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

$x$	$y$	Product ($xy$)
10	14	140
12	12	144 (Maximum)
14	10	140

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: $f(x) = x^2$

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

Feature	Value
Critical Point	$x = 0$
Local Minimum Value	$0$
Local Maximum	None

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: 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.

Point ($x$)	$f(x) = x + \sin 2x$	Calculated Value
$0$	$0 + \sin(0)$	$0$
$\frac{\pi}{3}$	$\frac{\pi}{3} + \sin(\frac{2\pi}{3}) = \frac{\pi}{3} + \frac{\sqrt{3}}{2}$	$\approx 1.913$
$\frac{2\pi}{3}$	$\frac{2\pi}{3} + \sin(\frac{4\pi}{3}) = \frac{2\pi}{3} - \frac{\sqrt{3}}{2}$	$\approx 1.230$
$\frac{4\pi}{3}$	$\frac{4\pi}{3} + \sin(\frac{8\pi}{3}) = \frac{4\pi}{3} + \frac{\sqrt{3}}{2}$	$\approx 5.055$
$\frac{5\pi}{3}$	$\frac{5\pi}{3} + \sin(\frac{10\pi}{3}) = \frac{5\pi}{3} - \frac{\sqrt{3}}{2}$	$\approx 4.372$
$2\pi$	$2\pi + \sin(4\pi)$	$2\pi \approx 6.283$

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 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

$x$	$y$	$S$	Conclusion
$0$	$0$	$25$	Farthest (relative)
$\pm 2\sqrt{2}$	$4$	$9$	Nearest

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

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

Extremum Type	Value	At Point ($x$)
Maximum Value	$3$	$-1$
Minimum Value	None	N/A

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 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:

Point	Calculation: $g(x) = x^3 - 3x$	Result
Local Maximum ($x = -1$)	$(-1)^3 - 3(-1) = -1 + 3$	$2$
Local Minimum ($x = 1$)	$(1)^3 - 3(1) = 1 - 3$	$-2$

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: Establishing the Theoretical Range of the Trigonometric Component

The function is defined as $h(x) = \sin(2x) + 5$. We begin by analyzing the fundamental properties of the sine function. [Per the definition of the range of trigonometric functions].

For any real number $\theta$, the range of $\sin(\theta)$ is given by:

$-1 \le \sin(\theta) \le 1$

In this specific case, $\theta = 2x$. Since $x \in \mathbb{R}$, $2x$ also spans all real values, thus:

$-1 \le \sin(2x) \le 1$

Step 2: Transforming the Inequality to match $h(x)$

To obtain the expression for $h(x)$, we perform a scalar addition of $5$ to all components of the established inequality. [Per the properties of inequality].

$-1 + 5 \le \sin(2x) + 5 \le 1 + 5$

$4 \le h(x) \le 6$

Step 3: Verification via First Derivative Test

To confirm the extrema, we compute the first derivative $h'(x)$ and identify critical points. [Per Fermat's Theorem on Stationary Points].

$h'(x) = \frac{d}{dx}(\sin(2x) + 5) = 2\cos(2x)$

Setting $h'(x) = 0$ for critical points:

$2\cos(2x) = 0 \implies \cos(2x) = 0$

$2x = n\pi + \frac{\pi}{2} \implies x = \frac{n\pi}{2} + \frac{\pi}{4}$

Evaluating $h(x)$ at critical points:

• When $\cos(2x) = 1$, $h(x) = 0 + 5 = 5$ (Not applicable here as we seek global extrema).
• When $\sin(2x) = 1$, $h(x) = 1 + 5 = 6$ (Local/Global Maximum).
• When $\sin(2x) = -1$, $h(x) = -1 + 5 = 4$ (Local/Global Minimum).

Visual Representation of the Function

Step 4: Conclusion of Extrema

Based on the derived range $[4, 6]$, the function oscillates between these bounds indefinitely for $x \in \mathbb{R}$.

Extremum Type	Value
Maximum Value	$6$
Minimum Value	$4$

Final Solution: The maximum value of $h(x)$ is $6$ and the minimum value of $h(x)$ is $4$.

