Solving Optimization Problems

AP Calculus · Analytical Applications of Differentiation

Solving Optimization Problems

Objective, constraint, one variable, justify — the calculus is the short part.

key idea$\text{Objective in ONE variable}\ \to\ \text{critical points}\ \to\ \text{justify}\ \to\ \text{answer the question asked}$

Setting up the model

[Figure: The classic open box: cut a square of side x from each corner of the sheet and fold. Reading which length becomes which dimension off the flat pattern is the whole setup.]

The classic open box: cut a square of side $x$ from each corner of the sheet and fold. Reading which length becomes which dimension off the flat pattern is the whole setup.

Optimisation is a set-up problem with a small calculus step in the middle. Nearly all the marks sit in building the right objective function and in justifying that your critical point is the extremum you claim.

The six moves

Name the variables and draw the situation. Decide what is being maximised or minimised — that is the objective.
Write the objective in terms of the variables. It usually starts with two.
Find the constraint — the other relationship the problem gives you (a fixed perimeter, a fixed volume, a fixed budget).
Use the constraint to eliminate one variable, so the objective is a function of a single variable.
State the domain. Physical quantities are non-negative, and the constraint usually caps them.
Differentiate, find critical points, justify, and answer the question that was asked — which is often not the variable you solved for.

Problem type	Objective	Constraint
largest area, fixed fencing	$A=xy$	perimeter fixed
cheapest can, fixed volume	surface area	$V=\pi r^{2}h$ fixed
closest point on a curve	distance (or its square)	the point lies on the curve
largest box from a sheet	$V=\ell wh$	cut dimensions from the sheet size
maximum revenue	$R=pq$	demand relates $p$ and $q$

Optimising the square of a distance gives the same location as optimising the distance, and avoids differentiating a square root. Use it whenever distance appears.

before you differentiate

Two variables in the objective means the constraint has not been used yet.
Distance problems: optimise $D^{2}$, never $D$.
A physical domain is almost always open — expect to justify with concavity.
Re-read the prompt before the final line.

Worked examples

EX 1

Fixed fencing, maximum area

$P=200$

A rectangular pen uses $200$ m of fencing. What dimensions maximise the area?

Name and draw

Let the sides be $x$ and $y$. Objective: maximise $A=xy$.

Write the constraint

$2x+2y=200$, so $y=100-x$.

Reduce to one variable

$A(x)=x(100-x)=100x-x^{2}$.

State the domain

$0\lt x\lt 100$ — both sides must be positive.

Optimise and justify

$A'(x)=100-2x=0$ gives $x=50$. $A''=-2\lt 0$, so this is a maximum. Dimensions $50\times50$ m, area $2500$ m$^{2}$.

EX 2

Closest point — optimise the square

$y=x^{2}$

Find the point on $y=x^{2}$ closest to $(0,3)$.

Write the objective

Distance $D=\sqrt{(x-0)^{2}+\left(x^{2}-3\right)^{2}}$.

Optimise $D^{2}$ instead

Minimising $D^{2}=x^{2}+\left(x^{2}-3\right)^{2}$ gives the same $x$ and avoids the root.

Expand and differentiate

$D^{2}=x^{4}-5x^{2}+9$, so $\dfrac{d\left(D^{2}\right)}{dx}=4x^{3}-10x$.

Solve

$2x\left(2x^{2}-5\right)=0$ gives $x=0$ or $x=\pm\sqrt{5/2}$.

Test and answer

$D^{2}(0)=9$; $D^{2}\left(\pm\sqrt{2.5}\right)=6.25-12.5+9=2.75$. The closest points are $\left(\pm\sqrt{2.5},\,2.5\right)$ — two of them, by symmetry.

EX 3

Cheapest can

fixed volume

A cylindrical can must hold $355$ cm$^{3}$. What radius minimises the surface area?

Objective and constraint

Minimise $S=2\pi r^{2}+2\pi rh$; constraint $\pi r^{2}h=355$.

Eliminate $h$

$h=\dfrac{355}{\pi r^{2}}$, so $S(r)=2\pi r^{2}+\dfrac{710}{r}$.

Domain

$r\gt 0$ — an open interval, so the Candidates Test does not apply.

Differentiate and solve

$S'(r)=4\pi r-\dfrac{710}{r^{2}}=0$ gives $r^{3}=\dfrac{710}{4\pi}$, so $r\approx3.84$ cm.

Justify on an open interval

$S''(r)=4\pi+\dfrac{1420}{r^{3}}\gt 0$ for all $r\gt 0$, so $S$ is concave up everywhere and this critical point is the absolute minimum.

EX 4

Answer the question asked

the last-step trap

In Example 3, the question asks for the height of the cheapest can. Finish it.

Recall what was solved

The critical point gave $r\approx3.84$ cm — a radius.

Return to the constraint

$h=\dfrac{355}{\pi r^{2}}$.

Substitute

$h=\dfrac{355}{\pi(3.84)^{2}}\approx7.67$ cm.

Note the relationship

$h\approx2r$ — the cheapest cylinder is as tall as it is wide. A pleasing result, and a useful check.

The trap

Stopping at $r$ answers a question that was not asked. Re-read the prompt before writing the final line.

Units & interpretation

Domain type	How to justify the extremum	Why
closed $[a,b]$	Candidates Test — compare $f$ at critical points and endpoints	EVT guarantees extrema exist
open $(a,b)$ or $(0,\infty)$	Second Derivative Test, or a sign chart on $f'$	no endpoints to compare
only one critical point on the domain	sign chart, or concavity throughout	a single interior extremum is absolute
any	state the units and the quantity	an unlabelled number is incomplete

Justification is worth real marks and is the step most often skipped. Say which test you used and what it showed.

Watch out

Reduce to one variable before differentiating

The constraint exists to eliminate a variable. Differentiating a two-variable objective is the standard failure.

Justify, do not assert

“$x=50$ so the area is maximised” earns nothing. Cite the Second Derivative Test or a sign chart and state what it showed.

Answer the question that was asked

Solving for $r$ when the question wants $h$, or giving dimensions when it wants the area, loses the final mark — Example 4.

State the domain and respect it

Negative lengths are not solutions. On an open domain the Candidates Test does not apply, so a different justification is needed.

Course map. Topic 5.11 · LO FUN-4.C — calculate minimum and maximum values in applied contexts (EK FUN-4.C.1).