Translating a description into an objective function, using a constraint to reduce it to a single variable, determining the physically meaningful domain, and applying the extreme-value machinery — worked through fencing, an open box, a minimal-surface can, a shortest distance and a revenue problem.
Subject: Calculus I · 65 slides · symbolic lesson
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Title
Calculus I · Chapter 4 — Applications of Derivatives
Applied Optimization Problems
Objectives
Five outcomes, and only the last is calculus. The other four are the translation, which is where these problems are won or lost.
OpenStax Calculus Volume 1, §4.7 Applied Optimization Problems §4.7, pp. 381-393 — the section these objectives are drawn from
Warm-up
Section 4.3 gave a complete method for finding absolute extrema on a closed interval, and Section 4.5 gave the tests for classifying critical points.
Discussion prompt
A farmer has 100 metres of fence for a rectangular pen. What is the largest area, and what makes this harder than the problems of Section 4.3?
Hint: You are not given a function to optimise; you have to construct one.
Answer:
\[ 2x + 2y = 100 \;\Longrightarrow\; y = 50 - x \;\Longrightarrow\; A(x) = x(50-x) \]
Once that function exists, Section 4.3 finishes it in three lines: the derivative 50 minus 2x vanishes at 25, and the area there is 625 square metres.
What is new is the two lines before it. The problem supplied a description, not a function, and constructing the right function from it is the entire difficulty of this section — the calculus that follows is already familiar.
Concept
An optimisation problem gives a quantity to make largest or smallest and a condition that limits the choices. Writing both as formulas and using the condition to eliminate variables produces a function of one variable, which the earlier machinery then handles.
objective and constraint — The objective is the quantity to be optimised; the constraint is the condition that must hold. The constraint is used to express the objective in terms of a single variable before any differentiation.
\[ \text{objective } Q(x,y), \; \text{constraint } g(x,y)=c \;\Longrightarrow\; Q(x) \]
Both formulas are always needed and they are easy to swap. The constraint is whichever quantity the problem holds fixed; the objective is whichever it asks you to make largest or smallest.
Figure (svg): The five steps of an optimisation problem, with the translation stage marked as the hard one
OpenStax Calculus Volume 1, §4.7 Applied Optimization Problems §4.7, pp. 381-384
Section
Section 1
Concept
Every optimisation problem contains two quantities. The constraint is fixed by the situation and provides an equation; the objective is what you are asked to maximise or minimise. Identifying which is which is the first and most important step.
identifying the two — The constraint is the quantity the problem states as given or fixed; the objective is the quantity the problem asks about. Reading the question's verb — maximise, minimise, largest, cheapest — identifies the objective.
\[ \text{'100 m of fence'} \to \text{constraint}; \quad \text{'largest area'} \to \text{objective} \]
Swapping them is not merely inconvenient but fatal. Maximising the perimeter subject to a fixed area gives an unbounded answer, since a long thin rectangle of fixed area has arbitrarily large perimeter.
Figure (svg): The objective and the constraint distinguished, on a fencing problem
OpenStax Calculus Volume 1, §4.7 Applied Optimization Problems §4.7, pp. 381-385 — setting up optimisation problems
Picture it
A fencing problem with both quantities labelled.
Figure (svg): The objective and the constraint distinguished, on a fencing problem
The perimeter is stated as 100 and cannot change; the area is what varies as the shape is chosen. That distinction determines which formula gets differentiated.
Worked example
Example 4.32. Construct, reduce, optimise.
\[ \text{With } 100 \text{ m of fence, find the rectangle of largest area.} \]
Name the variables
Why: Two side lengths.
Write the objective
Why: The quantity to maximise.
\[ A = x y \]
Write the constraint
Why: The fixed perimeter.
\[ 2 x + 2 y = 100 \]
Use the constraint to eliminate y
Why: Solve and substitute.
\[ y = 50 - x,\text{ so } A(x) = x(50 - x) \]
Optimise
Why: Differentiate and solve.
\[ \text{A' } = 50 - 2 x = 0\text{ at } x = 25 \]
Figure (svg): The physical domain of a fencing problem, showing where the model stops making sense
\[ x = y = 25, \quad A = 625 \text{ m}^{2} \]
Verify: check the endpoints and the shape of the answer
Why: At x equal to 0 or 50 the area is 0, so the interior critical point is indeed the maximum — the closed-interval method confirms it without any derivative test. The answer being a SQUARE is worth noticing: among all rectangles of fixed perimeter the square is largest, which is a general fact this computation has just demonstrated. An answer that turns out to be a familiar shape is usually a sign the setup was right.