Given Function & Domain

The function is defined as $h(x) = \sin x + \cos x$ within the open interval $x \in (0, \frac{\pi}{2})$.

Step 1: First Order Derivative and Critical Points

To find the critical points, we compute the first derivative $h'(x)$ with respect to $x$:
$h'(x) = \frac{d}{dx}(\sin x + \cos x)$
$h'(x) = \cos x - \sin x$ [Sum Rule of Differentiation]

Set $h'(x) = 0$ to identify stationary points:
$\cos x - \sin x = 0$
$\sin x = \cos x$
$\tan x = 1$ [Dividing by $\cos x$, since $\cos x \neq 0$ for $x \in (0, \pi/2)$]

Solving for $x$ in the interval $(0, \frac{\pi}{2})$:
$x = \frac{\pi}{4}$

Step 2: Second Order Derivative Test

To determine the nature of the critical point, we find the second derivative $h''(x)$:
$h''(x) = \frac{d}{dx}(\cos x - \sin x)$
$h''(x) = -\sin x - \cos x$

Evaluate $h''(x)$ at the critical point $x = \frac{\pi}{4}$:
$h''\left(\frac{\pi}{4}\right) = -\sin\left(\frac{\pi}{4}\right) - \cos\left(\frac{\pi}{4}\right)$
$h''\left(\frac{\pi}{4}\right) = -\frac{1}{\sqrt{2}} - \frac{1}{\sqrt{2}} = -\frac{2}{\sqrt{2}} = -\sqrt{2}$

Since $h''\left(\frac{\pi}{4}\right) < 0$, the function $h(x)$ attains a local maximum at $x = \frac{\pi}{4}$ [By Second Derivative Test].

Step 3: Calculation of Local Maximum Value

Substitute $x = \frac{\pi}{4}$ into the original function $h(x)$:
$h\left(\frac{\pi}{4}\right) = \sin\left(\frac{\pi}{4}\right) + \cos\left(\frac{\pi}{4}\right)$
$h\left(\frac{\pi}{4}\right) = \frac{1}{\sqrt{2}} + \frac{1}{\sqrt{2}} = \frac{2}{\sqrt{2}} = \sqrt{2}$

Step 4: Graphical Representation

Summary of Analysis

Point Type	Coordinate ($x$)	Value ($h(x)$)
Local Maxima	$\frac{\pi}{4}$	$\sqrt{2}$
Local Minima	None	N/A

Final Solution: The function has a local maximum at $x = \frac{\pi}{4}$ with a local maximum value of $\sqrt{2}$. There is no local minimum in the given interval.

Step 1: Establishing the Geometric Variables

Let the side length of the square to be cut from each corner be $x$ cm. When the four corners are removed and the flaps are folded upwards, the dimensions of the resulting rectangular box become:

• Length ($l$) = $45 - 2x$ cm
• Width ($w$) = $24 - 2x$ cm
• Height ($h$) = $x$ cm

[Constraint: Since dimensions must be positive, $2x < 24 \implies 0 < x < 12$].

Step 2: Formulating the Volume Function

The volume $V$ of a rectangular box is given by $V = l \cdot w \cdot h$. Substituting the variables:

$V(x) = (45 - 2x)(24 - 2x)x$

$V(x) = (1080 - 90x - 48x + 4x^2)x$

$V(x) = 4x^3 - 138x^2 + 1080x$

Step 3: Calculating Critical Points via Differentiation

To find the maximum volume, we differentiate $V(x)$ with respect to $x$ and set the derivative to zero [Per Fermat's Theorem on stationary points]:

$\frac{dV}{dx} = 12x^2 - 276x + 1080$

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

$12(x^2 - 23x + 90) = 0$

$x^2 - 18x - 5x + 90 = 0$ [Factoring the quadratic]

$(x - 18)(x - 5) = 0$

Thus, $x = 18$ or $x = 5$.

