Step 1: Define Variables and Establish Constraints

Let the two numbers be $x$ and $y$. Based on the problem statement, we have the sum constraint:

$x + y = 24 \implies y = 24 - x$ [Equation 1]

We aim to maximize the product $P$, which is defined as:

$P = x \cdot y$

Step 2: Express the Function in Terms of a Single Variable

Substitute Equation 1 into the product function to express $P$ solely as a function of $x$:

$P(x) = x(24 - x)$

$P(x) = 24x - x^2$

Step 3: First Derivative Analysis (Finding Critical Points)

To find the value of $x$ that maximizes $P(x)$, we calculate the first derivative with respect to $x$ [By the Power Rule for differentiation]:

$\frac{dP}{dx} = \frac{d}{dx}(24x - x^2)$

$\frac{dP}{dx} = 24 - 2x$

To locate the critical points, set the derivative to zero [Per Fermat's Theorem on Stationary Points]:

$24 - 2x = 0$

$2x = 24 \implies x = 12$

Step 4: Second Derivative Test (Verification of Maximum)

To confirm that $x = 12$ results in a maximum product, calculate the second derivative [By the Second Derivative Test]:

$\frac{d^2P}{dx^2} = \frac{d}{dx}(24 - 2x) = -2$

Since $\frac{d^2P}{dx^2} = -2 < 0$, the function $P(x)$ has a local maximum at $x = 12$.

Step 5: Determine the Second Number

Substitute $x = 12$ back into the constraint equation:

$y = 24 - 12 = 12$

Step 6: Comparative Analysis

Final Solution: The two numbers are 12 and 12, yielding the maximum product of 144.

Step 1: Establishing the Function and Derivative

Given the function $f(x) = x^2 - x + 1$. To determine the intervals of increase and decrease, we must calculate the first derivative with respect to $x$ [By the Power Rule of Differentiation, $\frac{d}{dx}(x^n) = nx^{n-1}$]:

$f'(x) = \frac{d}{dx}(x^2) - \frac{d}{dx}(x) + \frac{d}{dx}(1)$

$f'(x) = 2x - 1$

Step 2: Finding Critical Points

A function changes its monotonicity behavior at critical points where $f'(x) = 0$.

$2x - 1 = 0$

$2x = 1 \implies x = \frac{1}{2}$

Since $\frac{1}{2}$ lies within the interval $(-1, 1)$, we must evaluate the behavior of $f'(x)$ on either side of this critical point within the given domain.

Step 3: Interval Analysis

We partition the interval $(-1, 1)$ into two sub-intervals based on the critical point $x = \frac{1}{2}$:

Step 4: Logical Deduction

Per the Monotonicity Theorem, a function $f$ is strictly increasing on an interval if $f'(x) > 0$ for all $x$ in that interval, and strictly decreasing if $f'(x) < 0$ for all $x$ in that interval.

As demonstrated in Step 3:

• For $x \in (-1, 1/2)$, $f'(x) < 0$, therefore $f$ is strictly decreasing.
• For $x \in (1/2, 1)$, $f'(x) > 0$, therefore $f$ is strictly increasing.

Since the derivative changes sign within the interval $(-1, 1)$, the function fails the criteria for being strictly monotonic throughout the entire domain.

Final Solution: Because the derivative $f'(x) = 2x - 1$ takes both negative and positive values in the interval $(-1, 1)$, the function is neither strictly increasing nor strictly decreasing on the given interval.

Step 1: Identifying the Mathematical Relationship

The area $A$ of a circle as a function of its radius $r$ is given by the geometric formula: $$A = \pi r^2$$ [Per the definition of the area of a circle in Euclidean geometry]

Step 2: Differentiating with respect to $r$

To find the rate of change of the area with respect to the radius, we calculate the derivative $\frac{dA}{dr}$. Using the Power Rule for differentiation, $\frac{d}{dx}(x^n) = nx^{n-1}$:

Step 3: Evaluation at the specified radius

We are required to evaluate the rate of change at $r = 6$ cm. Substituting the value into the derived expression:

Visual Representation of the Rate of Change

Unit Analysis

Since the area $A$ is measured in $\text{cm}^2$ and the radius $r$ is measured in $\text{cm}$, the units of the rate of change $\frac{dA}{dr}$ are:

Thus, the final value is $12\pi \text{ cm}^2/\text{cm}$, which simplifies to $12\pi \text{ cm}$.