OpenStax Calculus Volume 1, §4.7 Applied Optimization Problems §4.7, pp. 383-384
Sorting
Which quantity is fixed and which is being chosen?
Sort into buckets
Sort each stated quantity.
The verb identifies the objective every time: maximise, minimise, largest, cheapest, shortest. Anything stated as available, given or exactly equal to something is a constraint.
Worked example
Checkpoint 4.32. The reversed problem has no answer.
\[ \text{Why can a rectangle of fixed area not have a largest perimeter?} \]
Write the constraint
Why: Now the area is fixed.
\[ x y = 100 \]
Write the objective
Why: Now the perimeter.
\[ P = 2 x + 2 y \]
Eliminate
Why: Substitute for y.
\[ P(x) = 2 x + \frac{200}{x} \]
Examine the behaviour
Why: As x shrinks toward zero.
Conclude
Why: No largest value exists.
Figure (svg): The solution to Worked example what swapping them does shown as a ladder of expressions, one row per legal move
\[ P(x) = 2x + \frac{200}{x} \;\to\; \infty \text{ as } x \to 0^{+} \]
Verify: notice what the problem DOES have
Why: There is no maximum, but there is a minimum: differentiating gives 2 minus 200 over x squared, which vanishes at x equal to 10 — a square again, with perimeter 40. So the reversed problem is not meaningless, it simply asks for the wrong extreme. This is the same square-is-optimal fact from the other side, and it shows that identifying the objective correctly determines whether the problem has an answer at all.
OpenStax Calculus Volume 1, §4.7 Applied Optimization Problems §4.7, pp. 384-385
Trap
\[ 2x + 2y = 100 \text{ with } y = 50-x \]
Differentiate the perimeter formula
Why: The student differentiates the equation that was given.
\[ P'(x) = 2 - 2 = 0 \text{ for every } x \quad \text{(no information)} \]
The perimeter is constant by assumption, so its derivative is zero everywhere and nothing has been learned.
\[ A(x) = x(50-x) \;\Longrightarrow\; A'(x) = 50 - 2x \]
Differentiate the OBJECTIVE, after using the constraint to reduce it
Why: The constraint is an input to the reduction, not the thing being optimised.
The symptom is unmistakable: differentiating a constraint gives a derivative that vanishes identically, because the quantity was fixed from the start. If your derivative carries no information, the wrong formula was differentiated.
Fill the middle
The fencing problem, with the constraint solved for y.
Fill in the blanks
y = 50 - x \;\Longrightarrow\; A(x) = x(50 - x)
Why: Substituting the constraint into the objective leaves a function of x alone, which is what the extreme-value machinery needs. The reduction always goes this way round: the constraint feeds into the objective.
Ranking
Turning a description into a function.
Put in order
Why: Step b before step c matters: writing formulas before deciding which is the objective is how the two get swapped. The drawing in step a is not decoration — it is what makes the relationships between the named quantities visible.
Prediction
Commit before reasoning.
Predict first
You differentiate and get a derivative that is zero for every x. What has happened?
Correct: The constraint was differentiated instead of the objective.
\[ 2x+2y = 100 \text{ always} \;\Longrightarrow\; \frac{dP}{dx} = 0 \text{ always} \]
Why: A constraint states that some quantity is fixed, so its derivative vanishes identically — that is what fixed means. An identically zero derivative therefore carries no information and signals that the wrong formula was differentiated. The fix is to go back and check which quantity the problem asked about, and differentiate that one instead after using the constraint to reduce it.
Section
Section 2
Concept
The reduced objective is usually defined for far more inputs than the situation allows. Lengths cannot be negative, a cut cannot exceed half the sheet, and a quantity sold cannot be negative — and the domain must be restricted accordingly.
the feasible domain — The set of inputs for which the physical situation makes sense. It is usually a closed or half-open interval, and its endpoints are frequently degenerate cases where the objective is zero.
\[ 0 \le x \le 50 \quad \text{for the fencing problem} \]
Determining the domain is not a formality. It decides whether the closed-interval method applies, it supplies the endpoints that must be checked, and it sometimes contains the optimum itself.