Step 4: Second Derivative Test for Maximization

We compute the second derivative $\frac{d^2V}{dx^2}$ to test for local maxima [Per Second Derivative Test]:

$\frac{d^2V}{dx^2} = 24x - 276$

Evaluating at $x = 5$:

$V''(5) = 24(5) - 276 = 120 - 276 = -156$

Since $V''(5) < 0$, the function achieves a local maximum at $x = 5$.

[Note: $x = 18$ is discarded because it exceeds the physical constraint $x < 12$].

Critical Point ($x$)	$V''(x)$	Nature of Extremum
5	-156	Maximum
18	156	Minimum (Infeasible)

Final Solution: The side of the square to be cut off is $5$ cm.

Step 1: Identifying the Function and Domain

Given the function: $f(x) = \sin x - \cos x$ within the open interval $x \in (0, 2\pi)$.

Step 2: Determining the First Derivative

To find the critical points, we compute the first derivative $f'(x)$ with respect to $x$ [By the Power and Trigonometric Differentiation Rules]:

$f'(x) = \frac{d}{dx}(\sin x) - \frac{d}{dx}(\cos x)$

$f'(x) = \cos x - (-\sin x) = \cos x + \sin x$

Step 3: Finding Critical Points

Critical points occur where $f'(x) = 0$ [Per the Fermat's Theorem for local extrema]:

$\cos x + \sin x = 0$

$\sin x = -\cos x$

$\tan x = -1$

Since $\tan x$ is negative in the second and fourth quadrants for the interval $(0, 2\pi)$, the solutions are:

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

$x = 2\pi - \frac{\pi}{4} = \frac{7\pi}{4}$

Step 4: Applying the Second Derivative Test

Compute the second derivative $f''(x)$ to determine the nature of the critical points [Per the Second Derivative Test]:

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

Testing $x = \frac{3\pi}{4}$:

$f''(\frac{3\pi}{4}) = -\sin(\frac{3\pi}{4}) + \cos(\frac{3\pi}{4})$

$f''(\frac{3\pi}{4}) = -(\frac{1}{\sqrt{2}}) + (-\frac{1}{\sqrt{2}}) = -\frac{2}{\sqrt{2}} = -\sqrt{2}$

Since $f''(\frac{3\pi}{4}) < 0$, $x = \frac{3\pi}{4}$ is a point of local maxima.

Local Maximum Value: $f(\frac{3\pi}{4}) = \sin(\frac{3\pi}{4}) - \cos(\frac{3\pi}{4}) = \frac{1}{\sqrt{2}} - (-\frac{1}{\sqrt{2}}) = \frac{2}{\sqrt{2}} = \sqrt{2}$.

Testing $x = \frac{7\pi}{4}$:

$f''(\frac{7\pi}{4}) = -\sin(\frac{7\pi}{4}) + \cos(\frac{7\pi}{4})$

$f''(\frac{7\pi}{4}) = -(-\frac{1}{\sqrt{2}}) + (\frac{1}{\sqrt{2}}) = \frac{2}{\sqrt{2}} = \sqrt{2}$

Since $f''(\frac{7\pi}{4}) > 0$, $x = \frac{7\pi}{4}$ is a point of local minima.

Local Minimum Value: $f(\frac{7\pi}{4}) = \sin(\frac{7\pi}{4}) - \cos(\frac{7\pi}{4}) = -\frac{1}{\sqrt{2}} - \frac{1}{\sqrt{2}} = -\frac{2}{\sqrt{2}} = -\sqrt{2}$.

Final Solution: The function has a local maximum value of $\sqrt{2}$ at $x = \frac{3\pi}{4}$ and a local minimum value of $-\sqrt{2}$ at $x = \frac{7\pi}{4}$.

Step 1: Defining Variables and Geometrical Constraints

Let the side of the square piece of tin be $S = 18\text{ cm}$. Let $x\text{ cm}$ be the side of the small square cut from each corner. After cutting the corners and folding the flaps, the dimensions of the resulting open-topped box are:

• Length ($l$) = $18 - 2x$
• Width ($w$) = $18 - 2x$
• Height ($h$) = $x$

[Constraint: $x > 0$ and $2x < 18$, therefore $0 < x < 9$.]