Final Solution: The rate of change of the area of the circle with respect to its radius at $r = 6$ cm is $12\pi \text{ cm}$.

Step 1: Establishing the Function and Derivative

Given the function: $f(x) = x^2$

To find the critical points, we must first determine the first derivative with respect to $x$ [Per the Power Rule of Differentiation, $\frac{d}{dx}(x^n) = nx^{n-1}$]:

$f'(x) = \frac{d}{dx}(x^2) = 2x$

Step 2: Identifying Critical Points

A critical point occurs where the first derivative is equal to zero or undefined. Since $f'(x) = 2x$ is a polynomial, it is defined for all $x \in \mathbb{R}$. We set $f'(x) = 0$:

$2x = 0 \implies x = 0$

The point $x = 0$ is the unique critical point of the function.

Step 3: Second Derivative Test

To classify the critical point as a local maximum or local minimum, we evaluate the second derivative $f''(x)$ [Per the Second Derivative Test for Local Extrema]:

$f''(x) = \frac{d}{dx}(2x) = 2$

Since $f''(0) = 2 > 0$, the function exhibits a local minimum at $x = 0$ because the curve is concave upward at the critical point.

Step 4: Calculating Local Minimum Value

We substitute the critical point back into the original function $f(x)$:

$f(0) = (0)^2 = 0$

Visual Representation

Summary of Analysis

Final Solution: The function $f(x) = x^2$ has a local minimum at $x = 0$ with a local minimum value of $0$. There is no local maximum.

Step 1: Identifying the Mathematical Relationship

The area $A$ of a circle as a function of its radius $r$ is given by the formula: $A = \pi r^2$ [Per the geometric definition of the area of a circle].

Step 2: Defining the Rate of Change

We are required to find the rate of change of the area $A$ with respect to the radius $r$. Mathematically, this is expressed as the derivative $\frac{dA}{dr}$. To find this, we differentiate $A$ with respect to $r$: $\frac{dA}{dr} = \frac{d}{dr}(\pi r^2)$

Step 3: Executing the Differentiation

Applying the Power Rule for differentiation, $\frac{d}{dx}(x^n) = nx^{n-1}$ [Fundamental Theorem of Calculus/Differentiation Rules]: $\frac{dA}{dr} = \pi \cdot (2r^{2-1})$ $\frac{dA}{dr} = 2\pi r$

Step 4: Substitution of Given Conditions

We must evaluate the rate of change when $r = 3$ cm. Substituting $r = 3$ into the expression derived in Step 3: $\left( \frac{dA}{dr} \right)_{r=3} = 2\pi(3)$ $\left( \frac{dA}{dr} \right)_{r=3} = 6\pi$

Visual Representation of the Rate Function

Unit Analysis

Since the area is measured in $\text{cm}^2$ and the radius is measured in $\text{cm}$, the rate of change $\frac{dA}{dr}$ is measured in units of: $\frac{[\text{Area}]}{[\text{Radius}]} = \frac{\text{cm}^2}{\text{cm}} = \text{cm}$

Final Solution: The rate of change of the area of the circle with respect to its radius when $r = 3$ cm is $6\pi \text{ cm}^2/\text{cm}$, or simply $6\pi \text{ cm}$.

Step 1: Definition of the Logarithmic Function

Let $f(x)$ be the natural logarithmic function defined on the domain $(0, \infty)$. The function is represented as:

$f(x) = \ln(x)$

Step 2: Differentiation of the Function

To determine the interval of increase, we must calculate the first derivative of the function with respect to $x$.

$f'(x) = \frac{d}{dx}[\ln(x)]$

[By the standard derivative rule for natural logarithms: $\frac{d}{dx}(\ln x) = \frac{1}{x}$]

$f'(x) = \frac{1}{x}$

Step 3: Analyzing the Sign of the Derivative

According to the First Derivative Test for monotonic functions, a function $f$ is strictly increasing on an interval if $f'(x) > 0$ for all $x$ within that interval.

• We are given the domain $x \in (0, \infty)$.
• For all $x > 0$, the value of $\frac{1}{x}$ is always positive.
• Mathematically, if $x > 0$, then $\frac{1}{x} > 0$.

[By the properties of real numbers and division, the reciprocal of a positive number is strictly positive.]

