Appendix I — Problem Set 7 | Extrema

Companion to §4.3 — finding maxima and minima with dy/dx = 0 and d²y/dx²

Purpose

Chemistry asks “where is the most?” constantly: the optimal temperature, the equilibrium bond length, the pH where an enzyme works best, the moment to stop a reaction. This set trains the complete method — and only the method:

  1. Differentiate and solve y' = 0 → the stationary points (the only interior candidates).
  2. Classify each with the second derivative: y'' < 0 → maximum (cap, \cap); y'' > 0 → minimum (cup, \cup); y'' = 0 → no verdict — check the sign of y' on each side.
  3. Check the endpoints of the domain; they obey no tangent condition.
  4. Interpret: one sentence of chemical meaning, with units.

The lesson behind this set is §4.3 Extrema — read it first if you have not; the two interactive figures there are worth ten minutes.

Worked example — read this before the problems

An enzyme’s best pH. A simple model for an enzyme’s activity (percent of maximum) as a function of pH, valid for 4 \le p \le 10: a(p) = -2p^{2} + 28p - 58. Find the pH of peak activity, the peak value, and confirm it is a maximum.

Step 1 — differentiate, set to zero. The power rule, term by term: a'(p) = -4p + 28. a'(p) = 0 gives 4p = 28, so p = 7. One stationary point, and it lies inside the domain.

Step 2 — classify. Differentiate again: a''(p) = -4 < 0 \quad\text{for every } p, a cap everywhere — so p = 7 is a maximum. (Note we did not need to plug p = 7 in: here a'' is a constant.)

Step 3 — endpoints. a(4) = -32 + 112 - 58 = 22; a(10) = -200 + 280 - 58 = 22; a(7) = -98 + 196 - 58 = 40. The interior maximum wins.

Step 4 — interpret. The enzyme works best at pH 7, at 40% of theoretical maximum activity; three pH units away in either direction, activity has fallen to 22%.

Check. a'(6) = +4 > 0 (still improving as pH rises toward 7) and a'(8) = -4 < 0 (past the peak): the slope changes from rising to falling across p = 7, exactly what a maximum demands. Units: a' is activity-percent per pH unit, zero at the peak — dimensionally consistent.

Every problem below is this same four-step rhythm on a different function. Show all four steps each time.

A. Reading before computing

1. The graph knows first

A curve rises from the left, flattens at x = 2, falls until x = 5, flattens again, then rises to the edge of the plot at x = 8.

  1. Where is y' = 0?
  2. Classify each of those points (maximum or minimum) from the story alone — no formulas.
  3. What is the sign of y'' at each of them?

2. Signs and units at a peak

A product concentration [\mathrm{P}](t) (in M, t in minutes) reaches its highest value at t = 12 min.

  1. What is the value of d[\mathrm{P}]/dt at t = 12, and what are its units?
  2. What is the sign of d^{2}[\mathrm{P}]/dt^{2} there, and what are its units?
  3. One sentence: physically, why must the rate be zero at the moment of the peak?

B. The method, start to finish

Use all four steps (differentiate/solve, classify, endpoints, interpret) on each.

3. The reactor from Problem Set 6, finished

n(t) = 3.0\,t - 0.25\,t^{2} (mol, hours, 0 \le t \le 8). In Set 6 you found where dn/dt = 0. Now classify that point with n'', check both endpoints, and report the maximum amount of product and when it occurs.

4. A yield curve with a cubic twist

y(t) = 9t^{2} - t^{3} (grams, hours, 0 \le t \le 9).

  1. Find both stationary points and classify each with y''.
  2. Check the endpoints and report the best stopping time and the yield there.
  3. One sentence: what is the destruction step doing to the curve after the peak?

5. Why a bond has a length

A simplified bond-energy model (reduced units): V(r) = \frac{1}{r^{2}} - \frac{2}{r}, \qquad r > 0.

  1. Write V = r^{-2} - 2r^{-1} and find V'(r). Solve V'(r) = 0.
  2. Compute V'' at that point and state what its sign tells you about the bond.
  3. Evaluate V at the equilibrium separation. One sentence: what does this (negative) number mean, compared with V \to 0 as r \to \infty?
  4. If the atoms are pushed slightly closer than equilibrium, which way does the force F = -dV/dr point? (Answer from the sign of V', then say it in words.)

6. Two stable isomers and the barrier between them

A reaction coordinate has potential energy E(x) = x^{4} - 2x^{2} \qquad (-2 \le x \le 2), where x is a structural coordinate and E is in reduced units.

  1. Find all three stationary points and classify each.
  2. Which two are stable species, and which one is the transition state between them?
  3. What is the barrier height — the energy climb from a minimum to the top?
  4. One sentence: why does a maximum of E deserve the name “transition state”?

C. Medium: logic, limits, and traps

7. The 50:50 mixing maximum (the differentiation is done for you)

The entropy of mixing two liquids at mole fraction x is S(x) = -R\,[\,x\ln x + (1-x)\ln(1-x)\,] for 0 < x < 1. Its derivatives are given (they need a rule you don’t have yet): S'(x) = -R\,\ln\!\frac{x}{1-x}, \qquad S''(x) = -R\left[\frac{1}{x} + \frac{1}{1-x}\right].

  1. Solve S'(x) = 0. (When is a logarithm zero?)
  2. Evaluate the sign of S'' at that composition and classify.
  3. One sentence: what does this say about which mixture is the most disordered?

8. The tempting conclusion (explain why it is wrong)

For y = x^{3}, a classmate computes y'(0) = 0 and announces “a minimum at x = 0.”

  1. Compute y''(0). What verdict does the second-derivative test actually deliver?
  2. Check the sign of y' at x = -0.1 and x = +0.1. Is x = 0 a maximum, a minimum, or neither?
  3. One sentence: state the rule for what you must do whenever y'' = 0 at a stationary point.

9. The endpoint trap

A culture’s biomass grows as m(t) = 0.50\,t (grams, 0 \le t \le 8 h).

  1. Try the method: solve m'(t) = 0. What happens?
  2. Where is the maximum biomass, and what is it?
  3. One sentence: what does this teach about relying on dy/dx = 0 alone?

10. Inverse direction: from the derivative back to the curve

You are told only this about a signal s(T) measured against temperature: s'(T) is positive for T < 340 K, zero at T = 340 K, and negative for T > 340 K; furthermore s''(340) = -0.04 units/K².

  1. What is happening to s at 340 K? How confident can you be, and why?
  2. Sketch (roughly) what s(T) must look like near 340 K.
  3. Propose one chemical system that could produce such a curve.

Expectations

  1. All four steps, every problem in Part B and C: differentiate/solve, classify, endpoints, interpret.
  2. Units on every derivative and every answer with dimensions.
  3. When the second-derivative test is silent, say so explicitly and use the sign-of-y' fallback.
  4. Submit on the course submission page as Set 7, or on paper.