Figure (svg): The physical domain of a fencing problem, showing where the model stops making sense
OpenStax Calculus Volume 1, §4.7 Applied Optimization Problems §4.7, pp. 384-388 — determining the domain
Picture it
The fencing objective plotted beyond its physical range.
Figure (svg): The physical domain of a fencing problem, showing where the model stops making sense
Outside the shaded band the parabola takes negative values, which would describe a rectangle with a negative side. The formula happily computes them and they mean nothing.
Worked example
Example 4.34. The geometry supplies the domain.
\[ \text{Squares of side } x \text{ are cut from a } 24 \text{-inch square sheet and the sides folded up. Maximise the volume.} \]
Write the dimensions in terms of x
Why: Two cuts are removed from each side.
\[ \text{base } 24 - 2 x,\text{ height } x \]
Write the objective
Why: Volume of a box.
\[ V(x) = x(24 - 2 x) ^{2} \]
Determine the domain
Why: The cut must be positive and the base must remain positive.
\[ 0 < x < 12 \]
Differentiate
Why: Expand first, or use the product rule.
\[ V' = 12(x - 4) (x - 12) \]
Find the critical point inside the domain
Why: Twelve is excluded.
\[ x = 4 \]
Figure (svg): An open box cut from a square sheet, with the cut size as the variable
\[ x = 4, \quad V = 4(16)^{2} = 1024 \text{ in}^{3} \]
Verify: check the endpoints and the discarded critical point
Why: As x approaches 0 the box has no height and the volume tends to 0; as x approaches 12 the base vanishes and the volume tends to 0 again. So the interior critical point at 4 is the maximum. The other root, x equal to 12, is exactly the excluded endpoint — it corresponds to cutting away the entire sheet, which is a degenerate case rather than a candidate. Determining the domain first is what made discarding it automatic.
OpenStax Calculus Volume 1, §4.7 Applied Optimization Problems §4.7, pp. 385-386
Matching
What the physics allows.
Match the pairs
Why: The first two are bounded by the fixed resource and the third is not, which changes which method applies. The fourth is the commonest shape in applied problems: non-negative with a practical ceiling that must be read from the wording.
Worked example
Checkpoint 4.34. Half-open intervals need care.
\[ \text{Minimise } P(x)=2x+\frac{200}{x} \text{ for } x > 0. \]
Note the domain
Why: A length must be positive; there is no upper bound.
\[ x > 0,\text{ an open ray} \]
Check whether the closed-interval method applies
Why: The interval is not closed.
Differentiate and find the critical point
Why: Set to zero.
\[ 2 - 200 / x ^{2} = 0\text{ at } x = 10 \]
Use the first derivative test instead
Why: Negative then positive.
Confirm it is absolute
Why: The only critical point on the ray.
Figure (svg): The solution to Worked example a domain that is not closed shown as a ladder of expressions, one row per legal move
\[ x = 10, \quad P = 40 \]
Verify: explain why this is an absolute minimum despite the open domain
Why: The Extreme Value Theorem does not apply, so existence is not automatic. But the first derivative test shows the function decreases on the whole interval to the left of 10 and increases on the whole interval to the right — so no value anywhere on the ray can be smaller. That argument replaces the closed-interval method when the domain is open, and it is worth stating explicitly rather than assuming the critical point wins.
OpenStax Calculus Volume 1, §4.7 Applied Optimization Problems §4.7, pp. 386-387
Error analysis
A student solves a fencing problem and reports both roots.
Annotate
On: \( A'(x) = 0 \text{ at } x = 25; \quad \text{also consider } x = -5 \text{ from a related equation} \)
The algebra will produce candidates the situation forbids, and it has no way of knowing they are absurd. Establishing the domain before differentiating makes the discard automatic rather than a judgement call.
Fill the middle
The box problem, where two cuts are removed from each side of a 24-inch sheet.
Fill in the blanks
24 - 2x > 0 \;\Longrightarrow\; x < 12
Why: The base's side must remain positive, which caps the cut at 12 inches. That upper limit is exactly where the second critical point sat, so establishing the domain first made discarding it automatic.
Sorting
Check against the physical domain.
Sort into buckets
Sort each candidate for the box problem, where 0 < x < 12.