Step 4: Geometric Visualization

Step 5: Conclusion based on Derivative Test

Since $f'(x) = \frac{1}{x} > 0$ for every $x \in (0, \infty)$, the slope of the tangent to the curve is always positive. Therefore, the function $f(x) = \ln(x)$ is strictly increasing on its entire domain $(0, \infty)$.

Final Solution: Since $f'(x) = \frac{1}{x} > 0$ for all $x \in (0, \infty)$, the logarithmic function is strictly increasing on the interval $(0, \infty)$.

Step 1: Defining the Function and Domain

Let the given function be $f(x) = x + \sin 2x$. The domain of the function is defined on the closed interval $[0, 2\pi]$.

Step 2: Finding the Critical Points

To find the critical points, we must calculate the first derivative $f'(x)$ and set it to zero [By Fermat's Theorem for local extrema].

$f'(x) = \frac{d}{dx}(x + \sin 2x)$

$f'(x) = 1 + 2\cos 2x$

Setting $f'(x) = 0$:

$1 + 2\cos 2x = 0 \implies \cos 2x = -\frac{1}{2}$

Given $x \in [0, 2\pi]$, we have $2x \in [0, 4\pi]$.

The values of $2x$ for which $\cos 2x = -\frac{1}{2}$ are:

• $2x = \frac{2\pi}{3}, \frac{4\pi}{3}, \frac{8\pi}{3}, \frac{10\pi}{3}$

Dividing by $2$, the critical points in the interval $(0, 2\pi)$ are:

$x = \frac{\pi}{3}, \frac{2\pi}{3}, \frac{4\pi}{3}, \frac{5\pi}{3}$

Step 3: Evaluating the Function at Critical and Boundary Points

Per the Extreme Value Theorem, the absolute maximum and minimum of a continuous function on a closed interval must occur at either critical points or boundaries.

Step 4: Graphical Representation of the Function

Step 5: Final Conclusion of Extrema

Comparing the values calculated in the table:

• Absolute Minimum Value: $f(0) = 0$
• Absolute Maximum Value: $f(2\pi) = 2\pi$

Final Solution: The absolute maximum value is $2\pi$ at $x = 2\pi$, and the absolute minimum value is $0$ at $x = 0$.

Step 1: Defining the Function and Domain

The given function is $f(x) = -(x - 1)^2 + 10$. Since no specific interval $[a, b]$ is provided for the domain, we assume the domain $D$ to be the set of all real numbers, $x \in \mathbb{R}$.

Step 2: Analytical Derivation of Extrema

We analyze the behavior of the function by examining its components. The term $(x - 1)^2$ is a squared real expression. By the property of real numbers, the square of any real number is non-negative [Axiom of Real Numbers]:

$(x - 1)^2 \geq 0$ for all $x \in \mathbb{R}$

Multiplying by $-1$ reverses the inequality sign [Property of Inequalities]:

$-(x - 1)^2 \leq 0$

Adding $10$ to both sides maintains the inequality:

$-(x - 1)^2 + 10 \leq 0 + 10$

$f(x) \leq 10$

Step 3: Identification of Maximum Value

The inequality $f(x) \leq 10$ indicates that the function never exceeds $10$. The value $10$ is achieved when the term $-(x - 1)^2$ equals $0$, which occurs at $x = 1$.

Therefore, $f(1) = -(1 - 1)^2 + 10 = 10$.

By definition, since $f(x) \leq f(1)$ for all $x \in \mathbb{R}$, $f(x)$ has a global maximum of $10$ at $x = 1$.

Step 4: Examination of Minimum Value

As $x \to \infty$, $-(x - 1)^2 \to -\infty$. Consequently, $f(x) \to -\infty$. As $x \to -\infty$, $-(x - 1)^2 \to -\infty$. Consequently, $f(x) \to -\infty$.

Since the function decreases without bound as $x$ increases or decreases, there exists no real number that serves as a global minimum. Thus, the function has no minimum value.

Step 5: Graphical Representation

Final Solution: The function has a maximum value of $10$ at $x = 1$, and it has no minimum value.

Step 1: Establishing the Function and Derivative

Given the function $f(x) = 2x^2 - 3x$, we proceed by finding the first derivative with respect to $x$, which represents the rate of change of the function [Per the Power Rule of Differentiation, $\frac{d}{dx}(x^n) = nx^{n-1}$].

