Maximum and Minimum Function Values with Applications

Tags: #◇Computer Science #◇CSMATH1 #◇Conceptual #◇Calculus #◇Optimization

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◇Overview

Finding maximum and minimum function values (extrema) is among the most consequential practical applications of differential calculus. By computing first and second derivatives and locating critical points, mathematicians and engineers determine optimal configurations across physical geometry, cost curves, and machine learning loss functions.

◇Key Definitions

• Absolute (Global) Extremum: The highest value (absolute maximum) or lowest value (absolute minimum) that a function attains across its entire domain or a specified interval $[a, b]$.
• Local (Relative) Extremum: A point $c$ where $f(c) \ge f(x)$ (local max) or $f(c) \le f(x)$ (local min) for all $x$ within an open neighborhood $(c - \delta, c + \delta)$.
• Critical Point: An interior point $c$ in the domain of $f$ where either the derivative is zero ($f'(c) = 0$) or the derivative does not exist ($f'(c) \text{ undefined}$).
• Fermat's Theorem on Stationary Points: If $f$ has a local extremum at $c$, and $f'(c)$ exists, then $f'(c) = 0$.
• Extreme Value Theorem (EVT): If $f$ is continuous on a closed interval $[a, b]$, then $f$ attains both an absolute maximum and an absolute minimum on $[a, b]$.

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◇Theoretical Framework: Optimization Procedure

1. Identify the Objective & Constraint Equations: State clearly what quantity is to be maximized or minimized.
2. Formulate a Single-Variable Objective Function: Use constraints to eliminate secondary variables: $f(x)$.
3. Establish the Feasible Domain: Determine physical and geometric boundary limits: $x \in [a, b]$ or $x \in (a, b)$.
4. Locate Critical Points: Solve $f'(x) = 0$ and identify where $f'(x)$ is undefined.
5. Classify Extrema:
   • First Derivative Test: Analyze the sign change of $f'(x)$ across critical points ($+ \to -$ gives local max; $- \to +$ gives local min).
   • Second Derivative Test: If $f'(c) = 0$:

• $f''(c) < 0 \implies$ Local Maximum (concave downward).
• $f''(c) > 0 \implies$ Local Minimum (concave upward).
• $f''(c) = 0 \implies$ Inconclusive (must revert to First Derivative Test).
  • Closed Interval Method: Compare $f(a)$, $f(b)$, and $f(c_i)$ for all critical points $c_i \in (a, b)$.

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◇Stepped Mathematical Proofs & Optimization Problems

Problem 1: Maximum Volume Open-Top Box

Problem Statement: An open-top box is constructed from a rectangular sheet of cardboard with dimensions $16\text{ cm} \times 30\text{ cm}$ by cutting congruent squares of side length $x$ from each of the four corners and folding up the resulting flaps. Determine the cut length $x$ that maximizes the interior volume, and compute the maximum volume.

Step 1: Formulate the Objective Function

• Box Height: $h = x$
• Box Width: $w = 16 - 2x$
• Box Length: $l = 30 - 2x$
• Volume Function:
  $V(x) = x(16 - 2x)(30 - 2x) = x(480 - 92x + 4x^2) = 4x^3 - 92x^2 + 480x$

Step 2: Determine the Physical Feasible Domain

All dimensions must be strictly positive:
$x > 0, \quad 16 - 2x > 0 \implies x < 8, \quad 30 - 2x > 0 \implies x < 15$
Thus, the feasible domain is the open interval $x \in (0, 8)$.

Step 3: Differentiate & Find Critical Points

$V'(x) = \frac{d}{dx}\left(4x^3 - 92x^2 + 480x\right) = 12x^2 - 184x + 480$
Set $V'(x) = 0$ and factor out the common scalar 4:
$4(3x^2 - 46x + 120) = 0$
Factoring the quadratic:
$(3x - 10)(x - 12) = 0 \implies x_1 = \frac{10}{3} \approx 3.33\text{ cm}, \quad x_2 = 12\text{ cm}$
Because $x_2 = 12 \notin (0, 8)$, it is physically extraneous (it would yield a negative width: $16 - 2(12) = -8$). The only feasible critical point is $x = \frac{10}{3}$.

Step 4: Classify Extrema via the Second Derivative Test

$V''(x) = \frac{d}{dx}(12x^2 - 184x + 480) = 24x - 184$
Evaluate at $x = \frac{10}{3}$:
$V''\left(\frac{10}{3}\right) = 24\left(\frac{10}{3}\right) - 184 = 80 - 184 = -104 < 0$
Because $V''\left(\frac{10}{3}\right) < 0$, the function is concave downward at this point, proving that $x = \frac{10}{3}\text{ cm}$ yields an absolute maximum on $(0, 8)$.