Step 2: Formulating the Volume Function

The volume $V$ of a rectangular box is given by $V = l \times w \times h$.

$V(x) = (18 - 2x)(18 - 2x)x$

$V(x) = x(18 - 2x)^2$

$V(x) = x(324 - 72x + 4x^2)$

$V(x) = 4x^3 - 72x^2 + 324x$

Step 3: Finding Critical Points via Differentiation

To maximize volume, we find the first derivative $V'(x)$ and set it to zero [Per the First Derivative Test for Extrema].

$V'(x) = \frac{d}{dx}(4x^3 - 72x^2 + 324x)$

$V'(x) = 12x^2 - 144x + 324$

Set $V'(x) = 0$:

$12(x^2 - 12x + 27) = 0$

$x^2 - 9x - 3x + 27 = 0$

$(x - 9)(x - 3) = 0$

The roots are $x = 9$ or $x = 3$.

Step 4: Second Derivative Test for Maximization

To confirm the nature of the critical point, calculate the second derivative $V''(x)$.

$V''(x) = \frac{d}{dx}(12x^2 - 144x + 324) = 24x - 144$

Evaluate at $x = 3$:

$V''(3) = 24(3) - 144 = 72 - 144 = -72$

[Since $V''(3) < 0$, the function achieves a local maximum at $x = 3$.]

[Note: $x = 9$ is rejected as it results in $l=0$ and $w=0$, leading to a volume of 0.]

Step 5: Conclusion

The side length of the square to be cut off to maximize the volume is $3\text{ cm}$.

Final Solution: The side of the square to be cut off is 3 cm.

Step 1: Identifying the Core Function and Domain

Given the function: $f(x) = |\sin 4x + 3|$

The domain of this function is $x \in \mathbb{R}$, as the sine function and the absolute value function are defined for all real numbers [By definition of elementary transcendental functions].

Step 2: Range Analysis of the Trigonometric Component

We analyze the behavior of the inner function $g(x) = \sin 4x$.

For any $x \in \mathbb{R}$, the value of $\sin \theta$ oscillates between $-1$ and $1$. Therefore:

$-1 \leq \sin 4x \leq 1$ [Property of the Sine function]

Step 3: Applying Linear Transformation and Absolute Value

We proceed to construct the expression $f(x) = |\sin 4x + 3|$ by performing compound inequalities:

Add 3 to all parts of the inequality:

$-1 + 3 \leq \sin 4x + 3 \leq 1 + 3$

$2 \leq \sin 4x + 3 \leq 4$

Since the entire expression is positive within this interval ($2$ to $4$), the absolute value operation $| \cdot |$ maintains these values:

$|2| \leq |\sin 4x + 3| \leq |4|$

$2 \leq f(x) \leq 4$

Step 4: Determining Extrema

Based on the inequality $2 \leq f(x) \leq 4$:

• The minimum value occurs when $\sin 4x = -1$.
  $4x = 2n\pi - \frac{\pi}{2} \implies x = \frac{n\pi}{2} - \frac{\pi}{8}$, for $n \in \mathbb{Z}$.
  Minimum value = $2$.
• The maximum value occurs when $\sin 4x = 1$.
  $4x = 2n\pi + \frac{\pi}{2} \implies x = \frac{n\pi}{2} + \frac{\pi}{8}$, for $n \in \mathbb{Z}$.
  Maximum value = $4$.

Visual Representation of the Function

Table: Summary of Results

Extremum Type	Functional Value	Condition
Maximum	4	$\sin 4x = 1$
Minimum	2	$\sin 4x = -1$

Final Solution: The maximum value of the function is 4 and the minimum value is 2.

Step 1: Establishing the Profit Function

The profit function is defined as a quadratic equation in terms of $x$, representing the number of units or a variable parameter:

$P(x) = 41 - 72x - 18x^2$

[The objective is to find the local maximum of this function using the First or Second Derivative Test.]