The candidate at 12 is the degenerate case where the whole sheet is cut away, and it emerged from the algebra as a genuine root of the derivative. Discarding it requires the domain, which the algebra alone cannot supply.
Prediction
Commit before reasoning.
Predict first
The domain of an optimisation problem is the open ray x greater than 0. What changes?
Correct: The closed-interval method does not apply, so the first derivative test is needed.
\[ f' < 0 \text{ on } (0,10), \; f' > 0 \text{ on } (10,\infty) \;\Longrightarrow\; \text{absolute minimum at } 10 \]
Why: Without a closed bounded interval the Extreme Value Theorem gives no guarantee, so existence must be argued rather than assumed. The first derivative test does it: showing the function decreases everywhere left of the critical point and increases everywhere right of it proves the value there is smallest on the whole ray. An extremum may well exist, as it does here — but the argument for it is different.
Section
Section 3
Concept
Once the objective is a function of one variable on a known domain, Section 4.3's method applies unchanged. Find the critical points inside the domain, evaluate there and at any endpoints, and compare.
applying the earlier method — The reduced objective is optimised exactly as in Section 4.3: candidates are the critical points inside the domain together with any endpoints the domain includes, and the extremum is found by comparing values.
\[ \text{candidates: critical points in } (a,b) \text{ and the endpoints } a, b \]
The endpoints frequently matter more here than in abstract problems, because a physical constraint often pushes the optimum against a boundary — a capacity limit, a budget, a legal maximum.
Figure (svg): Two optimisation problems, one with an interior optimum and one at a boundary
OpenStax Calculus Volume 1, §4.7 Applied Optimization Problems §4.7, pp. 386-390 — solving the optimisation problem
Picture it
Two objectives on the same domain.
Figure (svg): Two optimisation problems, one with an interior optimum and one at a boundary
The left objective turns and its maximum is interior; the right one climbs throughout and its maximum is at the right edge, where no derivative vanishes. Both are common in applications.
Worked example
Example 4.36. The answer is a shape, not a number.
\[ \text{A cylindrical can must hold } 355 \text{ cm}^{3}. \text{ Minimise the metal used.} \]
Write the constraint
Why: The fixed volume.
\[ \pi r ^{2} h = 355 \]
Write the objective
Why: Two circles and a rectangle.
\[ S = 2 \pi r ^{2} + 2 \pi r h \]
Eliminate h
Why: From the constraint.
\[ h = \frac{355}{\pi r ^{2}} \]
Substitute
Why: The surface in one variable.
\[ S(r) = 2 \pi r ^{2} + \frac{710}{r} \]
Differentiate and solve
Why: Set to zero.
\[ 4 \pi r - 710 / r ^{2} = 0 \]
Figure (svg): A cylindrical can with fixed volume, showing the surface area being minimised
\[ h = 2r \quad \text{at the minimum} \]
Verify: derive the shape relation in general and check against real cans
Why: Solving in general gives r cubed equal to the volume over 2 pi, and substituting back gives h equal to 2r exactly — the optimal can is as tall as it is wide, whatever the volume. Real drink cans are noticeably taller and thinner than this, which is not a failure of the mathematics: the model assumes metal cost dominates, whereas real cans must also fit a hand, stack on shelves and use thicker metal on the ends. A model's clean answer disagreeing with practice usually means the model omitted a real cost.
OpenStax Calculus Volume 1, §4.7 Applied Optimization Problems §4.7, pp. 387-389
Ranking
Once the objective is a function of one variable.
Put in order
Why: Step a comes first because it governs step c, and step e is what turns a number into an answer. Reporting 'x equals 4' when the question asked for a volume answers a different question.
Worked example
Checkpoint 4.36. The constraint decides.
\[ \text{Revenue is } R(x)=x(200-x) \text{ for } 0 \le x \le 80 \text{ units. Maximise it.} \]
Differentiate
Why: Expand first.
\[ R'(x) = 200 - 2 x \]
Find the critical point
Why: Set to zero.
\[ x = 100 \]
Check it against the domain
Why: The capacity is 80.
