4.3 Extrema: Where Chemistry Settles
The problem
A synthetic chemist runs a two-step reaction in a flask. The product grows for a while — then the second step, which slowly destroys it, catches up. Stop the reaction too early and you leave yield unmade; stop too late and you have burned it. The yield as a function of time, from a kinetic model of the two steps, is
y(t) = 6t^{2} - t^{3} \qquad (y \text{ in grams, } t \text{ in hours, } 0 \le t \le 6).
You know from the last problem set what the derivative of this is and what it means: y'(t) is the rate of yield growth, in grams per hour, right now. The question the chemist actually cares about is different: when is the yield largest, and how large is it? The derivative tells you how fast y is changing at any moment you ask. It does not — at first glance — tell you where y is biggest. This section is about the one observation that turns the first question into the second.
Where a maximum must hide
Watch the method unfold: Open the interactive extrema animation. Switch between a reaction-yield maximum and a molecular-energy minimum, then scrub through the curve or let it play.
For the next step, §4.4 Constrained Extrema finds the best point when two chemical quantities must add to a fixed total; its interactive studio is a good preview.
Follow the yield in your mind. Early on, the yield is rising: y' > 0. Late, the destruction step wins and the yield is falling: y' < 0. Somewhere in between the curve stopped rising and started falling — and at that turning moment, for one instant, the yield was doing neither. The tangent line there is flat.
That is the entire idea, and it deserves to be stated as a condition:
The first-derivative condition. If a smooth function y(x) has a maximum or a minimum at an interior point x = c of its domain, then y'(c) = 0. Points where y' = 0 are called stationary points (the tangent is horizontal — the function is momentarily stationary).
Read the logic direction carefully: every interior maximum or minimum is a stationary point, but not every stationary point is a maximum or minimum — we will meet the exception before the end of this section. The condition is a search warrant: it tells you the only interior places worth examining.
Telling tops from bottoms: the second derivative
A flat tangent happens at the top of a hill and at the bottom of a valley. The derivative alone cannot tell them apart — but the derivative of the derivative can. At a maximum the slope passes from positive through zero to negative: the slope is decreasing, so y'' < 0. At a minimum the slope passes from negative to positive: increasing, so y'' > 0. Geometrically, y'' < 0 means the curve is curving downward like a cap (\cap); y'' > 0 means curving upward like a cup (\cup). A cup holds its contents at a minimum; a cap sheds them from a maximum.
The second-derivative test. Suppose y'(c) = 0 at an interior point c. Then y''(c) < 0 \;\Rightarrow\; \text{local maximum at } c, \qquad y''(c) > 0 \;\Rightarrow\; \text{local minimum at } c. If y''(c) = 0, the test is silent — not “it’s a maximum,” not “it’s a minimum,” just no verdict — and you must examine the sign of y' on each side of c.
Worked example
Example 4.3.1 — the best time to stop the reaction. For y(t) = 6t^2 - t^3 on 0 \le t \le 6, find when the yield is largest and what it is.
Step 1 — differentiate and set to zero. y'(t) = 12t - 3t^{2} = 3t\,(4 - t) = 0 \quad\Rightarrow\quad t = 0 \;\text{ or }\; t = 4.
Step 2 — classify with the second derivative. y''(t) = 12 - 6t, \qquad y''(0) = +12 > 0 \;(\text{minimum}), \qquad y''(4) = -12 < 0 \;(\text{maximum}).
Step 3 — check the endpoints. The domain ends at t = 6: y(0) = 0, y(4) = 6(16) - 64 = 32 g, y(6) = 216 - 216 = 0 g. The interior maximum beats both endpoints.
Step 4 — answer the chemist’s question. Stop the reaction at t = 4 h, harvesting 32 g. Running the full 6 h destroys everything you made.
Check. Units: y' in g/h, zero at an instant measured in hours — consistent. Limiting behavior: y'(1) = +9 g/h (still rising, too early to stop) and y'(5) = -15 g/h (already losing product) bracket the answer with the right signs. And the number is plausible: a maximum of 32 g between two zero endpoints.
That four-step rhythm — differentiate, solve, classify, check the edges — is the entire method, and it does not change no matter what function chemistry hands you.
The deepest example in chemistry: why bonds have lengths
Two atoms approach each other. Too far apart, attraction pulls them in; too close, their electron clouds repel ferociously. The potential energy of the pair, in reduced units, is the Lennard-Jones curve V(r) = \frac{1}{r^{12}} - \frac{2}{r^{6}}, and the force between the atoms is F = -\dfrac{dV}{dr} — the system is pushed downhill in energy.
Where does the pair come to rest? Exactly where the force vanishes: \frac{dV}{dr} = -\frac{12}{r^{13}} + \frac{12}{r^{7}} = 0 \quad\Rightarrow\quad r^{6} = 1 \quad\Rightarrow\quad r = 1, with V(1) = -1. And is that rest stable? Differentiate again: \;V''(1) = 156 - 84 = 72 > 0 — a cup, a true minimum. Displace the atoms slightly and the force pushes them back.
The bond length of a molecule is the location of an energy minimum — a solved extremum problem. And the second derivative is not bookkeeping: V'' at the minimum is the stiffness of the bond, the force constant that sets its vibrational frequency in an IR spectrum. The first derivative tells you where the bond sits; the second tells you how it vibrates.
Remarks and cautions
A flat tangent is not always a top or a bottom. Try y = x^{3} at x = 0: y'(0) = 0 and y''(0) = 0 — the test is silent, and rightly so, because the curve just pauses and keeps rising (y' = 3x^2 > 0 on both sides). When y''(c) = 0, fall back to first principles: check the sign of y' on each side of c.
y' = 0 only searches the interior. On a closed interval the largest value can sit at an endpoint, where the tangent need not be flat at all — a reaction stopped by the clock, not by the chemistry. Always evaluate the endpoints alongside the stationary points, as in Step 3 above.
Remark. A maximum of y is a zero of y'. You have been using this backwards all course: when a plotted signal levels off, its derivative has dropped to zero. “Leveling off,” “turning over,” “settling” — chemistry’s vocabulary for equilibrium is the vocabulary of vanishing derivatives.
Summary
Chemistry’s “where is the most?” and “where does it settle?” questions are answered by a two-derivative interrogation. The first derivative provides the search warrant: interior maxima and minima can hide only where y' = 0, so differentiate, set to zero, and solve. The second derivative provides the verdict: a negative y'' means a cap and a maximum, a positive y'' a cup and a minimum, and a zero y'' means no verdict at all — examine the slope’s sign on either side. Endpoints obey no tangent condition and must be checked by hand. The same four steps located a reaction’s optimal quench time and a molecule’s bond length; in the second case the curvature itself was physical, the stiffness that an IR spectrometer reads as a vibrational frequency. What the method does not yet answer: a bond vibrates about its minimum — how do we describe motion that endlessly trades position for speed? That is a differential equation, and it is where CHANGE goes next.
Practice until it sticks
This section’s exercise set is Problem Set 7 — Extrema, which opens with a second fully worked example before the problems begin.