Step 2: Calculating the First Derivative

To determine the critical points, we compute the first derivative of $P(x)$ with respect to $x$ using the power rule $\frac{d}{dx}(x^n) = nx^{n-1}$:

$P'(x) = \frac{d}{dx}(41) - \frac{d}{dx}(72x) - \frac{d}{dx}(18x^2)$

$P'(x) = 0 - 72 - 36x$

$P'(x) = -72 - 36x$

Step 3: Identifying the Critical Point

We set the first derivative equal to zero to find the stationary point [By Fermat's Theorem on Stationary Points]:

$-72 - 36x = 0$

$-36x = 72$

$x = \frac{72}{-36}$

$x = -2$

Step 4: Verification using Second Derivative Test

We compute the second derivative $P''(x)$ to determine the nature of the critical point:

$P''(x) = \frac{d}{dx}(-72 - 36x)$

$P''(x) = -36$

[Since $P''(x) < 0$ for all $x$, the function is concave downwards, implying that the critical point $x = -2$ corresponds to a local maximum.]

Step 5: Calculating Maximum Profit

Substitute the critical value $x = -2$ back into the original profit function $P(x)$:

$P(-2) = 41 - 72(-2) - 18(-2)^2$

$P(-2) = 41 + 144 - 18(4)$

$P(-2) = 41 + 144 - 72$

$P(-2) = 185 - 72$

$P(-2) = 113$

Final Solution: The maximum profit the company can make is 113 units.

Step 1: Geometric Setup and Variable Definition

Let the sphere have a fixed radius $R$. Let the cone inscribed in the sphere have a height $h$ and a base radius $r$. Let the center of the sphere be $O$. If we consider a vertical cross-section through the center of the sphere, the cone appears as an isosceles triangle inscribed in a circle of radius $R$.

Let $x$ be the distance from the center of the sphere $O$ to the base of the cone. Thus, the height of the cone is $h = R + x$ (where $-R \leq x \leq R$).

By applying the Pythagorean theorem in the right triangle formed by the radius of the cone base ($r$), the distance $x$, and the sphere radius $R$:

$r^2 + x^2 = R^2 \implies r^2 = R^2 - x^2$ [By Pythagorean Theorem]

Step 2: Defining the Objective Function

The volume $V$ of a cone is given by $V = \frac{1}{3}\pi r^2 h$. Substituting $r^2$ and $h$ in terms of $R$ and $x$:

$V(x) = \frac{1}{3}\pi (R^2 - x^2)(R + x)$ $V(x) = \frac{1}{3}\pi (R^3 + R^2x - Rx^2 - x^3)$

Step 3: Differentiation and Critical Points

To find the maximum volume, differentiate $V$ with respect to $x$ and set the derivative to zero:

$\frac{dV}{dx} = \frac{\pi}{3} \frac{d}{dx} (R^3 + R^2x - Rx^2 - x^3)$ $\frac{dV}{dx} = \frac{\pi}{3} (R^2 - 2Rx - 3x^2)$ [By Power Rule of Differentiation]

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

$3x^2 + 2Rx - R^2 = 0$ $(3x - R)(x + R) = 0$

The roots are $x = \frac{R}{3}$ and $x = -R$. Since $x = -R$ results in a cone of height 0 (minimum), we select $x = \frac{R}{3}$.

Step 4: Verification of Maxima

Using the second derivative test:

$\frac{d^2V}{dx^2} = \frac{\pi}{3} (-2R - 6x)$

At $x = \frac{R}{3}$:

$\frac{d^2V}{dx^2} = \frac{\pi}{3} (-2R - 6(\frac{R}{3})) = \frac{\pi}{3}(-4R) < 0$

Since the second derivative is negative, $x = \frac{R}{3}$ yields the maximum volume. [By Second Derivative Test for Local Maxima]

Step 5: Final Volume Calculation and Comparison

Substituting $x = \frac{R}{3}$ into $h$ and $r^2$:

$h = R + \frac{R}{3} = \frac{4R}{3}$ $r^2 = R^2 - (\frac{R}{3})^2 = R^2 - \frac{R^2}{9} = \frac{8R^2}{9}$

Max Volume $V_{cone} = \frac{1}{3}\pi (\frac{8R^2}{9})(\frac{4R}{3}) = \frac{32\pi R^3}{81}$

Volume of the Sphere $V_{sphere} = \frac{4}{3}\pi R^3 = \frac{108\pi R^3}{81}$

Ratio = $\frac{V_{cone}}{V_{sphere}} = \frac{32\pi R^3 / 81}{4\pi R^3 / 3} = \frac{32}{81} \times \frac{3}{4} = \frac{8}{27}$

Final Solution: The volume of the largest inscribed cone is $\frac{8}{27}$ of the volume of the sphere.

