Solving Optimization Problems

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Textbook Reference

Opening Scenario

This lesson picks up where the setup lesson ended. You have a single-variable function $f(x)$ on a domain $D$, and you want to find the value of $x$ in $D$ where $f$ is largest (or smallest). The method is just the Closed Interval Method or the derivative tests, applied to the setup function.

The new challenge is that optimization problems often have open or half-open domains (not closed intervals), so the Closed Interval Method does not always apply directly.

Quick Reference

If the domain is a closed interval $[a, b]$: use the Closed Interval Method.

1. Find all critical numbers in $(a, b)$.
2. Evaluate $f$ at the critical numbers and at $a$ and $b$.
3. The largest value is the absolute maximum; the smallest is the absolute minimum.

If the domain is open (e.g., $(0, \infty)$) with only one critical number $c$:

• If $f''(c) < 0$ (concave down): $c$ gives an absolute maximum.
• If $f''(c) > 0$ (concave up): $c$ gives an absolute minimum.
• Or use the First Derivative Test: if $f'$ goes from $+$ to $-$, the function has a global max at $c$ (assuming $f \to$ smaller values at the boundaries of the domain).

Key Concepts

1. Completing the Box Problem

From the setup lesson: maximize $V(x) = 4x^3 - 48x^2 + 144x$ on $(0, 6)$.

Find critical numbers.

$V'(x) = 12x^2 - 96x + 144 = 12(x^2 - 8x + 12) = 12(x - 2)(x - 6)$.

$V'(x) = 0$ at $x = 2$ and $x = 6$.

Only $x = 2$ is in the open interval $(0, 6)$; $x = 6$ is the endpoint.

Classify $x = 2$.

The domain is open, so the Closed Interval Method does not apply directly. Use the Second Derivative Test (or reason about end behavior).

$V''(x) = 24x - 96$. $V''(2) = 48 - 96 = -48 < 0$.

Concave down at $x = 2$: the critical number is a local maximum.

Is it the absolute maximum? As $x \to 0^+$ and $x \to 6^-$: $V(0) = 0$ and $V(6) = 4(216) - 48(36) + 144(6) = 864 - 1728 + 864 = 0$. The volume approaches $0$ at both ends.

Since $V$ is positive on the interior (try $x = 2$: $V(2) = 4(8) - 48(4) + 144(2) = 32 - 192 + 288 = 128 > 0$) and the function vanishes at both endpoints, the single interior local maximum at $x = 2$ is the absolute maximum on $(0, 6)$.

Compute the maximum volume.

$V(2) = 2(12 - 4)^2 = 2 \cdot 64 = 128$ cubic inches.

Answer the question. Cut $x = 2$-inch squares from each corner. The maximum volume is $128$ cubic inches. The base dimensions are $8 \times 8$ inches and the height is $2$ inches.

Boxed answer: Maximum volume of $128 \text{ in}^3$ when $x = 2$ inches.

2. The Farmer’s Fence (Closed Domain)

From the setup lesson: maximize $A(w) = 150w - w^2$ on $(0, 150)$.

Note: the domain is open because the dimensions must be strictly positive. However, $A$ is a polynomial, so we can analyze it on the closed interval $[0, 150]$ (including the endpoints, even though they give zero area) and then interpret.

$A'(w) = 150 - 2w = 0 \Rightarrow w = 75$.

$A''(w) = -2 < 0$: the critical number $w = 75$ is a local (and absolute) maximum.

$A(75) = 150(75) - 75^2 = 11{,}250 - 5{,}625 = 5{,}625 \text{ m}^2$.

$l = 150 - 75 = 75$ m. The optimal enclosure is a square.

Boxed answer: Maximum area of $5{,}625 \text{ m}^2$ when $l = w = 75$ meters.

Recap. This problem’s objective is a downward-opening parabola with a single interior maximum. Many “maximize with a fixed perimeter” problems have this structure: the optimum is a square or circle (the most symmetric shape).

3. Verifying the Answer Makes Sense

After computing, always check:

• Are the dimensions positive? (Here: $x = 2 > 0$ and $12 - 2(2) = 8 > 0$. Yes.)
• Does the answer actually maximize (not minimize)? (Check $V''(2) < 0$, confirming a local max. Also, the endpoints give $V = 0 < 128$.)
• Is the answer in the right units? (Volume in cubic inches. Yes.)
• Does the answer feel reasonable? ($128 \text{ in}^3 = 128/1728 \approx 0.074 \text{ ft}^3$. A box 8 in by 8 in by 2 in. Reasonable for a 12-inch sheet.)

4. Using the First Derivative Test on an Open Domain

An alternative to the Second Derivative Test when the domain is open:

Reasoning. For the box problem, $V'(x) = 12(x-2)(x-6)$.

$V'$ goes from $+$ to $-$ at $x = 2$: the function increases then decreases. So $x = 2$ gives a local maximum, and since the function vanishes at both endpoints, it is the absolute maximum.

The First Derivative Test approach is often cleaner than the second derivative approach, especially when $f''$ is complicated.

Common Errors Summary

Leveled Practice

Level 1 -- Finish a Given Setup

Problem 1. The objective function from the setup problem on the can is $C(r) = 1000/r + 2\pi r^2$ (with $c = 1$), for $r > 0$. Find the value of $r$ that minimizes $C$.

Level 2 -- Full Problem

Problem 2. A rancher wants to fence a rectangular pasture with an area of $800 \text{ m}^2$. The north and south sides use premium fencing at \$8 per meter; the east and west sides use standard fencing at \$5 per meter. Find the dimensions that minimize the total fencing cost.

Mastery Checklist

• Apply the Closed Interval Method when the domain is a closed interval
• Use the Second Derivative Test or First Derivative Test when the domain is open
• Check that the critical number gives the correct type of extremum (max vs min)
• Verify the answer physically (positive dimensions, correct units, sensible magnitude)
• State the answer as: “The [quantity] is maximized/minimized at [input] = [value], giving [objective] = [value] [units]”

Mental Model

The solving step is the payoff for the setup. You have one function and one domain. The Extreme Value Theorem (for closed domains) or the derivative tests (for open domains) tell you where the maximum or minimum is.

The one extra step compared to abstract calculus is the physical check: the answer must describe a real object. Negative lengths, zero volumes, and units that do not match are signs that something went wrong in the setup. If the answer fails the physical check, return to the setup step.

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