5.5 Applied Optimization Problems

5.5 Applied Optimization Problems

Core CLEP Concept: Applied optimization represents one of the most practical and heavily weighted problem categories on the CLEP Calculus examination. The goal is to maximize desirable quantities (such as enclosed area, box volume, or business profit) or minimize undesirable quantities (such as material surface area, construction cost, or physical travel time) subject to explicit geometric and algebraic constraints.

---

1. The 6-Step Master Optimization Framework

Every applied optimization problem on the CLEP exam—regardless of whether it involves cardboard boxes, fencing wire, or corporate pricing—can be resolved through this systematic protocol.

                           6-Step Master Optimization Strategy
                                            │
     1. Read & Sketch Diagram ──────────────┼───> Draw figure and label all constant & variable terms
     2. Formulate Objective Function ───────┼───> Equation for quantity to maximize/minimize: Q = f(x, y)
     3. Formulate Constraint Equation ──────┼───> Geometric/physical relationship: g(x, y) = Constant
     4. Reduce to Single Variable ──────────┼───> Solve constraint for y and substitute: Q(x)
     5. Establish Feasible Domain ──────────┼───> Physical bounds: a ≤ x ≤ b or (0, ∞)
     6. Differentiate, Solve & Justify ─────┴───> Find Q'(x) = 0 and justify absolute extremum (EVT or 2nd Deriv)

Detailed Breakdown of the Steps

1. Read & Sketch: Draw a clear geometric diagram. Assign algebraic variables to unknown quantities and record all numerical constants.
2. Identify the Objective Function: Formulate the primary formula representing the quantity $Q$ to be maximized or minimized (e.g., Area $A = xy$, Volume $V = lwh$, Cost $C = 2x + 5y$, Profit $P = R - C$).
3. Identify the Constraint Equation: Formulate the secondary equation connecting the variables based on given limitations (e.g., fixed perimeter $2x + 2y = 100$, fixed volume $\pi r^2 h = 500$).
4. Reduce to a Single Variable: Solve the constraint equation for one variable (e.g., $y = 50 - x$) and substitute into the objective function to obtain $Q(x)$.
5. Determine the Feasible Domain: Determine the allowable physical interval for $x$. (Dimensions cannot be negative; cutouts cannot exceed total material dimensions).
6. Differentiate, Solve & Rigorously Justify: Compute $Q'(x)$, set $Q'(x) = 0$, and find critical numbers. Prove the extremum is global using the Closed Interval Method (if domain is $[a, b]$) or the First/Second Derivative Test (if domain is open $(0, \infty)$).

---

2. Core Archetype 1: Geometric Enclosures & Cutouts

Archetype A: The Open-Top Cardboard Box

Problem: A rectangular sheet of cardboard measures $24\text{ inches}$ by $15\text{ inches}$. Equal squares of side length $x$ are cut from each of the four corners, and the remaining flaps are folded upward to form an open-top box. Find the value of $x$ that maximizes the volume of the box.

      +---------+-----------------------+---------+
      |  x | x  |                       |  x | x  |
      | - - + - -                       - - + - - |
      |  x | x  |                       |  x | x  |
   15 +---------+                       +---------+
      |         |                       |         |  Fold along dashed lines
      |         |     Base of Box       |         |  Height = x
      |         | (24 - 2x) by (15 - 2x)|         |  Length = 24 - 2x
   15 +---------+                       +---------+  Width  = 15 - 2x
      |  x | x  |                       |  x | x  |
      | - - + - -                       - - + - - |
      |  x | x  |                       |  x | x  |
      +---------+-----------------------+---------+
                            24

1. Objective Function: $V = \text{length} \times \text{width} \times \text{height} = (24 - 2x)(15 - 2x)(x)$.
2. Expand algebraically: $V(x) = x(360 - 48x - 30x + 4x^2) = x(4x^2 - 78x + 360) = 4x^3 - 78x^2 + 360x$
3. Feasible Domain: Height must be positive ($x > 0$), and width must be positive ($15 - 2x > 0 \implies x < 7.5$). Thus, domain is $[0, 7.5]$. (At $x = 0$ or $x = 7.5$, $V = 0$).
4. Differentiate and set $V'(x) = 0$: $V'(x) = 12x^2 - 156x + 360 = 0$ Divide by $12$: $x^2 - 13x + 30 = 0 \implies (x - 3)(x - 10) = 0$
5. Evaluate critical numbers:
   • $x = 10$ is outside the feasible domain $[0, 7.5]$ (rejected).
   • $x = 3$ is inside the feasible domain.
6. Justify Maximum: Using the Second Derivative Test: $V''(x) = 24x - 156 \implies V''(3) = 24(3) - 156 = 72 - 156 = -84 < 0$ (Concave down $\implies$ Absolute Maximum). $\text{Optimal Cutout: } x = 3\text{ inches} \quad (V_{\text{max}} = (18)(9)(3) = 486\text{ in}^3)$

Archetype B: The Optimal Cylindrical Container

Problem: A closed cylindrical beverage can is designed to hold a fixed volume of $V_0 = 16\pi\text{ cm}^3$. Find the dimensions (radius $r$ and height $h$) that minimize the total surface area of metal required.