Evaluate at the endpoints
Why: Both ends of the interval.
\[ R(0) = 0, R(80) = 9600 \]
Compare
Why: Only the endpoints remain.
\[ \text{maximum at } x = 80 \]
Figure (svg): Two optimisation problems, one with an interior optimum and one at a boundary
\[ \max R = 9600 \text{ at } x = 80 \]
Verify: interpret what the discarded critical point means
Why: The unconstrained optimum is 100 units, but capacity caps production at 80 — so the firm is constrained rather than optimising freely, and the marginal revenue at 80 is still positive at 40 per unit. That is genuinely useful information: it says expanding capacity would be profitable, which a bare answer of 'produce 80' would hide. Whenever a critical point falls outside the domain, the optimum is at a boundary and the constraint is binding.
OpenStax Calculus Volume 1, §4.7 Applied Optimization Problems §4.7, pp. 389-390
Trap
\[ R'(x) = 200 - 2x = 0 \text{ at } x = 100 \]
Report 100 units as the answer
Why: The student solves and stops.
\[ \text{but the capacity is } 80 \]
A hundred units cannot be produced, so the answer describes something the firm cannot do.
\[ 100 \notin [0,80] \;\Longrightarrow\; \text{discard; compare the endpoints} \]
Check every critical point against the physical domain before evaluating
Why: A candidate outside the domain is not a candidate.
The constrained answer is also more informative than it looks. That the unconstrained optimum lies beyond the capacity says the constraint is binding and relaxing it would help — which is exactly what a manager wants to know.
Fill the middle
A revenue problem whose unconstrained optimum exceeds the capacity.
Fill in the blanks
R'(x)=0 \text80 x=100, \text___ [0,80] \;\Longrightarrow\; \text___ x = ___
Why: With the only critical point outside the domain, the endpoints are the only candidates and the right one wins. The constraint is binding, which itself is useful information.
Two truths and a lie
All three are about applied optima.
Eliminate the wrong options
Two of these are true. Knock those out and keep the false one.
Survives elimination: B
Why: The survivor is the false one. When the objective is monotone on the feasible domain, or when the unconstrained optimum lies outside it, the best available value is at an endpoint where no derivative vanishes. This is Section 4.3's endpoint rule in an applied setting, and constraints make it far more common here than in abstract problems.
Prediction
Commit before reasoning.
Predict first
The unconstrained optimum is 100 units but capacity is 80. What does that say?
Correct: The constraint is binding, and relaxing it would help.
\[ R'(80) = 40 > 0: \text{ still climbing when the capacity binds} \]
Why: At 80 units the marginal revenue is 200 minus 160, which is 40 per unit — still positive, so each additional unit would earn money if it could be produced. That is precisely the information a manager needs: the capacity limit is costing revenue, and investment in capacity would pay. A bare answer of 'produce 80' conceals it, which is why interpreting the result matters as much as computing it.
Section
Section 4
Concept
A distance involves a square root, which makes differentiation awkward. Since squaring is increasing on the non-negative numbers, the distance and its square are smallest at the same input — so minimise the square instead.
monotone substitution — Replacing an objective by a monotone increasing function of it. Because such a function preserves order, the two are optimised at the same input, and the substitution can remove roots or logarithms.
\[ d \text{ minimal} \iff d^{2} \text{ minimal}, \quad d \ge 0 \]
The same idea handles products and quotients: taking logarithms turns them into sums and differences, and the logarithm is increasing, so the optimum is unchanged.
Figure (svg): Minimising a distance by minimising its square, avoiding the root
OpenStax Calculus Volume 1, §4.7 Applied Optimization Problems §4.7, pp. 388-392 — simplifying before optimising
Picture it
Finding the point on a curve nearest a fixed point.
Figure (svg): Minimising a distance by minimising its square, avoiding the root
The distance formula carries a square root and its square does not, so differentiating the square is far quicker — and because squaring preserves order on non-negative values, the minimising input is identical.
