -2 - 1 + 2x + f(x) \Rightarrow f'(x) - 2f(x) = 2x - 3$.
This is a linear differential equation with integrating factor $I.F. = e^{\int -2 dx} = e^{-2x}$.
$f(x)e^{-2x} = \int (2x-3)e^{-2x} dx = (2x-3)\frac{e^{-2x}}{-2} - \int 2 \cdot \frac{e^{-2x}}{-2} dx = -\frac{2x-3}{2}e^{-2x} - \frac{1}{2}e^{-2x} + C$.
$f(x) = -x + \frac{3}{2} - \frac{1}{2} + Ce^{2x} = 1-x + Ce^{2x}$.
Since $f(0) = 1-2(0) + 0 = 1$, we have $1 = 1-0 + C \Rightarrow C=0$.
So, $f(x) = 1-x$.
Now, $g(x) = \int_{0}^{x} (1-t+2)^{15}(t-4)^6(t+12)^{17} dt = \int_{0}^{x} (3-t)^{15}(t-4)^6(t+12)^{17} dt$.
$g'(x) = (3-x)^{15}(x-4)^6(x+12)^{17} = -(x-3)^{15}(x-4)^6(x+12)^{17}$.
Analyzing the sign of $g'(x)$ around critical points $-12, 3, 4$:
For $x < -12$, $g'(x) < 0$.
For $-12 < x < 3$, $g'(x) > 0$.
For $3 < x < 4$, $g'(x) < 0$.
For $x > 4$, $g'(x) < 0$.
Local minimum at $p = -12$ and local maximum at $q = 3$.
Thus, $|p+q| = |-12+3| = |-9| = 9$.

(C) Given $f(t) = \frac{|t+1|}{t^2}$ for $t < 0$.
For $t \in (-1, 0)$, $f(t) = \frac{t+1}{t^2} = \frac{1}{t} + \frac{1}{t^2}$. Then $f'(t) = -\frac{1}{t^2} - \frac{2}{t^3} = -\frac{t+2}{t^3}$. Since $t < 0$, $t^3 < 0$, so $f'(t) > 0$ for $t \in (-1, 0)$.
For $t \in (-\infty, -1)$, $f(t) = \frac{-(t+1)}{t^2} = -\frac{1}{t} - \frac{1}{t^2}$. Then $f'(t) = \frac{1}{t^2} + \frac{2}{t^3} = \frac{t+2}{t^3}$.
For $f(t)$ to be strictly decreasing, $f'(t) < 0$. Since $t^3 < 0$, we need $t+2 > 0$, i.e.,$t > -2$. Thus, $f(t)$ is strictly decreasing on $(-2, -1)$.
Comparing $(-2, -1)$ with $(2\alpha, \alpha)$, we get $\alpha = -1$.
Now, $g(x) = 2\log_e(x-2) - x^2 + 4x + 1$ for $x > 2$.
$g'(x) = \frac{2}{x-2} - 2x + 4 = \frac{2 - 2x(x-2) + 4(x-2)}{x-2} = \frac{2 - 2x^2 + 4x + 4x - 8}{x-2} = \frac{-2x^2 + 8x - 6}{x-2} = \frac{-2(x^2 - 4x + 3)}{x-2} = \frac{-2(x-3)(x-1)}{x-2}$.
For $x > 2$, $g'(x) = 0$ at $x = 3$.
For $x \in (2, 3)$, $g'(x) > 0$ and for $x > 3$, $g'(x) < 0$. Thus, $g(x)$ has a local maximum at $x = 3$.
The local maximum value is $g(3) = 2\log_e(3-2) - 3^2 + 4(3) + 1 = 2\log_e(1) - 9 + 12 + 1 = 0 - 9 + 13 = 4$.