Step 5: Compute Maximum Volume

$V\left(\frac{10}{3}\right) = \left(\frac{10}{3}\right) \left(16 - \frac{20}{3}\right) \left(30 - \frac{20}{3}\right) = \left(\frac{10}{3}\right) \left(\frac{28}{3}\right) \left(\frac{70}{3}\right) = \frac{19600}{27} \approx 725.93\text{ cm}^3$

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Problem 2: Minimizing Surface Area of a Cylindrical Can (Fixed Volume)

Problem Statement: A manufacturing plant produces closed cylindrical cans holding a fixed volume $V_0 = 1000\text{ cm}^3$ ($1\text{ L}$). Determine the ratio of height $h$ to radius $r$ that minimizes the total surface area of metal required.

Step 1: Equations & Constraints

• Volume Constraint: $V_0 = \pi r^2 h \implies h = \frac{V_0}{\pi r^2}$
• Surface Area (Objective): $S = 2\pi r^2 + 2\pi r h$

Step 2: Express Objective in Single Variable $r$

$S(r) = 2\pi r^2 + 2\pi r \left(\frac{V_0}{\pi r^2}\right) = 2\pi r^2 + \frac{2V_0}{r}, \quad r \in (0, \infty)$

Step 3: Differentiate & Solve for Critical Radius

$S'(r) = 4\pi r - \frac{2V_0}{r^2}$
Setting $S'(r) = 0$:
$4\pi r = \frac{2V_0}{r^2} \implies 4\pi r^3 = 2V_0 \implies r^3 = \frac{V_0}{2\pi} \implies r = \sqrt[3]{\frac{V_0}{2\pi}}$

Step 4: Optimal Aspect Ratio Derivation

Substitute $r^3 = \frac{V_0}{2\pi}$ back into the height constraint equation:
$h = \frac{V_0}{\pi r^2} = \frac{2\pi r^3}{\pi r^2} = 2r$
Key Engineering Finding: The optimal can has a height equal to its diameter ($h = 2r$), minimizing manufacturing material for any given volume.

Step 5: Second Derivative Test Verification

$S''(r) = 4\pi + \frac{4V_0}{r^3}$
Since $V_0 > 0$ and $r > 0$, $S''(r) > 0$ everywhere on $(0, \infty)$. The objective function is strictly convex, confirming that $r = \sqrt[3]{\frac{V_0}{2\pi}}$ is a unique global minimum.

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Problem 3: Closest Point on a Parabola (Distance Minimization)

Problem Statement: Find the points on the parabola $y = x^2$ that are closest to the point $(0, 2)$.

Step 1: Distance Formulation

The Euclidean distance from any point $(x, x^2)$ to $(0, 2)$ is:
$d(x) = \sqrt{(x - 0)^2 + (x^2 - 2)^2}$
Because the square root function is monotonically increasing for non-negative arguments, minimizing $d(x)$ is equivalent to minimizing the squared distance $f(x) = [d(x)]^2$:
$f(x) = x^2 + (x^4 - 4x^2 + 4) = x^4 - 3x^2 + 4, \quad x \in (-\infty, \infty)$

Step 2: Differentiate & Find Critical Points

$f'(x) = 4x^3 - 6x = 2x(2x^2 - 3) = 0$
Critical points:
$x = 0 \quad \text{or} \quad 2x^2 - 3 = 0 \implies x = \pm\sqrt{\frac{3}{2}}$

Step 3: Second Derivative Test

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

• At $x = 0$: $f''(0) = -6 < 0 \implies$ Local Maximum distance!
• At $x = \pm\sqrt{\frac{3}{2}}$:
  $f''\left(\pm\sqrt{\frac{3}{2}}\right) = 12\left(\frac{3}{2}\right) - 6 = 18 - 6 = 12 > 0 \implies \text{Local Minima!}$

Step 4: Closest Coordinate & Minimal Distance

Substitute $x = \pm\sqrt{\frac{3}{2}}$ into $y = x^2$:
$y = \left(\pm\sqrt{\frac{3}{2}}\right)^2 = \frac{3}{2}$
The two closest points are $\left(-\sqrt{\frac{3}{2}}, \frac{3}{2}\right)$ and $\left(\sqrt{\frac{3}{2}}, \frac{3}{2}\right)$.
The minimum distance is:
$d_{\min} = \sqrt{\left(\sqrt{\frac{3}{2}}\right)^2 + \left(\frac{3}{2} - 2\right)^2} = \sqrt{\frac{3}{2} + \frac{1}{4}} = \sqrt{\frac{7}{4}} = \frac{\sqrt{7}}{2} \approx 1.323$
(Note: At $x = 0$, $d(0) = \sqrt{0 + (0-2)^2} = 2 > \frac{\sqrt{7}}{2}$, confirming $(0, 0)$ is a local maximum relative to the parabola's arms).

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◇References & Navigation

• Previous Topic: ◇Linear Approximations and Differentials
• Next Topic: ◇Absolute Extremum
• Related Notes: ◇Absolute Extremum, ◇Derivative as Rate of Change and Related Rates Part 1