Step 1: Given Variables & Initial Setup

We are given the function $f(x) = (x-1)^2 + 3$ defined on the closed interval $I = [-3, 1]$. To find the absolute extrema, we must evaluate the function at its critical points within the interval and at the boundaries of the interval [Per the Extreme Value Theorem].

Step 2: Finding Critical Points

To locate critical points, we compute the first derivative $f'(x)$ and equate it to zero:

$f(x) = (x-1)^2 + 3$

Differentiating with respect to $x$ using the Chain Rule:

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

Setting $f'(x) = 0$:

$2(x-1) = 0 \implies x = 1$

Since $x = 1$ is a boundary point of our closed interval $[-3, 1]$, we include it in our analysis.

Step 3: Evaluating at Critical and Boundary Points

We evaluate $f(x)$ at the critical point $x = 1$ and the boundary point $x = -3$:

Point ($x$)	Calculation $f(x) = (x-1)^2 + 3$	Value $f(x)$
$x = 1$	$f(1) = (1-1)^2 + 3 = 0 + 3$	$3$
$x = -3$	$f(-3) = (-3-1)^2 + 3 = (-4)^2 + 3 = 16 + 3$	$19$

Step 4: Visualizing the Function

Step 5: Comparison and Conclusion

Comparing the values obtained:

• Maximum value = $19$ at $x = -3$
• Minimum value = $3$ at $x = 1$

[Per the Extreme Value Theorem, the absolute maximum and minimum of a continuous function on a closed interval must occur at either a critical point or an endpoint].

Final Solution: The absolute maximum value is $19$ at $x = -3$, and the absolute minimum value is $3$ at $x = 1$.

Step 1: Establishing the Function and Derivative

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

To determine the existence of local maxima or minima, we compute the first derivative with respect to $x$ [Per the Power Rule for Exponential Functions, $\frac{d}{dx}(e^u) = e^u \cdot \frac{du}{dx}$]:
$f'(x) = \frac{d}{dx}(e^x) = e^x$

Step 2: Identifying Critical Points

Critical points occur where $f'(x) = 0$ or where $f'(x)$ is undefined [Per the First Derivative Test for Extrema].
Setting the derivative to zero:
$e^x = 0$

By the property of the exponential function $e^x$, there is no real value of $x$ for which $e^x$ equals zero. The range of the function $f(x) = e^x$ is $(0, \infty)$. Therefore, $f'(x) \neq 0$ for all $x \in \mathbb{R}$.

Step 3: Monotonicity Analysis

Since $e^x > 0$ for all $x \in \mathbb{R}$, the slope of the tangent to the curve is strictly positive at every point in its domain [By the Monotonicity Theorem].

Step 4: Conclusion via First Derivative Test

For a function to possess a local maximum or minimum, the derivative must change sign at a critical point $c$ where $f'(c) = 0$ or $f'(c)$ is undefined.

• Since $f'(x) > 0$ for all $x$, the function is strictly increasing on the interval $(-\infty, \infty)$.
• There are no stationary points ($f'(x) = 0$) where the slope changes from positive to negative or vice versa.

Consequently, the function $f(x) = e^x$ exhibits monotonic behavior throughout its entire domain and cannot satisfy the conditions for local extrema.

Final Solution: Since $f'(x) = e^x > 0$ for all $x \in \mathbb{R}$, the function $f(x)$ is strictly increasing and possesses no critical points, thereby confirming that $f(x) = e^x$ has no maxima or minima.

Step 1: Identifying the Function and the Closed Interval

The given function is defined as $f(x) = 4x - \frac{1}{2}x^2$.
The domain of the function is the closed interval $I = [-2, \frac{9}{2}]$.

