Appendix J — Problem Set 8 | Constrained Extrema

Companion to §4.4 — optimizing with a budget

Purpose

Chemistry fixes totals: total concentration, total length, total energy. When two quantities must share a budget, the best split is a constrained extremum. This set trains one method, with the §4.3 steps inside it:

Name the objective f (what you want largest or smallest) and the constraint g (the budget). Count the free variables — one budget removes one.
Eliminate: solve the constraint for one variable and substitute, so f becomes a function of a single variable. (Equivalent check: \nabla f = \lambda\nabla g.)
Solve and classify: set the derivative to zero, classify with the second derivative, and check the endpoints of the allowed range.
Interpret: units, one sentence of chemistry, and — when asked — what \lambda means.

Read §4.4 first; the two interactive figures and the constrained-extrema studio are worth ten minutes before you start.

Worked example — read this before the problems

The best feed for a termolecular step. The elementary reaction 2\mathrm{A}+\mathrm{B}\to\mathrm{P} has initial rate k[\mathrm{A}]^{2}[\mathrm{B}]. With the total concentration capped at [\mathrm{A}]+[\mathrm{B}] = 0.90 M, find the feed that maximizes the rate, and the value of q = [\mathrm{A}]^{2}[\mathrm{B}] there.

Step 1 — name and count. Objective q = [\mathrm{A}]^{2}[\mathrm{B}] (units M³). Constraint g = [\mathrm{A}]+[\mathrm{B}] = 0.90 M. Two variables, one constraint: one free variable. Domain 0 \le [\mathrm{A}] \le 0.90.

Step 2 — eliminate. [\mathrm{B}] = 0.90 - [\mathrm{A}], so q([\mathrm{A}]) = [\mathrm{A}]^{2}\,(0.90 - [\mathrm{A}]) = 0.90\,[\mathrm{A}]^{2} - [\mathrm{A}]^{3}.

Step 3 — solve and classify. By the power rule, q'([\mathrm{A}]) = 1.80\,[\mathrm{A}] - 3\,[\mathrm{A}]^{2} = 3\,[\mathrm{A}]\,(0.60 - [\mathrm{A}]) = 0, so [\mathrm{A}] = 0 or 0.60 M. Then q'' = 1.80 - 6[\mathrm{A}]: q''(0.60) = -1.80 < 0, a cap and a maximum; q''(0) = +1.80 > 0, a minimum (where q = 0). Endpoints: q(0) = 0 and q(0.90) = 0. The interior maximum wins: q_{\max} = (0.60)^{2}(0.30) = 0.108\ \text{M}^{3}.

Step 4 — interpret. The best feed is 0.60 M A and 0.30 M B — a 2\!:\!1 ratio, which is exactly the ratio in which 2\mathrm{A}+\mathrm{B} consumes the reagents.

Check. Test a neighbouring split: 0.50/0.40 would be 0.25 \times 0.40 = 0.100 < 0.108 ✓. Gradient check: \nabla q = (2[\mathrm{A}][\mathrm{B}],\,[\mathrm{A}]^{2}) = (0.36,\,0.36) M² = \lambda\,(1,1) with \lambda = 0.36 M² — parallel to \nabla g = (1,1) ✓, and the units of \lambda are M³ per M, as the “objective per budget” rule predicts.

Every problem below uses this four-step rhythm. Show all four steps each time.

A. Reading before computing

1. The contour story

On a contour map of an objective q, a constraint line crosses the contour q=1.0 at two points, touches the contour q = 1.7 at exactly one point, and never reaches the contour q = 2.5.

Where on the line is the maximum of q?
What is the slope of q along the line at that point?
Why can the maximum not be at either point where the line merely crosses the q=1.0 contour?

2. Objective, constraint, free variables

For each system, name the objective, the constraint, and how many free variables remain:

Initial rate k[\mathrm{A}][\mathrm{B}] of \mathrm{A}+\mathrm{B}\to\mathrm{P} with [\mathrm{A}]+[\mathrm{B}] = 0.50 M.
Strain energy of two bonds in series when the two ends of the chain are held a fixed distance apart.
Total entropy of two bodies that share a fixed total energy.
One sentence: what do (a)–(c) have in common that lets the §4.3 method apply?