1. Objective Function: Total Surface Area $A = 2\pi r^2 + 2\pi r h$ (two circular ends + side wall).
2. Constraint Equation: Volume $V_0 = \pi r^2 h = 16\pi \implies h = \frac{16\pi}{\pi r^2} = \frac{16}{r^2}$.
3. Express $A$ in terms of $r$ alone: $A(r) = 2\pi r^2 + 2\pi r \left(\frac{16}{r^2}\right) = 2\pi r^2 + \frac{32\pi}{r} = 2\pi r^2 + 32\pi r^{-1}, \quad r \in (0, \infty)$
4. Differentiate and find critical numbers: $A'(r) = 4\pi r - 32\pi r^{-2} = 4\pi r - \frac{32\pi}{r^2} = 0$ $4\pi r = \frac{32\pi}{r^2} \implies 4\pi r^3 = 32\pi \implies r^3 = 8 \implies r = 2\text{ cm}$
5. Find optimal height: $h = \frac{16}{r^2} = \frac{16}{(2)^2} = \frac{16}{4} = 4\text{ cm}$
6. Universal Proportionality Principle: Notice that $h = 4 = 2(2) = 2r$. For any closed cylinder with fixed volume, material surface area is minimized when the height equals the diameter ($h = 2r$).

---

3. Core Archetype 2: Distance Minimization

Finding the point $(x, y)$ on a specified curve $y = f(x)$ that is closest to a given external point $(x_0, y_0)$ is a classic CLEP archetype.

The Squared Distance Simplification Rule

The Euclidean distance formula between $(x, y)$ and $(x_0, y_0)$ is:

$d = \sqrt{(x - x_0)^2 + (y - y_0)^2}$

Calculus Simplification Rule: Because distance $d \ge 0$, minimizing $d$ is mathematically identical to minimizing the squared distance: $S(x) = d^2 = (x - x_0)^2 + [f(x) - y_0]^2$ Minimizing $S(x)$ completely eliminates the square root, avoiding messy chain rule and radical algebra while yielding the exact same critical numbers!

Step-by-Step Worked Example: Closest Point on a Parabola

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

1. Set up squared distance function: $S(x) = (x - 0)^2 + (y - 2)^2 = x^2 + (x^2 - 2)^2$
2. Expand algebraically: $S(x) = x^2 + (x^4 - 4x^2 + 4) = x^4 - 3x^2 + 4, \quad x \in (-\infty, \infty)$
3. Differentiate and set $S'(x) = 0$: $S'(x) = 4x^3 - 6x = 2x(2x^2 - 3) = 0$ $x = 0 \quad \text{or} \quad 2x^2 = 3 \implies x = \pm\sqrt{\frac{3}{2}} = \pm\frac{\sqrt{6}}{2}$
4. Evaluate distances:
   • At $x = 0$: $y = 0^2 = 0 \implies S(0) = 4 \implies d = \sqrt{4} = 2$
   • At $x = \pm\sqrt{3/2}$: $y = (\pm\sqrt{3/2})^2 = 3/2 \implies S = (3/2)^2 - 3(3/2) + 4 = 9/4 - 18/4 + 16/4 = 7/4 \implies d = \sqrt{7/4} = \frac{\sqrt{7}}{2} \approx 1.323$
5. Conclusion: Since $\frac{\sqrt{7}}{2} < 2$, the closest points are $\left(-\sqrt{\frac{3}{2}}, \frac{3}{2}\right)$ and $\left(\sqrt{\frac{3}{2}}, \frac{3}{2}\right)$.

---

4. Core Archetype 3: Economic & Business Optimization

Economic models optimize monetary functions using marginal calculus analysis.

Fundamental Economic Definitions

• Demand (Price) Function $p(x)$: The price per unit when $x$ units are produced and sold.
• Total Revenue $R(x)$: Total incoming cash: $R(x) = x \cdot p(x)$
• Total Cost $C(x)$: Total production cost to manufacture $x$ units.
• Total Profit $P(x)$: Net earnings: $P(x) = R(x) - C(x) = x\cdot p(x) - C(x)$
• Marginal Revenue $MR(x)$: $R'(x) = \frac{dR}{dx}$
• Marginal Cost $MC(x)$: $C'(x) = \frac{dC}{dx}$

The Maximum Profit Theorem

To maximize profit, set $P'(x) = 0$:

$P'(x) = R'(x) - C'(x) = 0 \implies R'(x) = C'(x) \iff MR = MC$

Economic Golden Rule: Profit is maximized at the production volume $x$ where Marginal Revenue equals Marginal Cost ($MR = MC$), provided $P''(x) < 0$ ($R''(x) < C''(x)$).

Average Cost Minimization

The average cost per unit is $\bar{C}(x) = \frac{C(x)}{x}$. Differentiating using the Quotient Rule:

$\bar{C}'(x) = \frac{x C'(x) - C(x)(1)}{x^2} = 0 \implies x C'(x) = C(x) \implies C'(x) = \frac{C(x)}{x} = \bar{C}(x)$

Average Cost Rule: Average cost is minimized at the production level where Marginal Cost equals Average Cost ($MC = \bar{C}$).

---

5. Summary of Common Optimization Archetypes