Step 2: Finding the Critical Points

To find the critical points, we compute the first derivative of $f(x)$ with respect to $x$ [Per the Fermat's Theorem on Stationary Points]:

$f'(x) = \frac{d}{dx} \left( 4x - \frac{1}{2}x^2 \right)$

$f'(x) = 4 - \frac{1}{2}(2x) = 4 - x$

Set $f'(x) = 0$ to identify stationary points:

$4 - x = 0 \implies x = 4$

Since $x = 4$ lies within the interval $[-2, 4.5]$, it is a valid critical point.

Step 3: Evaluating the Function at Critical Points and Endpoints

According to the Extreme Value Theorem, the absolute extrema of a continuous function on a closed interval must occur at the critical points or the endpoints of the interval.

Point ($x$)	Value $f(x) = 4x - 0.5x^2$
$x = -2$	$f(-2) = 4(-2) - 0.5(-2)^2 = -8 - 2 = -10$
$x = 4$	$f(4) = 4(4) - 0.5(4)^2 = 16 - 8 = 8$
$x = \frac{9}{2} = 4.5$	$f(4.5) = 4(4.5) - 0.5(4.5)^2 = 18 - 0.5(20.25) = 18 - 10.125 = 7.875$

Visual Representation

Step 4: Comparison and Conclusion

Comparing the values obtained: $-10$, $8$, and $7.875$.

• The largest value is $8$, which occurs at $x = 4$.
• The smallest value is $-10$, which occurs at $x = -2$.

Final Solution: The absolute maximum value is $8$ and the absolute minimum value is $-10$.

Step 1: Defining the Function and Domain

Let the given function be denoted by $f(x)$ on the closed interval $[0, 1]$:

$f(x) = [x(x-1) + 1]^{\frac{1}{3}}$

Expanding the quadratic expression within the brackets: $f(x) = [x^2 - x + 1]^{\frac{1}{3}}$

Step 2: Finding Critical Points via Differentiation

To determine the local extrema, we calculate the first derivative $f'(x)$ using the chain rule [Per the Chain Rule of Calculus, $\frac{d}{dx}g(h(x)) = g'(h(x)) \cdot h'(x)$]:

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

$f'(x) = \frac{1}{3}(x^2 - x + 1)^{-\frac{2}{3}} \cdot (2x - 1)$

Step 3: Setting the Derivative to Zero

Critical points occur where $f'(x) = 0$ or is undefined. Since $(x^2 - x + 1)$ is always positive for all real $x$ (the discriminant $D = (-1)^2 - 4(1)(1) = -3 < 0$), the derivative is defined for all $x \in [0, 1]$.

$f'(x) = 0 \implies 2x - 1 = 0 \implies x = \frac{1}{2}$

Since $\frac{1}{2} \in [0, 1]$, this is a valid critical point.

Step 4: Evaluating the Function at Boundaries and Critical Points

Per the Extreme Value Theorem, the absolute maximum of a continuous function on a closed interval occurs at either a critical point or the boundaries.

Point ($x$)	Calculation $f(x) = (x^2 - x + 1)^{1/3}$	Result
$x = 0$	$(0^2 - 0 + 1)^{1/3}$	$1$
$x = 1/2$	$((\frac{1}{2})^2 - \frac{1}{2} + 1)^{1/3} = (\frac{1}{4} - \frac{2}{4} + \frac{4}{4})^{1/3}$	$(\frac{3}{4})^{1/3}$
$x = 1$	$(1^2 - 1 + 1)^{1/3}$	$1$

Step 5: Visual Representation of the Function

Step 6: Comparative Analysis

Comparing the values calculated: $1$, $(\frac{3}{4})^{1/3}$, and $1$.

Since $(\frac{3}{4}) < 1$, it follows that $(\frac{3}{4})^{1/3} < 1$.

Therefore, the maximum value attained at the endpoints $x=0$ and $x=1$ is $1$.

Final Solution: The maximum value of the function on the interval $[0, 1]$ is 1.