$f'(x) = \frac{d}{dx}(2x^2 - 3x) = 4x - 3$

Step 2: Determining Critical Points

To find the intervals of increase or decrease, we identify the critical points by setting the first derivative equal to zero [By Fermat’s Theorem on stationary points].

$4x - 3 = 0$

$x = \frac{3}{4}$

Step 3: Interval Analysis

The critical point $x = \frac{3}{4}$ divides the real number line into two distinct intervals: $(-\infty, \frac{3}{4})$ and $(\frac{3}{4}, \infty)$. We analyze the sign of $f'(x)$ in each interval to determine monotonicity [By the First Derivative Test for Monotonicity].

Step 4: Logical Deduction of Increasing Interval

A function $f$ is strictly increasing on an interval if $f'(x) > 0$ for all $x$ within that interval. Based on the analysis in Step 3, $f'(x) > 0$ when $x \in (\frac{3}{4}, \infty)$.

Final Solution: The function $f(x) = 2x^2 - 3x$ is increasing on the interval $(\frac{3}{4}, \infty)$.

Step 1: Establishing the Objective Function

Let the coordinates of any point $P$ on the curve $x^2 = 2y$ be $(x, y)$. Since $x^2 = 2y$, we can express the point as $(x, \frac{x^2}{2})$. We are given a fixed point $A(0, 5)$. The objective is to minimize the distance $D$ between $P(x, \frac{x^2}{2})$ and $A(0, 5)$.

The squared distance $S = D^2$ is given by the distance formula [Per the Cartesian Distance Formula]:

$S = (x - 0)^2 + (\frac{x^2}{2} - 5)^2$

$S = x^2 + (\frac{x^4}{4} - 5x^2 + 25)$

$S = \frac{x^4}{4} - 4x^2 + 25$

Step 2: Determining Critical Points

To find the minimum distance, we differentiate $S$ with respect to $x$ and set the derivative to zero [Per the First Derivative Test for Extrema]:

$\frac{dS}{dx} = \frac{4x^3}{4} - 8x = x^3 - 8x$

Setting $\frac{dS}{dx} = 0$:

$x(x^2 - 8) = 0$

$x = 0$ or $x^2 = 8 \implies x = \pm 2\sqrt{2}$

Step 3: Evaluating Extrema via Second Derivative Test

We find the second derivative to verify local minima [Per the Second Derivative Test]:

$\frac{d^2S}{dx^2} = 3x^2 - 8$

• For $x = 0$: $\frac{d^2S}{dx^2} = 3(0)^2 - 8 = -8 < 0$ (Local Maximum).
• For $x = \pm 2\sqrt{2}$: $\frac{d^2S}{dx^2} = 3(8) - 8 = 16 > 0$ (Local Minimum).

Step 4: Calculating Coordinates

Using the local minimum points $x = \pm 2\sqrt{2}$ and the curve equation $y = \frac{x^2}{2}$:

If $x^2 = 8$, then $y = \frac{8}{2} = 4$.

The points on the curve are $(2\sqrt{2}, 4)$ and $(-2\sqrt{2}, 4)$.

Step 5: Tabulation of Analytical Result

Final Solution: The points on the curve nearest to $(0, 5)$ are $(2\sqrt{2}, 4)$ and $(-2\sqrt{2}, 4)$.

Step 1: Given Variables and Rates of Change

Based on the problem statement, we define the variables and their time-dependent derivatives as follows:

• Length of the rectangle: $x = 8 \text{ cm}$
• Width of the rectangle: $y = 6 \text{ cm}$
• Rate of change of length: $\frac{dx}{dt} = -5 \text{ cm/min}$ [Negative sign indicates decrease]
• Rate of change of width: $\frac{dy}{dt} = 4 \text{ cm/min}$ [Positive sign indicates increase]

Step 2: Defining the Perimeter Function

The perimeter $P$ of a rectangle is given by the formula:

$P = 2(x + y)$

[By the geometric definition of a rectangle perimeter]

Step 3: Differentiation with respect to time ($t$)

To find the rate of change of the perimeter, we differentiate both sides of the equation with respect to $t$ using the Chain Rule [Per the derivative operator properties]:

$\frac{dP}{dt} = \frac{d}{dt} [2(x + y)]$

$\frac{dP}{dt} = 2 \left( \frac{dx}{dt} + \frac{dy}{dt} \right)$

Step 4: Substituting Values and Calculation

Substitute the given rates $\frac{dx}{dt} = -5$ and $\frac{dy}{dt} = 4$ into the derived expression:

$\frac{dP}{dt} = 2 (-5 + 4)$

$\frac{dP}{dt} = 2 (-1)$

$\frac{dP}{dt} = -2 \text{ cm/min}$

Visual Representation of Rates

Step 5: Interpretation

The negative result signifies that the perimeter is decreasing at the specified moment. [Per the interpretation of first-order derivatives in physical applications].