B. The method, start to finish

3. The even split, with a new budget

For q = [\mathrm{A}][\mathrm{B}] with [\mathrm{A}]+[\mathrm{B}] = 1.50 M: find the optimum split, the maximum q (with units), classify it with the second derivative, check the endpoints, and find \lambda from \nabla q = \lambda\nabla g.

4. When the exponents differ

The elementary step \mathrm{A}+3\mathrm{B}\to\mathrm{P} has initial rate k[\mathrm{A}][\mathrm{B}]^{3}, with [\mathrm{A}]+[\mathrm{B}] = 1.20 M.

Find the feed that maximizes q = [\mathrm{A}][\mathrm{B}]^{3}. (Hint: after eliminating, use the product rule and the chain rule on (1.20-[\mathrm{A}])^{3} — or check against the gradient condition a/[\mathrm{A}] = b/[\mathrm{B}] from §4.4.)
Classify it and give the maximum q with units.
Compare with the even split [\mathrm{A}]=[\mathrm{B}]=0.60 M. How much smaller is the rate there?
One sentence: what is the feed ratio, and how does it relate to the reaction stoichiometry?

5. Who stretches more?

Two bonds in a chain have strain energy E = 3x_1^{2} + x_2^{2} (reduced units), where x_1,x_2 \ge 0 are their extensions. The chain’s ends are held so that x_1 + x_2 = 4.

Find x_1, x_2 and the minimum E. Classify with E'' and check both endpoints.
Compute the force in each bond, F_1 = dE_1/dx_1 and F_2 = dE_2/dx_2. What do you notice, and what is its relation to \lambda?
Which bond stretches more? Compare the ratio x_1\!:\!x_2 with the ratio of the stiffnesses (3\!:\!1 in the coefficients of E).
One sentence of chemistry: which bond of a loaded chain carries most of the extension?

6. The multiplier is a price

For Problem 3 the best value is q^{*}(C) = C^{2}/4. Compute dq^{*}/dC at C = 1.50 M and compare with your \lambda.
Estimate \big[q^{*}(1.60) - q^{*}(1.50)\big]/0.10. Is it close to \lambda? Why is it slightly different?
For Problem 5 the best strain is E^{*}(\Delta) = \tfrac34\Delta^{2} with \Delta = x_1+x_2. Compute dE^{*}/d\Delta at \Delta = 4 and compare with the force in (5b).
One sentence for each part: what does \lambda tell the experimentalist?

C. Medium: traps, physics, and reasoning backwards

7. The tempting error (explain why it is wrong)

A classmate maximizes q = [\mathrm{A}][\mathrm{B}] subject to [\mathrm{A}]+[\mathrm{B}] = C by setting both partial derivatives to zero: \partial q/\partial[\mathrm{A}] = 0 and \partial q/\partial[\mathrm{B}] = 0.

Carry this out. What point do you get, and what is q there?
Explain, in terms of the constraint, why that is not the answer.
State the condition that replaces “\nabla q = 0” for a constrained problem.

8. Why heat flows until temperatures match

Two bodies share a fixed total energy E_1+E_2 = 10 (arbitrary units, k_B = 1). Their total entropy is S = 3\ln E_1 + 2\ln E_2, and each body’s temperature obeys 1/T_i = dS_i/dE_i.

Eliminate E_2 and find the E_1 that maximizes S.
Check the second derivative is negative.
Evaluate 3/E_1 and 2/E_2 at the optimum. What do the two numbers represent, and what is their common value \lambda in terms of temperature?
Find T_1 = E_1/3 and T_2 = E_2/2 at the optimum. What do you notice?
Suppose the bodies start at E_1 = 8. Compute dS/dE_1 there. Which way does energy flow, and which body is hotter?
One sentence: why does heat flow from hot to cold, in terms of entropy?

9. Inverse direction: from the optimum back to the gradient

A function f, restricted to the line x+y=4, has its maximum at a point where \partial f/\partial x = 5. What must \partial f/\partial y be there?
Now the constraint is 2x + y = 6 and, at the constrained optimum, \partial f/\partial x = 8. What is \partial f/\partial y?
One sentence: how must \nabla f be oriented relative to the constraint line at an optimum, and why?

Expectations

All four steps in every Part B and C problem: name and count, eliminate, solve and classify and check endpoints, interpret.
Units on every derivative, every optimum, and every \lambda.
A multiplier is always (objective units) per (budget units); say what it means physically whenever asked.
Submit on the course submission page as Set 8, or on paper.