(A) Given $f(x) = x^{2025} - x^{2000}$.
To find the minimum value, we find the derivative $f'(x) = 2025x^{2024} - 2000x^{1999}$.
Setting $f'(x) = 0$, we get $x^{1999}(2025x^{25} - 2000) = 0$.
Since $x \in [0, 1]$, the critical point is $x = (\frac{2000}{2025})^{1/25} = (\frac{80}{81})^{1/25} = \alpha$.
Evaluating $f(\alpha) = (\frac{80}{81})^{2025/25} - (\frac{80}{81})^{2000/25} = (\frac{80}{81})^{81} - (\frac{80}{81})^{80}$.
$f(\alpha) = (\frac{80}{81})^{80} (\frac{80}{81} - 1) = (\frac{80}{81})^{80} (-\frac{1}{81}) = 80^{80} \cdot 81^{-80} \cdot (-81)^{-1} = 80^{80} \cdot (-81)^{-81}$.
Comparing this with $(80)^{80}(n)^{-81}$, we get $n = -81$.

(A) To find the absolute minimum value of the function $f(x) = x^3 - 18x^2 + 96x$ on the interval $[0, 9]$, we first find the derivative $f'(x)$.
$f'(x) = 3x^2 - 36x + 96$.
Setting $f'(x) = 0$ to find critical points:
$3(x^2 - 12x + 32) = 0$
$3(x - 4)(x - 8) = 0$
So, the critical points are $x = 4$ and $x = 8$.
Now, we evaluate the function $f(x)$ at the critical points and the endpoints of the interval $[0, 9]$:
$f(0) = 0^3 - 18(0)^2 + 96(0) = 0$
$f(4) = 4^3 - 18(4)^2 + 96(4) = 64 - 288 + 384 = 160$
$f(8) = 8^3 - 18(8)^2 + 96(8) = 512 - 1152 + 768 = 128$
$f(9) = 9^3 - 18(9)^2 + 96(9) = 729 - 1458 + 864 = 135$
Comparing these values $(0, 160, 128, 135)$, the absolute minimum value is $0$.

(C) We know that the absolute value function $|x+1|$ is always non-negative, meaning $|x+1| \ge 0$ for all $x \in R$.
Multiplying by $-1$ reverses the inequality: $-|x+1| \le 0$.
Adding $3$ to both sides gives: $-|x+1| + 3 \le 0 + 3$, which simplifies to $f(x) \le 3$.
The maximum value is attained when the term $-|x+1|$ is equal to $0$, which happens at $x = -1$.
Therefore, the maximum value of the function is $3$.

(C) The height function is $h(t) = -5t^2 + 14t + 3$.
To find the maximum height, we find the derivative with respect to $t$ and set it to zero: $\frac{dh}{dt} = -10t + 14$.
Setting $\frac{dh}{dt} = 0$, we get $-10t + 14 = 0$, which implies $t = 1.4 \text{ s}$.
Substitute $t = 1.4$ into the height equation: $h(1.4) = 3 + 14(1.4) - 5(1.4)^2$.
$h(1.4) = 3 + 19.6 - 5(1.96) = 22.6 - 9.8 = 12.8 \text{ m}$.
Thus, the maximum height is $12.8 \text{ m}$.

(C) Let $f(x) = (1/x)^x = x^{-x}$.
Taking the natural logarithm on both sides: $\ln(f(x)) = -x \ln(x)$.
To find the critical points, differentiate with respect to $x$: $\frac{d}{dx}(\ln(f(x))) = -[\ln(x) + x \cdot \frac{1}{x}] = -(\ln(x) + 1)$.
Set the derivative to $0$: $-(\ln(x) + 1) = 0 \implies \ln(x) = -1 \implies x = e^{-1} = 1/e$.
Check the second derivative: $\frac{d^2}{dx^2}(\ln(f(x))) = -\frac{1}{x}$. At $x = 1/e$, this is $-e < 0$, so it is a local maximum.
The maximum value is $f(1/e) = (1/(1/e))^{1/e} = e^{1/e}$.

(D) Let $f(x) = \frac{\log x}{x}$.
Find the derivative: $f'(x) = \frac{x(\frac{1}{x}) - \log x(1)}{x^2} = \frac{1 - \log x}{x^2}$.
Set $f'(x) = 0$ to find critical points: $1 - \log x = 0 \implies \log x = 1 \implies x = e$.
Since $e \approx 2.718$, $x = e$ lies in the interval $(2, \infty)$.
Check the sign of $f'(x)$ around $x = e$: for $x < e$, $f'(x) > 0$ (increasing); for $x > e$, $f'(x) < 0$ (decreasing).
Thus, $x = e$ is a point of local maximum.
As $x \to 2^+$, $f(x) \to \frac{\log 2}{2} \approx 0.346$.
As $x \to \infty$, $f(x) \to 0$.
Since the function increases until $x = e$ and then decreases towards $0$ as $x \to \infty$, the infimum of the function on $(2, \infty)$ is $0$, but it is never attained. Therefore, the minimum value does not exist.

