12. Integrated Models and Solution Workshop

Purpose and preparation

This workshop connects elementary methods through hypothetical business and social-science models. A complete solution states a domain, chooses a mathematical representation, derives its conclusion and checks what the conclusion means in the model. Chapters 1–4 and Chapters 5–11 provide the underlying teaching.

12.1 One question, several legitimate representations

Suppose a fictional project's price rule is $p(q)=12-q$, variable cost is $2q$ and fixed cost is $9$. Permit real production levels $0\le q\le10$ as a continuous planning approximation. Revenue is $R(q)=12q-q^2$ and profit is

$$\Pi(q)=R(q)-(2q+9)=10q-q^2-9.$$

Factoring gives $\Pi(q)=-(q-1)(q-9)$, so nonnegative profit occurs on $[1,9]$. Completing the square gives $\Pi(q)=16-(q-5)^2$, establishing a global maximum of $16$ at $q=5$ on the feasible interval. Differentiation independently gives $\Pi'=10-2q$, with positive sign before $5$ and negative sign afterward. Endpoints give $\Pi(0)=\Pi(10)=-9$, consistent with the global conclusion.

The representations answer different questions efficiently: factors locate break-even levels, the square locates the optimum, and the derivative describes local change. They are not competing answers. If only whole quantities are meaningful, the integer feasibility must be checked separately; here $5$ is already integral. The revenue maximum is instead at $q=6$, with $R(6)=36$, showing why “maximize revenue” and “maximize profit” cannot be interchanged.

12.2 A rate, its accumulated total, and a non-elementary inverse

Consider a hypothetical processing rate $r(t)=(t+1)e^{-t}$ for $t\ge0$, with $t$ in hours and compatible rate units. The total processed by time $T$ is

$$A(T)=\int_0^T(t+1)e^{-t}\,dt=2-(T+2)e^{-T}.$$

To derive the formula, parts gives $\int t e^{-t}dt=-(t+1)e^{-t}$; adding $\int e^{-t}dt=-e^{-t}$ produces $-(t+2)e^{-t}$. Endpoint subtraction gives the displayed total, and differentiation returns $A'=r$ as a check.

The rate is positive, so $A$ is strictly increasing. Since $(T+2)e^{-T}>0$, $A(T)<2$. The exponential dominates the linear factor, as justified by the bound in Chapter 11 Exercise 11, so $A(T)\to2$. Thus the range is $[0,2)$ and there is a unique inverse time for every total in that range. A target total of $2$ has no finite solution.

There is an inverse, but it need not have an expression in the elementary functions taught here. Solving $A(T)=1$ yields $(T+2)e^{-T}=1$, whose logarithm is $\ln(T+2)-T=0$. The variable occurs in both a logarithm and a linear term; taking logs has not isolated it. Evaluating $A(1)\approx0.896$ and $A(2)\approx1.459$ brackets the unique solution between one and two hours by continuity and monotonicity. An exact formula is not necessary to establish existence, uniqueness and a useful interval.

12.3 A periodic level versus a periodic rate

For a hypothetical weekly workload index observed over months, let

$$Q(t)=100+20\cos(\pi t/6),\qquad0\le t\le12.$$

The period is $12$ months, the range is $[80,120]$, and

$$Q'(t)=-\frac{10\pi}{3}\sin(\pi t/6),\qquad Q''(t)=-\frac{5\pi^2}{9}\cos(\pi t/6).$$

The level decreases from $0$ to $6$ months and increases from $6$ to $12$. It has a minimum of $80$ at $6$ and maxima of $120$ at the two interval endpoints. Its net change is

$$Q(12)-Q(0)=\int_0^{12}Q'(t)\,dt=0.$$

That is not the same as $\int_0^{12}Q(t)dt=1200$, whose units are workload-index times months. If the intended quantity were a rate measured in tasks per month, that latter integral would instead have units tasks. Algebra cannot determine which interpretation was intended; the model definition must do so.

12.4 Graded integrative exercises

Foundation-to-core transfer

1. An arithmetic schedule begins with $10$ tasks and adds $2$ each week. A geometric schedule begins with $10$ and multiplies the weekly number by $1.1$. Find each fifth-week level and each five-week total. Which comparisons are about levels and which about accumulation?
2. Use the binomial theorem to expand $(1+h)^4$, then prove $(1+h)^4\ge1+4h$ for every $h\ge-1$ by Bernoulli's inequality. Does discarding all but the first two binomial terms prove the latter inequality for negative $h$?
3. Let $S(t)=60(1-e^{-t/3})$ on $t\ge0$. Find its range, inverse on that range, first two derivatives and the time at which $S=30$.
4. A reported indicator is $f(x)=x^2$ for $x\le0$ and $f(x)=\sin x$ for $x>0$. Determine continuity and differentiability at zero from the definitions.

