Applied optimization

Every applied optimization problem — however different the story — reduces to the same seven moves.

Set up. First name the quantity to optimize, $Q$, and write it in terms of whatever variables the geometry naturally offers; usually there are two variables at this stage, linked by one constraint — a fixed total (fencing, surface area, sheet perimeter) that the problem hands you for free.

Reduce and solve. Use the constraint to eliminate one variable, so $Q$ becomes a function of a single variable $x$. Differentiate and solve $Q'(x)=0$ for critical points, then restrict those points to the domain where the setup makes physical sense — a length or an area cannot be negative or zero.

where: $Q(x)$ is the quantity being optimized written as a function of the single surviving variable $x$, $Q'(x)$ is its derivative, and $x_0$ is a critical point; the sign of the second derivative $Q''(x_0)$ classifies it without needing to check neighbouring values by hand. On a closed domain, any candidate critical point must also be compared against the endpoint values of $Q$, since the true maximum or minimum can sit at a boundary rather than at a critical point.

Report. Finally, substitute $x_0$ back to state the optimal dimensions — not just the value of $x$ — and give the optimized value of $Q$ itself, in the problem's own units.

An open-top rectangular box is made from a square cardboard sheet of side $12$ cm, by cutting equal squares of side $x$ from each corner and folding up the sides. Find the value of $x$ that maximizes the volume.

Step 1. After folding, the base is a square of side $12-2x$ and the height is $x$, so the volume is $V(x) = x(12-2x)^2$, already a function of the single variable $x$ because the sheet's fixed size is built directly into the formula.

Step 2. Determine the domain. Both $x>0$ and $12-2x>0$ are required for the box to exist, so $0<x<6$.

Step 3. Expand and differentiate. $V(x) = 144x - 48x^2 + 4x^3$, so $V'(x) = 144-96x+12x^2 = 12(x-2)(x-6)$.

Step 4. Solve $V'(x)=0$ within the domain: $x=2$ or $x=6$. Since $x=6$ makes the base side zero and lies outside the open domain $(0,6)$, only $x=2$ is a valid critical point.

Step 5. Confirm a maximum. $V''(x) = -96+24x$, so $V''(2) = -48 < 0$, confirming $x=2$ is a maximum.

Step 6. Compute the maximum volume. $V(2) = 2(12-4)^2 = 2(64) = 128$. Cutting squares of side $x=2$ cm gives base $8\times8$ cm and height $2$ cm, so the maximum volume is $\boxed{128\ \text{cm}^3}$.

Competitive stretch

A closed cylindrical can of fixed volume $V_0$ minimizes surface area — and hence material — when its height equals its diameter, $h=2r$. The setup is the same recipe: write surface area $S = 2\pi r^2 + 2\pi rh$ as the quantity to minimize, use the constraint $V_0 = \pi r^2 h$ to eliminate $h = V_0/(\pi r^2)$, reduce $S$ to a single variable $S(r) = 2\pi r^2 + 2V_0/r$, and solve $S'(r)=0$. This is exactly the kind of constraint-elimination step drilled above, just with a volume constraint instead of a fencing or sheet-size constraint.

• Differentiating a two-variable expression directly, without first using the constraint to reduce it to one variable — the single most common setup error.
• Accepting a critical point that is outside the physically meaningful domain, such as $x=6$ in the box problem, which forces a dimension to zero.
• Skipping the first- or second-derivative test and assuming the only critical point found must automatically be the answer.
• Reporting the optimal value of $x$ alone and forgetting to compute the actual optimized quantity — the maximum area or volume — that the question asked for.

Ask the tutor: stuck setting up the constraint for a new optimization story, like minimizing the material for a closed cylindrical can — ask the AI tutor to walk through the seven-step recipe with you on that exact problem.