(A) Let the radius of the base be $r$ and the height be $h$. The volume $V = \pi r^2 h = 125\pi$, so $h = \frac{125}{r^2}$.
The surface area $S$ of a cylinder without a top lid is $S = \pi r^2 + 2\pi rh$.
Substitute $h$: $S = \pi r^2 + 2\pi r \left( \frac{125}{r^2} \right) = \pi r^2 + \frac{250\pi}{r}$.
To find the minimum, differentiate with respect to $r$: $\frac{dS}{dr} = 2\pi r - \frac{250\pi}{r^2}$.
Set $\frac{dS}{dr} = 0$: $2\pi r = \frac{250\pi}{r^2} \implies r^3 = 125 \implies r = 5 \text{ cm}$.
Substitute $r = 5$ into $S$: $S = \pi(5)^2 + \frac{250\pi}{5} = 25\pi + 50\pi = 75\pi \text{ cm}^2$.

(D) Let the two parts be $x$ and $28-x$, where $x > 0$ and $28-x > 0$.
Let $f(x) = x^3 + (28-x)^2$.
To find the minimum, differentiate $f(x)$ with respect to $x$: $f'(x) = 3x^2 - 2(28-x) = 3x^2 + 2x - 56$.
Set $f'(x) = 0$: $3x^2 + 2x - 56 = 0$.
Using the quadratic formula $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$: $x = \frac{-2 \pm \sqrt{4 - 4(3)(-56)}}{6} = \frac{-2 \pm \sqrt{4 + 672}}{6} = \frac{-2 \pm \sqrt{676}}{6} = \frac{-2 \pm 26}{6}$.
Since $x > 0$, $x = \frac{24}{6} = 4$.
The two parts are $4$ and $28-4 = 24$.
The absolute difference is $|24 - 4| = 20$.

(B) Let a point on the curve $4y = x^2$ be $(x, y)$, where $y = \frac{x^2}{4}$.
The distance $D$ between $(x, \frac{x^2}{4})$ and $(0, 5)$ is given by $D^2 = (x - 0)^2 + (\frac{x^2}{4} - 5)^2$.
Let $f(x) = D^2 = x^2 + \frac{x^4}{16} - \frac{5x^2}{2} + 25 = \frac{x^4}{16} - \frac{3x^2}{2} + 25$.
To find the minimum distance, differentiate $f(x)$ with respect to $x$ and set it to $0$:
$f'(x) = \frac{4x^3}{16} - 3x = \frac{x^3}{4} - 3x = 0$.
$x(\frac{x^2}{4} - 3) = 0$, which gives $x = 0$ or $x^2 = 12 \implies x = \pm 2\sqrt{3}$.
For $x = \pm 2\sqrt{3}$, $y = \frac{(\pm 2\sqrt{3})^2}{4} = \frac{12}{4} = 3$.
Thus, the points are $(2\sqrt{3}, 3)$ and $(-2\sqrt{3}, 3)$.

(C) Let the length be $x \text{ m}$ and width be $y \text{ m}$.
Volume $V = x \cdot y \cdot 4 = 36 \implies xy = 9 \implies y = \frac{9}{x}$.
Cost $C = 100(xy) + 50(2 \cdot 4x + 2 \cdot 4y) = 100(9) + 400(x + y) = 900 + 400(x + \frac{9}{x})$.
To minimize $C$, find $\frac{dC}{dx} = 400(1 - \frac{9}{x^2}) = 0 \implies x^2 = 9 \implies x = 3$.
Since $x=3$, $y = \frac{9}{3} = 3$.
Minimum cost $C = 900 + 400(3 + 3) = 900 + 400(6) = 900 + 2400 = 3300$.