Synthesis

5. For the profit model of §12.1, add a capacity restriction $0\le q\le4$. Find the best feasible profit and explain why setting the derivative equal to zero alone is insufficient. Repeat with capacity $q\le4$ and nonnegative integer $q$.
6. For $g(x)=(x^2+2x)/(x^2+x-2)$, find its maximal real domain, intercepts, holes, vertical asymptotes and horizontal asymptote. Give one-sided signs at each vertical asymptote.
7. Evaluate $\int_0^1x\ln(1+x)dx$ using parts followed by algebra. Give a simple upper bound from $\ln(1+x)\le x$ and check your exact answer against it.
8. Explain and repair all three errors: (a) “the inverse of $x^2$ on $\mathbb R$ is $\sqrt x$”; (b) “$\lim_{h\to0}\sin(5h)/h=1$”; (c) “$\int_0^1x e^{x^2}dx=e+C$.”

12.5 Full solutions with method commentary

1. The arithmetic fifth term is $10+4(2)=18$, and the total is $5(10+18)/2=70$. The geometric fifth term is $10(1.1)^4=14.641$, while its total is $10[(1.1)^5-1]/0.1=61.051$. Comparing $18$ with $14.641$ compares a single week's levels; comparing $70$ with $61.051$ compares five-week accumulation. A geometric schedule need not initially exceed an arithmetic one just because its growth is proportional.

2. The expansion is $1+4h+6h^2+4h^3+h^4$. Bernoulli's inequality from Chapter 6 directly proves the bound on $h\ge-1$. For negative $h$, the cubic term is negative, so termwise deletion of all the extra terms without reasoning is not justified. A correct direct check groups the remainder as $h^2(h^2+4h+6)=h^2[(h+2)^2+2]\ge0$, in fact proving this particular fourth-power inequality for every real $h$.

3. The range is $[0,60)$. Solving $s=60(1-e^{-t/3})$ gives $t=-3\ln(1-s/60)$ for $0\le s<60$. The derivatives are $S'=20e^{-t/3}>0$ and $S''=-(20/3)e^{-t/3}<0$. At $s=30$, $t=3\ln2$. The level increases while its rate decreases; the limiting level $60$ is not attained in finite time.

4. The left limit is zero and the right limit is zero, using sine continuity; $f(0)=0$, so it is continuous. At zero the difference quotient is $h$ for $h<0$ and $\sin h/h$ for $h>0$. Their limits are $0$ and $1$, so no two-sided derivative exists.

5. On $[0,4]$, $\Pi'=10-2q\ge2>0$, so profit is increasing. The maximum is at $q=4$ with value $40-16-9=15$. The stationary input $5$ is not feasible. Integer feasibility changes nothing in this particular case because $4$ is an admissible integer endpoint. In a different model a noninteger capacity or stationary input would require a separate integer comparison.

6. Factor to get $g(x)=x(x+2)/[(x+2)(x-1)]$. The original domain excludes $-2,1$ and the simplified expression is $x/(x-1)$ there. The only intercept is $(0,0)$; $-2$ is not a root because it is excluded. There is a hole at $(-2,2/3)$ and a vertical asymptote at $1$. Since $g=1+1/(x-1)$, its left limit at $1$ is $-\infty$ and right limit $+\infty$. The horizontal asymptote at both ends is $y=1$.

7. Let $u=\ln(1+x)$, $dv=x\,dx$. Parts gives

   $$\int_0^1x\ln(1+x)dx=\tfrac12\ln2-\tfrac12\int_0^1\frac{x^2}{1+x}dx.$$

   Divide: $x^2/(1+x)=x-1+1/(1+x)$. Its integral is $1/2-1+\ln2=\ln2-1/2$. The exact answer is therefore $1/4$. For $x\ge0$, the area definition gives $\ln(1+x)\le x$, so the original integral is at most $\int_0^1x^2dx=1/3$. The positive exact value $1/4$ passes this bound.

8. (a) Squaring is not injective on $\mathbb R$. On the restricted domain $[0,\infty)$ it has inverse $\sqrt x$ on $[0,\infty)$; on $(-\infty,0]$ it has inverse $-\sqrt x$. (b) Rewrite the quotient as $5[\sin(5h)/(5h)]$, giving limit $5$ with radian inputs. (c) Substitution $u=x^2$ contributes a factor $1/2$ and changes bounds from $0,1$ to $0,1$; the value is $(e-1)/2$. A definite integral is a number and has no arbitrary $C$.

Summary: a reliable solution protocol

Translate before calculating: identify input set, units, model assumptions and the requested output. Choose representations deliberately: factors for roots, squares for global quadratic bounds, inverse functions for reverse questions, derivatives for rates, integrals for accumulated change. Then check domain, endpoints, signs and scale. A continuous model may require integer feasibility afterward. A named inverse may exist without an elementary formula, and a finite limiting level may lie outside the actual range.