Two equations in, one function out AB & BC

Two sentences to write before touching a derivative.

1. The objective is the quantity the question asks to make largest or smallest. Area, volume, cost, distance, revenue.
2. The constraint is the relationship that limits the choices. A fixed amount of fence, a fixed volume, a fixed budget, a curve a point has to lie on.

They are told apart by the question, not by the algebra. "Maximize the area" makes area the objective, and whatever the problem says is fixed is the constraint. Getting them the wrong way round produces a well-formed calculation of the wrong thing.

A farmer has 1200 feet of fence for three sides of a rectangular plot, the fourth side being a river, and wants the largest possible area. Let $x$ be each of the two sides perpendicular to the river and $y$ the side parallel to it.

1. Objective: $A = xy$.
2. Constraint: $2x + y = 1200$.

Without a constraint the area could be made as large as you like, and the problem would have no answer. The constraint is what makes it a question.

The objective has two variables and the tools of this unit take one. So solve the constraint for whichever variable is easier and substitute.

$$y = 1200 - 2x \;\Longrightarrow\; A(x) = x(1200 - 2x) = 1200x - 2x^{2}.$$

Now there is a single function of a single variable, and everything from 5.3 through 5.5 applies to it.

The failure to avoid here has a specific shape. Starting from $A = xy$ and computing $\frac{dA}{dx} = y$, on the grounds that $y$ is a different letter and therefore a constant, is wrong because $y$ is not free: choosing $x$ determines it. Treating it as a constant asserts that lengthening one side leaves the other alone, which is exactly what the fixed amount of fence forbids.

Two practical notes on doing the substitution:

1. Solve for the variable that appears once, or linearly. In $2x + y = 1200$ that is $y$. Substituting for $x$ would work and would be messier.
2. Substitute into the objective, not into the constraint. Substituting the constraint into itself produces $1200 = 1200$, which is true and useless.

A cylindrical can holding 355 cubic centimetres with the least metal works the same way. Objective $S = 2\pi r^{2} + 2\pi rh$, constraint $\pi r^{2}h = 355$. Solving the constraint gives $h = \frac{355}{\pi r^{2}}$, and

$$S(r) = 2\pi r^{2} + 2\pi r \cdot \frac{355}{\pi r^{2}} = 2\pi r^{2} + \frac{710}{r}.$$

$A(x) = 1200x - 2x^{2}$ is a polynomial, so as an abstract function its domain is every real number. As a description of a field it is not, and using the wrong one is how an answer of $x = -300$ survives to the end of a solution.

Two questions produce the domain: what makes each quantity non-negative, and what makes the constraint satisfiable.

1. $x > 0$, since a side has positive length.
2. $y = 1200 - 2x > 0$, so $x < 600$.

So $0 < x < 600$, and every candidate outside it is discarded on sight rather than tested.

Take a $12 \times 12$ sheet with squares of side $x$ cut from the corners and the sides folded up. Then $V(x) = x(12 - 2x)^{2}$, and the domain is $0 < x < 6$: below zero there is nothing to cut, and at $x = 6$ the two cuts meet and the base has no width left. Trying $x = 8$ gives $V = 8 \cdot 16 = 128$, a perfectly respectable number describing a box that cannot be built.

Three kinds of restriction worth watching for beyond positivity:

1. Counting. A number of people or items has to be a whole number, so the calculus answer may need rounding, and both neighbours checked.
2. A stated cap. "At most 60 passengers" is part of the domain.
3. Where a formula stops applying. A discount rule that runs out is an endpoint.

Everything above, in the order to write it, with a picture first whenever there is anything to draw.

1. Draw and label. Give every varying quantity a letter, and mark what is fixed.
2. Objective. Write the quantity to be optimized as an equation.
3. Constraint. Write the fixed relationship as an equation.
4. Reduce. Solve the constraint and substitute into the objective.
5. Domain. State it, with a reason for each end.

One more setup move worth knowing now, because it saves work in 5.11. When the objective is a distance, minimize the square of the distance instead. Squaring is increasing on non-negative numbers, so $d$ and $d^{2}$ are smallest at exactly the same input, and $d^{2}$ has no square root to differentiate through.

For the point on $y = x^{2}$ closest to $(0, 3)$: the objective is $d = \sqrt{x^{2} + (y - 3)^{2}}$ and the constraint is that the point lies on the curve, so $y = x^{2}$. Reducing gives

$$d^{2}(x) = x^{2} + (x^{2} - 3)^{2},$$

with domain all real $x$, since every real $x$ names a genuine point of the parabola. Not every problem restricts its domain; what every problem requires is that you check.