Final Solution: The rate of change of the perimeter is $-2 \text{ cm/min}$, which indicates the perimeter is decreasing at a rate of $2 \text{ cm/min}$.

Step 1: Establishing the Derivative

Given the function $f(x) = x^2 + ax + 1$, we must first determine the condition under which the function is increasing on the interval $[1, 2]$. A function $f(x)$ is increasing on an interval if its first derivative, $f'(x)$, is greater than or equal to zero for all $x$ in that interval [Per the First Derivative Test for Monotonicity].

Calculating the derivative with respect to $x$:

$f'(x) = \frac{d}{dx}(x^2 + ax + 1)$

$f'(x) = 2x + a$

Step 2: Defining the Condition for Monotonicity

For $f(x)$ to be increasing on $[1, 2]$, we require:

$f'(x) \geq 0$ for all $x \in [1, 2]$

$2x + a \geq 0$

$a \geq -2x$

Step 3: Determining the Range of $a$ via Interval Analysis

Since the inequality $a \geq -2x$ must hold for all $x$ in the interval $[1, 2]$, the value of $a$ must be greater than or equal to the maximum possible value of the expression $-2x$ on the interval $[1, 2]$.

Let $g(x) = -2x$. Since $g(x)$ is a strictly decreasing function (as the coefficient of $x$ is negative), its maximum value on the closed interval $[1, 2]$ occurs at the lower bound, $x = 1$.

Calculating the maximum value of $-2x$ on $[1, 2]$:

$g(1) = -2(1) = -2$

Thus, for $a \geq -2x$ to hold true for all $x \in [1, 2]$, $a$ must satisfy:

$a \geq -2$

Step 4: Geometric Representation of the Derivative

Step 5: Logical Synthesis

By evaluating the inequality at the extreme points of the domain, we ensure the derivative remains non-negative throughout the entire specified interval. If $a \geq -2$, then $2x + a \geq 2(1) + (-2) = 0$ for all $x \geq 1$. Therefore, the condition $f'(x) \geq 0$ is satisfied for all $x \in [1, 2]$.

Final Solution: $a \in [-2, \infty)$

Step 1: Theoretical Foundation of Marginal Revenue

In economic mathematics, the total revenue function $R(x)$ represents the total income from selling $x$ units. Marginal Revenue ($MR$) is defined as the instantaneous rate of change of the total revenue with respect to the number of units sold. Mathematically, this is the first derivative of the revenue function: $MR = \frac{dR}{dx}$ [Per the definition of Marginal Analysis in calculus].

Step 2: Defining the Given Revenue Function

The total revenue function is provided as: $R(x) = 3x^2 + 36x + 5$

Step 3: Differentiation of the Revenue Function

To find the marginal revenue function $MR(x)$, we differentiate $R(x)$ with respect to $x$ using the Power Rule of differentiation, which states $\frac{d}{dx}(x^n) = nx^{n-1}$ [By the Power Rule of Calculus]:

$\frac{dR}{dx} = \frac{d}{dx}(3x^2) + \frac{d}{dx}(36x) + \frac{d}{dx}(5)$

$\frac{dR}{dx} = 3(2x^{2-1}) + 36(1x^{1-1}) + 0$

$MR(x) = 6x + 36$

Step 4: Evaluation at the Specified Quantity

The question requires the calculation of marginal revenue when $x = 15$. We substitute $x = 15$ into the derived expression for $MR(x)$:

$MR(15) = 6(15) + 36$

$MR(15) = 90 + 36$

$MR(15) = 126$

Visual Representation of the Rate of Change

Step 5: Analysis of Results

The marginal revenue represents the revenue generated by selling one additional unit when 15 units have already been produced. The value $126$ indicates the slope of the revenue curve at that specific point.

Final Solution: The Marginal Revenue at $x = 15$ is ₹ 126.

Step 1: Given Function and Derivative Calculation

Let the given function be $f(x) = -2x^3 - 9x^2 - 12x + 1$.