(B) Step $1$: Find the derivative $f'(x)$ using the product rule.
$f'(x) = 25x^{24}(1 - x)^{75} + x^{25} \cdot 75(1 - x)^{74}(-1)$
Step $2$: Simplify $f'(x)$.
$f'(x) = 25x^{24}(1 - x)^{74} [(1 - x) - 3x] = 25x^{24}(1 - x)^{74} (1 - 4x)$
Step $3$: Set $f'(x) = 0$ to find critical points.
$25x^{24}(1 - x)^{74} (1 - 4x) = 0$
The critical points are $x = 0$, $x = 1$, and $x = 1/4$.
Step $4$: Evaluate $f(x)$ at critical points and endpoints.
$f(0) = 0$, $f(1) = 0$, $f(1/4) = (1/4)^{25}(3/4)^{75} > 0$.
Since $f(x) \ge 0$ on $[0, 1]$ and the value at $x = 1/4$ is positive, the maximum occurs at $x = 1/4$.

(C) Let the length and breadth of the rectangle be $l$ and $b$ respectively.
The perimeter of the rectangle is $2(l + b) = 64 \text{ m}$, so $l + b = 32 \text{ m}$, which implies $b = 32 - l$.
The area $A$ is given by $A = l \times b = l(32 - l) = 32l - l^2$.
To find the maximum area, we differentiate $A$ with respect to $l$: $\frac{dA}{dl} = 32 - 2l$.
Setting $\frac{dA}{dl} = 0$, we get $32 - 2l = 0$, so $l = 16 \text{ m}$.
Then $b = 32 - 16 = 16 \text{ m}$.
The maximum area is $A = 16 \times 16 = 256 \text{ m}^2$.

(D) Step $1$: Find the derivative $f'(x) = 6x^2 - 30x + 36$.
Step $2$: Set $f'(x) = 0$ to find critical points: $6(x^2 - 5x + 6) = 0 \implies 6(x-2)(x-3) = 0$. Thus, $x = 2$ and $x = 3$.
Step $3$: Calculate the local maximum value at $x = 2$: $f(2) = 2(8) - 15(4) + 36(2) + 40 = 16 - 60 + 72 + 40 = 68$.
Step $4$: Calculate the local minimum value at $x = 3$: $f(3) = 2(27) - 15(9) + 36(3) + 40 = 54 - 135 + 108 + 40 = 67$.
Step $5$: The difference between the local extreme values is $|68 - 67| = 1$.

(A) Let the length of the piece for the square be $x$ and the length of the piece for the circle be $40 - x$.
Side of the square $s = \frac{x}{4}$. Area of square $A_s = (\frac{x}{4})^2 = \frac{x^2}{16}$.
Circumference of the circle $2\pi r = 40 - x$, so $r = \frac{40 - x}{2\pi}$. Area of circle $A_c = \pi r^2 = \pi (\frac{40 - x}{2\pi})^2 = \frac{(40 - x)^2}{4\pi}$.
Total area $A(x) = \frac{x^2}{16} + \frac{(40 - x)^2}{4\pi}$.
To minimize, find $A'(x) = \frac{2x}{16} + \frac{2(40 - x)(-1)}{4\pi} = \frac{x}{8} - \frac{40 - x}{2\pi} = 0$.
$\frac{x}{8} = \frac{40 - x}{2\pi} \implies \pi x = 160 - 4x \implies x(\pi + 4) = 160$.
$x = \frac{160}{\pi + 4}$.
The other piece is $40 - \frac{160}{\pi + 4} = \frac{40\pi + 160 - 160}{\pi + 4} = \frac{40\pi}{\pi + 4}$.

(A) Step $1$: Find the first derivative of the function $f(t) = -2t^3 + 3t^2 + 5$ with respect to $t$.
$f'(t) = \frac{d}{dt}(-2t^3 + 3t^2 + 5) = -6t^2 + 6t$.
Step $2$: Set the first derivative to zero to find the critical points.
$-6t^2 + 6t = 0 \Rightarrow -6t(t - 1) = 0$.
This gives $t = 0$ and $t = 1$.
Step $3$: Find the second derivative to determine the nature of the critical points.
$f''(t) = \frac{d}{dt}(-6t^2 + 6t) = -12t + 6$.
Step $4$: Evaluate the second derivative at the critical points.
For $t = 1$, $f''(1) = -12(1) + 6 = -6$.
Since $f''(1) < 0$, the function has a local maximum at $t = 1$ hour.