Solve it, justify it, then answer the question AB & BC

The fence problem, all the way through. Objective $A = xy$, constraint $2x + y = 1200$, reduced to $A(x) = x(1200 - 2x)$ on $0 < x < 600$.

1. Differentiate. $A'(x) = 1200 - 4x$.
2. Solve. $1200 - 4x = 0$ gives $x = 300$, which is inside the domain.
3. Justify. $A''(x) = -4 < 0$ everywhere, so the graph is concave down and the single critical point is the absolute maximum.
4. Back-substitute. $y = 1200 - 2(300) = 600$.
5. Answer. Dimensions $300$ by $600$ feet; maximum area $180{,}000$ square feet.

The box problem, same shape. $V(x) = x(12 - 2x)^{2}$ on $0 < x < 6$, and

$$V'(x) = (12 - 2x)^{2} + x \cdot 2(12 - 2x)(-2) = (12 - 2x)(12 - 6x).$$

The roots are $x = 6$ and $x = 2$. Only $x = 2$ is inside the domain, and $V(2) = 2 \cdot 8^{2} = 128$ cubic units. The other root sits exactly at the excluded endpoint, where the volume is zero, which is a useful reminder that a root of the derivative is not automatically an answer.

The can: $S(r) = 2\pi r^{2} + \frac{710}{r}$ on $r > 0$, so $S'(r) = 4\pi r - \frac{710}{r^{2}}$, and setting it to zero gives $4\pi r^{3} = 710$, so $r = \sqrt[3]{\frac{710}{4\pi}} \approx 3.837$ centimetres. Then $h = \frac{355}{\pi r^{2}} \approx 7.675$, and $S \approx 277.5$ square centimetres.

Finding a critical number is not finding a maximum. The argument required depends on the domain, and there are three standard ones.

1. Closed interval. Use the Candidates Test from 5.5: evaluate at the critical numbers inside and at both endpoints, and compare. The table is the justification.
2. Open interval or a ray, one critical number. Use the First or Second Derivative Test, and say that since it is the only critical number, the relative extremum is absolute. For the can, $S''(r) = 4\pi + \frac{1420}{r^{3}} > 0$ for every $r > 0$, so $S$ is concave up throughout and the single critical point is the absolute minimum.
3. Open interval, several critical numbers. Run the sign chart, then compare the values at the relative extrema against the behaviour at the ends.

Three phrasings that do not justify anything: "$A'(300) = 0$", which makes it a candidate; "it is the only critical number", which is true and does not say which kind of extremum, or that there is one; and "from the graph on my calculator", which is evidence for you and not an argument on the page.

When the domain is closed, note that $V(0) = V(6) = 0$ for the box, so the interior candidate wins by inspection. When the domain is open, an endpoint that is not included can still matter: if $A$ increased all the way to a boundary it never reaches, there would be no maximum at all, and saying why that does not happen is part of the argument.

The fence problem produces at least four correct numbers, and a question asks for one of them.

1. $x = 300$ feet, the critical input.
2. $y = 600$ feet, the other dimension.
3. Dimensions $300$ by $600$ feet, if the question asks for both.
4. $180{,}000$ square feet, the maximum area.

Every one of these is a true statement about the solved problem, and three of them are wrong answers to any given prompt. The critical input is the most tempting, because it is the number that came out of the equation, and it is the one the question asks for least often.

The same split runs through the others. The can problem asks for a radius, a height, a surface area, or dimensions; the distance problem asks for a point, a value of $x$, or a distance. Note that the distance problem has one extra trap of its own: minimizing $d^{2}$ produces the smallest value of $d^{2}$, and the distance is its square root. For the point on $y = x^{2}$ nearest $(0, 3)$, the minimum of $d^{2}$ is $2.75$, and the distance is $\sqrt{2.75} \approx 1.658$.

A discipline that costs nothing: after solving, write the question's own last clause at the top of the answer line, then fill in the number underneath it.

Some answers recur often enough to be worth knowing, both as a check on the arithmetic and as a warning against applying them where they do not hold.

1. Fixed perimeter, four sides: the square wins. A rectangle of perimeter $P$ has maximum area when it is a square of side $\frac{P}{4}$.
2. Fixed volume cylinder, least surface area: $h = 2r$. The optimal can is exactly as tall as it is wide. The numbers above confirm it: $2(3.837) = 7.674 \approx h$.

Now the warning, because both of these get over-applied. The three-sided fence problem does not give a square. With $1200$ feet on three sides the answer is $300$ by $600$, and a square using the same fence would be $400$ by $400$, enclosing $160{,}000$ square feet against the optimum's $180{,}000$. The general pattern for these fence problems is that half the fence goes on the side parallel to the river, which is worth noticing and is not worth substituting for the calculation.

Similarly $h = 2r$ holds for a can with a top and a bottom. An open-topped can has one fewer disc, the surface area is $\pi r^{2} + \frac{710}{r}$, and the optimum comes out at $h = r$ instead. A remembered result applied to the wrong configuration is harder to catch than an arithmetic slip, because it looks like knowledge.