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Calculus BC · Unit 5 · 5.10
AP Calculus · Analytical Applications of Differentiation
Introduction to Optimization Problems
Two sentences in the prompt: one names what you optimise, the other states what limits you.
key idea$\text{Objective}=\text{what you optimise};\quad \text{Constraint}=\text{what limits you}$
Setting up the model
Before any calculus, an optimisation problem has to be translated. Two sentences in the prompt do all the work: one names the thing being maximised or minimised, the other states the limitation. Telling them apart is this topic.
Reading the prompt
- Find the objective. Look for “largest”, “smallest”, “maximum”, “minimum”, “cheapest”, “most efficient”. The noun after that word is what you optimise.
- Find the constraint. Look for a fixed quantity — “$200$ m of fencing”, “must hold $355$ cm$^{3}$”, “a budget of $500”. It is always an equation.
- Write both as equations in the same variables.
- Identify the domain from physical sense: lengths positive, quantities non-negative, and the constraint caps the rest.
| Phrase in the prompt | Role | Becomes |
|---|
| “maximise the area” | objective | $A=\dots$ — the function to optimise |
| “using 200 m of fencing” | constraint | $2x+2y=200$ |
| “must hold 355 cm$^{3}$” | constraint | $\pi r^{2}h=355$ |
| “minimise the cost” | objective | $C=\dots$ |
| “the field borders a river” | constraint modifier | only three sides need fencing |
| “an open-top box” | objective modifier | surface area omits the lid |
The last two rows change the equations without announcing themselves. Read the whole sentence, not just the numbers.
the distinction
- The objective is what the question asks about.
- The constraint is what the situation fixes.
- Two variables plus one constraint gives one free variable.
- Sketching the situation catches most translation errors.
Worked examples
EX 1
Identify both, do not solve
translation only“A farmer has $400$ m of fencing to enclose a rectangular field. What dimensions give the largest area?” Name the objective and the constraint.
1
Find the optimisation word
“largest area”, so area is the objective.
2
Write the objective
$A=xy$ for a rectangle with sides $x$ and $y$.
3
Find the fixed quantity
$400$ m of fencing is fixed — that is the constraint.
4
Write the constraint
$2x+2y=400$.
5
State the domain
$0\lt x\lt 200$, since $x$ and $y=200-x$ must both be positive.
EX 2
A modifier changes the constraint
riverSame farmer, same $400$ m — but one side of the field borders a straight river and needs no fence. Rewrite the constraint.
1
Identify what changed
Only three sides are fenced, not four.
2
Choose the labelling
Let $x$ be the two sides perpendicular to the river and $y$ the side parallel to it.
3
Write the new constraint
$2x+y=400$ — one $y$, not two.
4
Keep the objective
$A=xy$ is unchanged; the shape is still a rectangle.
5
Note the effect
The optimal field is now $100\times200$ rather than square. A single unfenced side changes the answer entirely.
EX 3
Objective and constraint swapped
diagnoseA student writes: “objective $2x+2y=400$, constraint $A=xy$”. What went wrong, and does it matter?
1
Check the question
It asks for the largest area, so area is what is being optimised.
2
Name the error
The student has the roles reversed: the fencing equation is the fixed limitation, not the thing to maximise.
3
Ask whether it matters
Yes. Maximising $2x+2y$ subject to $xy$ fixed is a different problem with a different answer.
4
Give the test
The objective is what the question asks about; the constraint is what the situation fixes at a stated number.
5
A quick check
The constraint always contains a specific number from the prompt. The objective usually does not.
EX 4
Reduce to one variable
the mechanical stepFor the original field, reduce the objective to a single variable and state the domain.
1
Solve the constraint
$2x+2y=400$ gives $y=200-x$.
2
Substitute into the objective
$A(x)=x(200-x)=200x-x^{2}$.
3
Check the variable count
One variable, $x$. This is now an ordinary max-min problem.
4
State the domain
$x\gt 0$ and $y=200-x\gt 0$, so $0\lt x\lt 200$.
5
Note what comes next
Everything from here is the standard critical-point procedure. The translation was the part specific to this topic.
Units & interpretation
| Quantity | Typical units | Domain restriction |
|---|
| length, width | m | strictly positive |
| area | m$^{2}$ | positive |
| volume | cm$^{3}$ or m$^{3}$ | positive |
| cost | dollars | non-negative |
| number of items | items | non-negative integer — note the extra restriction |
Physical domains are almost always open intervals, because a zero length collapses the object. That affects how the extremum must later be justified.
Watch out
Objective and constraint are not interchangeable
Reversing them produces a different problem with a different answer — Example 3.
Read the modifiers
“Borders a river”, “open-top”, “two identical pens” all silently change an equation. They rarely come with a number attached.
Physical domains exclude zero
A rectangle with a side of $0$ is not a rectangle. Use open intervals, and remember the Candidates Test then does not apply.
Sketch it
Most translation errors — a missing side, a wrong labelling — become obvious the moment the situation is drawn.