To determine the intervals of strictly increasing or decreasing behavior, we must find the first derivative of $f(x)$ with respect to $x$ [Per the Derivative Power Rule, $\frac{d}{dx}(ax^n) = anx^{n-1}$]:

$f'(x) = \frac{d}{dx}(-2x^3 - 9x^2 - 12x + 1)$

$f'(x) = -6x^2 - 18x - 12$

Step 2: Determining Critical Points

Critical points occur where $f'(x) = 0$ or where $f'(x)$ is undefined. Since $f'(x)$ is a polynomial, it is defined everywhere. We solve for $f'(x) = 0$:

$-6x^2 - 18x - 12 = 0$

Divide the entire equation by $-6$ [Algebraic simplification]:

$x^2 + 3x + 2 = 0$

Factor the quadratic equation:

$(x + 1)(x + 2) = 0$

Thus, the critical points are $x = -1$ and $x = -2$.

Step 3: Interval Analysis

The critical points divide the real number line into three distinct intervals: $(-\infty, -2)$, $(-2, -1)$, and $(-1, \infty)$. We test the sign of $f'(x) = -6(x+1)(x+2)$ in each interval:

Visual Representation

Step 4: Conclusion based on the First Derivative Test

• [Per the First Derivative Test for Monotonicity]: If $f'(x) > 0$, the function is strictly increasing; if $f'(x) < 0$, the function is strictly decreasing.
• The function is strictly increasing on the interval $(-2, -1)$.
• The function is strictly decreasing on the intervals $(-\infty, -2) \cup (-1, \infty)$.

Final Solution: The function is strictly increasing in $(-2, -1)$ and strictly decreasing in $(-\infty, -2) \cup (-1, \infty)$.

Step 1: Theoretical Foundation of Absolute Value Functions

The function is defined as $g(x) = -|x + 1| + 3$. By the definition of the absolute value function, $f(x) = |x|$, we know that $|x| \geq 0$ for all $x \in \mathbb{R}$. [By the property of non-negativity of the modulus function].

Step 2: Determining the Bound for the Modulus Term

Since $|x + 1| \geq 0$ for all $x \in \mathbb{R}$, we apply the scalar multiplication of $-1$ to the inequality, which reverses the inequality sign:
$-|x + 1| \leq 0$ [By the property of inequality multiplication].

Step 3: Deriving the Function Range

To obtain $g(x)$, we add $3$ to both sides of the inequality derived in Step 2:
$-|x + 1| + 3 \leq 0 + 3$
$g(x) \leq 3$

This demonstrates that the maximum value of the function is $3$, occurring when $-|x + 1| = 0$, which implies $x = -1$.

Step 4: Analyzing for Minimum Values

As $x \to \infty$, $g(x) = -|x + 1| + 3 \to -\infty$. Similarly, as $x \to -\infty$, $g(x) \to -\infty$. Because the function is unbounded below, there is no minimum value for $g(x)$. [Per the definition of global extrema in an unbounded interval].

Step 5: Visual Representation

Step 6: Summary of Extrema

Final Solution: The function $g(x) = -|x + 1| + 3$ has a maximum value of $3$ at $x = -1$ and has no minimum value.

Step 1: Identifying Given Variables and Geometric Relationships

Let $V$ be the volume of the conical pile of sand, $r$ be the radius of its base, and $h$ be its height at any time $t$.

• The rate of change of volume is given as $\frac{dV}{dt} = 12 \text{ cm}^3/\text{s}$.
• The geometric relationship provided is $h = \frac{1}{6}r$, which implies $r = 6h$.
• We are required to find the rate of change of height $\frac{dh}{dt}$ when $h = 4 \text{ cm}$.

Step 2: Establishing the Volume Formula

The volume $V$ of a cone is given by the formula [Geometric Formula for Volume of a Right Circular Cone]:

$V = \frac{1}{3}\pi r^2 h$

Substituting $r = 6h$ into the volume equation to express $V$ as a function of $h$ solely:

$V = \frac{1}{3}\pi (6h)^2 h$

$V = \frac{1}{3}\pi (36h^2) h$

$V = 12\pi h^3$

Step 3: Differentiation with respect to time $t$

Differentiating both sides with respect to $t$ using the [Chain Rule of Calculus]:

$\frac{dV}{dt} = \frac{d}{dt} (12\pi h^3)$

$\frac{dV}{dt} = 12\pi \cdot 3h^2 \cdot \frac{dh}{dt}$

$\frac{dV}{dt} = 36\pi h^2 \frac{dh}{dt}$

Step 4: Substitution and Solving for $\frac{dh}{dt}$

Substitute the known values $\frac{dV}{dt} = 12$ and $h = 4$ into the differentiated equation:

$12 = 36\pi (4)^2 \frac{dh}{dt}$

$12 = 36\pi (16) \frac{dh}{dt}$

$12 = 576\pi \frac{dh}{dt}$

Isolating $\frac{dh}{dt}$:

$\frac{dh}{dt} = \frac{12}{576\pi}$

Reducing the fraction by $12$:

$\frac{dh}{dt} = \frac{1}{48\pi} \text{ cm/s}$

Final Solution: The height of the sand cone is increasing at the rate of $\frac{1}{48\pi} \text{ cm/s}$ when the height is $4 \text{ cm}$.

Step 1: Identifying the Function and First Derivative

Given the function: $g(x) = x^3 - 3x$

To find the critical points, we must determine the first derivative $g'(x)$ with respect to $x$ [By the Power Rule of Differentiation: $\frac{d}{dx}(x^n) = nx^{n-1}$]:

$g'(x) = \frac{d}{dx}(x^3) - \frac{d}{dx}(3x)$

$g'(x) = 3x^2 - 3$

Step 2: Locating Critical Points

Critical points occur where $g'(x) = 0$ or where the derivative is undefined. Since $g'(x)$ is a polynomial, it is defined for all $x \in \mathbb{R}$.

$3x^2 - 3 = 0$

$3(x^2 - 1) = 0$

$3(x - 1)(x + 1) = 0$

Thus, the critical points are $x = 1$ and $x = -1$.

Step 3: Second Derivative Test for Local Extrema

We compute the second derivative $g''(x)$ to determine the nature of the critical points [Per the Second Derivative Test for Local Extrema]:

$g''(x) = \frac{d}{dx}(3x^2 - 3) = 6x$

• For $x = 1$:

  $g''(1) = 6(1) = 6$. Since $g''(1) > 0$, the function has a local minimum at $x = 1$.

• For $x = -1$:

  $g''(-1) = 6(-1) = -6$. Since $g''(-1) < 0$, the function has a local maximum at $x = -1$.

Step 4: Calculating Local Extrema Values

To find the local maximum and minimum values, we evaluate $g(x)$ at the critical points:

Visualization of the Function

Final Solution: The function $g(x) = x^3 - 3x$ has a local maximum value of $2$ at $x = -1$ and a local minimum value of $-2$ at $x = 1$.

Step 1: Establishing the Function and Interval

Let the given function be defined as $f(x) = x^4 - 62x^2 + ax + 9$ over the closed interval $[0, 2]$. We are given that $f(x)$ attains its maximum value at $x = 1$.

Step 2: Deriving the First Derivative

To identify critical points where the slope of the tangent is zero (a necessary condition for local maxima/minima in differentiable functions), we compute the first derivative with respect to $x$:

$f'(x) = \frac{d}{dx}(x^4 - 62x^2 + ax + 9)$

$f'(x) = 4x^3 - 124x + a$ [By the Power Rule of Differentiation, $\frac{d}{dx}x^n = nx^{n-1}$]

Step 3: Applying the Condition for Local Maxima

If $f(x)$ attains a maximum at $x=1$, and given that $f(x)$ is a polynomial (and therefore differentiable everywhere), then $x=1$ must be a critical point. By Fermat's Theorem on Stationary Points, we set $f'(1) = 0$:

$4(1)^3 - 124(1) + a = 0$

$4 - 124 + a = 0$

$-120 + a = 0$

$a = 120$

Step 4: Verification via Second Derivative Test

To confirm that $x=1$ is indeed a maximum, we compute the second derivative $f''(x)$:

$f''(x) = \frac{d}{dx}(4x^3 - 124x + 120) = 12x^2 - 124$

Evaluating at $x=1$:

$f''(1) = 12(1)^2 - 124 = 12 - 124 = -112$

Since $f''(1) < 0$, by the Second Derivative Test, $x=1$ is confirmed to be a point of local maximum.