Worked example
Example 4.37. Square the distance first.
\[ \text{Find the point on } y=\sqrt{x} \text{ nearest to } (4,0). \]
Write the distance
Why: Between a general curve point and the fixed one.
\[ d = \sqrt{(x - 4) ^{2} + x} \]
Minimise the square instead
Why: Squaring preserves order.
\[ D(x) = (x - 4) ^{2} + x \]
Expand
Why: A quadratic.
\[ D = x ^{2} - 7 x + 16 \]
Differentiate and solve
Why: Set to zero.
\[ 2 x - 7 = 0,\text{ so } x = 3.5 \]
Find the point
Why: Substitute into the curve.
\[ (3.5, \sqrt{3.5}) \]
Figure (svg): Minimising a distance by minimising its square, avoiding the root
\[ (3.5, \sqrt{3.5}) \approx (3.5, 1.871) \]
Verify: check that the shortcut cost nothing and the answer is sensible
Why: Differentiating the distance itself would have needed the chain rule on a root and produced the same critical point after considerably more algebra. The answer is also geometrically plausible: the nearest point is slightly to the left of directly below the fixed point, because the curve climbs away as x increases. The minimum distance is the root of 3.75, about 1.94. Note the domain is x at least 0, and 3.5 is comfortably inside.
OpenStax Calculus Volume 1, §4.7 Applied Optimization Problems §4.7, pp. 390-391
Fill the middle
The distance from a point on the square root curve to a fixed point, squared and expanded.
Fill in the blanks
D(x) = (x-4)^16+x = x^___-7x+___
Why: Expanding gives x squared minus 8x plus 16 plus x, which is x squared minus 7x plus 16. Differentiating this quadratic is far easier than differentiating a square root.
Worked example
Checkpoint 4.37. Order-preserving is the whole condition.
\[ \text{Explain why minimising } d^{2} \text{ minimises } d. \]
Note that a distance is non-negative
Why: By definition.
\[ d \ge 0 \]
Note that squaring is increasing there
Why: Larger non-negative inputs give larger squares.
\[ a < b\text{ implies } a ^{2} < b ^{2} \]
Deduce the order is preserved
Why: Both directions.
\[ d\text{ smallest iff } d ^{2}\text{ smallest} \]
Conclude
Why: The minimising input is the same.
Note the condition that matters
Why: Non-negativity.
Figure (svg): The solution to Worked example why the substitution is legitimate shown as a ladder of expressions, one row per legal move
\[ d \ge 0 \text{ and } t \mapsto t^{2} \text{ increasing on } [0,\infty) \]
Verify: test what goes wrong without non-negativity
Why: For a quantity that can be negative the substitution fails: negative 5 is smaller than 1, but its square 25 is larger. So minimising the square would pick the wrong input entirely. Distances are always non-negative, which is why the trick is safe here — but the condition must be checked rather than assumed, and the same caution applies to taking logarithms, which requires positivity.
OpenStax Calculus Volume 1, §4.7 Applied Optimization Problems §4.7, pp. 391-392
Error analysis
A student finds the nearest point on a curve.
Annotate
On: \( D(x) = x^{2}-7x+16 \text{ is minimised at } x = 3.5, \text{ so the distance is } 3.5 \)
Two separate slips are available here: reporting the input as the answer, and forgetting that the optimised quantity was a square. Naming what each symbol means before reporting anything prevents both.
Sorting
The replacement must be increasing on the range involved.
Sort into buckets
Sort each proposed simplification.
The last is worth noting: negating turns a maximum into a minimum, which is a useful trick deliberately used but a disaster applied accidentally. Every substitution must be checked for monotonicity on the range actually involved.
Two truths and a lie
All three are about simplifying the objective.
Eliminate the wrong options
Two of these are true. Knock those out and keep the false one.
Survives elimination: B
Why: The survivor is the false one. Only an INCREASING substitution preserves the optimising input, and even then the optimal value changes and must be converted back. Squaring a quantity that can be negative reverses order on the negatives and gives the wrong answer entirely.
Prediction
Commit before reasoning.
Predict first
You minimise D = d^2 and find its minimum is 3.75 at x = 3.5. What is the answer to the original question?
Correct: The distance is the root of 3.75, about 1.94, at x equal to 3.5.
\[ x = 3.5 \text{ (input)}, \; D = 3.75 \text{ (square)}, \; d = \sqrt{3.75} \approx 1.94 \]
Why: The minimising INPUT survives the substitution unchanged, but the minimum VALUE does not — it is the square of the distance and must be rooted to recover the distance itself. And 3.5 is the input, not either value. All three quantities are different and the question determines which is wanted, which is why naming each symbol before reporting is worth the few seconds.
