16.3 Optimization Problems and Curve Sketching with Derivatives

From Local Extrema to Global Answers

By this point you can already find where $f'(x) = 0$ and classify local maxima and minima using the first-derivative sign test. Selective-compulsory optimization work asks a sharper question: on a closed interval $[a, b]$, or in a real-world constrained scenario, what is the actual largest or smallest value $f$ can take? Local extrema are candidates for that answer, but they are not the whole story — the interval's endpoints matter too.

The Closed-Interval Extreme Value Procedure

For $f$ continuous on $[a, b]$ and differentiable on $(a, b)$, the global maximum and minimum are guaranteed to exist and must occur either at a critical point (where $f'(x) = 0$ or $f'$ is undefined) or at an endpoint. The procedure is mechanical:

1. Find $f'(x)$ and solve $f'(x) = 0$ to get all critical points inside $(a, b)$.
2. Evaluate $f$ at every critical point found in step 1.
3. Evaluate $f$ at the two endpoints, $f(a)$ and $f(b)$.
4. Compare all the values from steps 2–3: the largest is the global maximum, the smallest is the global minimum.

Worked Example 1. Find the global maximum and minimum of $f(x) = x^3 - 3x^2 + 1$ on $[-1, 3]$.

$f'(x) = 3x^2 - 6x = 3x(x - 2)$. Critical points: $x = 0$ and $x = 2$, both inside $(-1, 3)$.

• $f(-1) = -1 - 3 + 1 = -3$
• $f(0) = 1$
• $f(2) = 8 - 12 + 1 = -3$
• $f(3) = 27 - 27 + 1 = 1$

Comparing $\{-3, 1, -3, 1\}$: the global maximum is $1$ (attained at both $x = 0$ and $x = 3$), and the global minimum is $-3$ (attained at both $x = -1$ and $x = 2$). The two local extrema are not automatically the global extrema — you must check the endpoints to be sure.

Applied Optimization Word Problems

Most Gaokao optimization problems hide the interval procedure inside a physical setup: maximize area, volume, revenue, or minimize cost, material, or distance, subject to a fixed constraint. A reliable workflow:

• Name the variable. Choose the single free variable (often a length $x$) and write its natural domain (e.g., $0 < x < 10$) from the physical constraint.
• Build the objective function. Use the constraint to eliminate all other variables so you have one function of that one variable, e.g. Volume $V(x)$ or Cost $C(x)$.
• Differentiate and solve $V'(x) = 0$ for critical points inside the domain.
• Classify with the interval procedure (or a first-derivative sign change if the domain is open) to confirm the critical point is a genuine maximum or minimum, not just a stationary point.
• Answer the original question, translating back from $x$ into the requested quantity and stating units.

Worked Example 2. A rectangular box with a square base and open top is to be built from $108\ \mathrm{m}^2$ of material. Find the dimensions that maximize its volume.

Let the base side be $x$ and the height be $h$. Surface area constraint: $x^2 + 4xh = 108$, so $h = \frac{108 - x^2}{4x}$, valid for $0 < x < \sqrt{108}$.

Volume: $V(x) = x^2 h = x^2 \cdot \frac{108 - x^2}{4x} = \frac{108x - x^3}{4} = 27x - \frac{x^3}{4}$.

$V'(x) = 27 - \frac{3}{4}x^2$. Setting $V'(x) = 0$: $x^2 = 36$, so $x = 6$ (rejecting the negative root since $x > 0$).

Since $V'(x) > 0$ for $x < 6$ and $V'(x) < 0$ for $x > 6$, $V$ has a maximum at $x = 6$.

Then $h = \frac{108 - 36}{24} = 3$. Maximum volume: $V(6) = 27(6) - \frac{216}{4} = 162 - 54 = 108\ \mathrm{m}^3$, at base side $6\ \mathrm{m}$ and height $3\ \mathrm{m}$.

A Systematic Curve-Sketching Checklist

Curve sketching packages everything you know about a function's derivative into a single accurate picture. Work through these checks in order:

Worked Example 3. Sketch the key features of $f(x) = x^3 - 3x$.

• Domain: all real $x$.
• Intercepts: $f(0) = 0$; solving $x^3 - 3x = 0$ gives $x(x^2 - 3) = 0$, so $x = 0, \pm\sqrt{3}$.
• Symmetry: $f(-x) = -x^3 + 3x = -f(x)$, so $f$ is odd.
• Monotonicity: $f'(x) = 3x^2 - 3 = 3(x-1)(x+1)$, positive outside $[-1, 1]$ and negative inside it, so $f$ increases on $(-\infty, -1)$, decreases on $(-1, 1)$, and increases again on $(1, \infty)$.
• Local extrema: local maximum at $x = -1$ ($f(-1) = 2$), local minimum at $x = 1$ ($f(1) = -2$).
• No asymptotes, since $f$ is a polynomial.

This checklist output is enough to draw an accurate rise–fall–rise cubic without plotting dozens of points.

A Note on Scope: Integrals Are Not Required

It is tempting, once you can sketch a curve, to ask "what is the area under it?" and reach for integration. Under the current national Gaokao selective-compulsory curriculum, the required content is 一元函数导数及其应用 (derivatives of functions of one variable and their applications) — definite integrals and the Fundamental Theorem of Calculus are not part of the nationally examined syllabus. Spend preparation time mastering the derivative-based procedures above (interval extrema, applied optimization, and the sketching checklist), since these are what the exam actually tests; do not budget review time for computing definite integrals as an exam technique.