Step 5: Visual Representation

Step 6: Analysis of Boundaries

In the interval $[0, 2]$, we compare the values at the endpoints and the critical point:

• $f(0) = 0 - 0 + 0 + 9 = 9$
• $f(1) = 1 - 62 + 120 + 9 = 68$
• $f(2) = 16 - 62(4) + 120(2) + 9 = 16 - 248 + 240 + 9 = 17$

Since $68 > 17$ and $68 > 9$, the global maximum on the interval $[0, 2]$ is indeed at $x=1$ for $a=120$.

Final Solution: $a = 120$

Step 1: Given Variables and Rates of Change

Based on the problem statement, we define the following variables and their temporal derivatives (rates of change with respect to time $t$):

• Length of the rectangle: $x = 8\text{ cm}$
• Width of the rectangle: $y = 6\text{ cm}$
• Rate of change of length: $\frac{dx}{dt} = -5\text{ cm/min}$ [Negative sign denotes decreasing rate]
• Rate of change of width: $\frac{dy}{dt} = 4\text{ cm/min}$ [Positive sign denotes increasing rate]

Step 2: Defining the Area Function

The area $A$ of a rectangle is defined as the product of its length and width:

$A = x \cdot y$

Step 3: Differentiation with Respect to Time

To find the rate of change of the area with respect to time $\frac{dA}{dt}$, we apply the Product Rule of differentiation, which states that for a product of two functions $u$ and $v$, $\frac{d}{dt}(uv) = u\frac{dv}{dt} + v\frac{du}{dt}$ [Per the Leibniz Product Rule]:

$\frac{dA}{dt} = \frac{d}{dt}(x \cdot y)$

$\frac{dA}{dt} = x\left(\frac{dy}{dt}\right) + y\left(\frac{dx}{dt}\right)$

Step 4: Substitution and Calculation

Substitute the given values into the derived differential equation:

$\frac{dA}{dt} = (8\text{ cm}) \cdot (4\text{ cm/min}) + (6\text{ cm}) \cdot (-5\text{ cm/min})$

$\frac{dA}{dt} = 32\text{ cm}^2\text{/min} - 30\text{ cm}^2\text{/min}$

$\frac{dA}{dt} = 2\text{ cm}^2\text{/min}$

Visual Representation of the Rate Vectors

Final Solution: The area of the rectangle is increasing at the rate of $2\text{ cm}^2\text{/min}$.

Step 1: Establishing the Derivative Function

To determine the intervals of monotonicity for the function $f(x) = \sin x$, we must examine the sign of its first derivative, $f'(x)$, with respect to $x$.

Given: $f(x) = \sin x$

Differentiating with respect to $x$:
$f'(x) = \frac{d}{dx}(\sin x) = \cos x$ [By the derivative rule for trigonometric functions]

Step 2: Analyzing the Critical Points in the Interval $(0, \pi)$

A function is strictly increasing if $f'(x) > 0$ and strictly decreasing if $f'(x) < 0$. To analyze the behavior of $f'(x) = \cos x$ in the interval $(0, \pi)$, we must identify the point where the derivative changes sign.

Setting $f'(x) = 0$:
$\cos x = 0$
In the interval $(0, \pi)$, the only solution is $x = \frac{\pi}{2}$.

Step 3: Evaluating Monotonicity via Sub-intervals

We partition the interval $(0, \pi)$ into two sub-intervals based on the critical point $x = \frac{\pi}{2}$:

Step 4: Geometric Visualization

Step 5: Synthesis of Results

Per the definition of monotonic functions [Definition of Monotonicity], a function is called increasing on an interval $I$ if $f'(x) \geq 0$ for all $x \in I$. Conversely, it is decreasing if $f'(x) \leq 0$ for all $x \in I$.

• Since $f'(x) > 0$ on $(0, \frac{\pi}{2})$, the function is strictly increasing on this sub-interval.
• Since $f'(x) < 0$ on $(\frac{\pi}{2}, \pi)$, the function is strictly decreasing on this sub-interval.

Because the derivative $f'(x)$ changes sign from positive to negative at $x = \frac{\pi}{2}$, the function does not maintain a consistent monotonic behavior across the entire interval $(0, \pi)$. Therefore, the function cannot be classified as globally increasing or globally decreasing on the specified domain.

Final Solution: Since the derivative $f'(x) = \cos x$ is positive in $(0, \pi/2)$ and negative in $(\pi/2, \pi)$, the function $f(x) = \sin x$ is neither increasing nor decreasing in $(0, \pi)$.