Section
Section 5
Concept
An optimisation problem asks about a real situation, so the answer must be stated in the situation's own terms with units. The optimising input is rarely what was wanted on its own.
interpretation — Converting the mathematical result back into the problem's terms: the dimensions rather than the variable, the volume rather than the input, with units and a statement of which quantity is optimised.
\[ x = 4 \;\Longrightarrow\; \text{cut } 4 \text{ in, giving } V = 1024 \text{ in}^{3} \]
A clean answer is often a shape relation rather than a number, and that is more informative. The optimal can having height equal to diameter holds for every volume, which no single numerical answer would reveal.
Figure (svg): A cylindrical can with fixed volume, showing the surface area being minimised
OpenStax Calculus Volume 1, §4.7 Applied Optimization Problems §4.7, pp. 387-393 — interpreting optimisation results
Picture it
The minimal-surface can.
Figure (svg): A cylindrical can with fixed volume, showing the surface area being minimised
The answer h equals 2r holds for every fixed volume, which is a stronger statement than any particular set of dimensions. Solving in general before substituting is what makes such relations visible.
Worked example
Every part of the question, with units.
\[ \text{For the open box, state a complete answer.} \]
State the optimising input
Why: The cut size.
\[ x = 4\text{ inches} \]
State the resulting dimensions
Why: Base and height.
\[ 16\text{ by } 16\text{ by } 4\text{ inches} \]
State the optimised quantity
Why: With units.
\[ \text{volume } 1024\text{ cubic inches} \]
State which extreme it is
Why: Maximum.
Note the comparison
Why: Against the alternatives.
\[ \text{endpoints give } 0 \]
Figure (svg): The solution to Worked example reporting a complete answer shown as a ladder of expressions, one row per legal move
\[ 16 \times 16 \times 4, \quad V_{\max} = 1024 \text{ in}^{3} \]
Verify: check the dimensions multiply back correctly
Why: Sixteen times 16 times 4 is 1024, confirming both the dimensions and the volume. Also, 16 plus two 4-inch cuts is 24, matching the original sheet — a check that the dimensions are consistent with the geometry rather than merely with the algebra. Reporting only 'x equals 4' would answer a question about a variable the problem never mentioned.
OpenStax Calculus Volume 1, §4.7 Applied Optimization Problems §4.7, pp. 386-387
Matching
Read the verb and the noun.
Match the pairs
Why: Three different quantities emerge from one computation and the question decides which is wanted. Reporting the wrong one is the commonest way a correct calculation earns no credit.
Worked example
Checkpoint 4.37. Solve symbolically before substituting.
\[ \text{Show that the minimal-surface can has } h=2r \text{ for any fixed volume } V. \]
Write the constraint generally
Why: Volume fixed at V.
\[ \pi r ^{2} h = V \]
Write the objective and eliminate
Why: Substitute for h.
\[ S(r) = 2 \pi r ^{2} + 2 V / r \]
Differentiate and set to zero
Why: Solve for r.
\[ 4 \pi r = 2 V / r ^{2},\text{ so } r ^{3} = \frac{V}{2 \pi} \]
Express h in terms of r
Why: From the constraint.
\[ h = \frac{V}{\pi r ^{2}} \]
Substitute and simplify
Why: Using the value of r cubed.
\[ h = 2 r \]
Figure (svg): The solution to Worked example a general relation shown as a ladder of expressions, one row per legal move
\[ h = 2r \quad \text{for every } V \]
Verify: check the relation against the specific case
Why: With V equal to 355 the general formula gives r cubed equal to 355 over 2 pi, so r is about 3.84 and h about 7.67 — and 7.67 is indeed twice 3.84. The general result reproduces the specific one and says far more: no arithmetic is needed to know the shape of the optimal can for any volume. Solving symbolically before substituting is usually worth the small extra effort for exactly this reason.
OpenStax Calculus Volume 1, §4.7 Applied Optimization Problems §4.7, pp. 388-389
Trap
\[ V'(x) = 0 \text{ at } x = 4 \]
Report 4 as the answer
Why: The student gives the optimising input.
The question asked for the largest volume, which is 1024 cubic inches. Four is the cut size that achieves it.
\[ x = 4 \;\Longrightarrow\; V = 1024 \text{ in}^{3}, \text{ from a } 16\times 16\times 4 \text{ box} \]
Convert the optimising input back into the quantity asked about
Why: State it with units and dimensions.
Both pieces are usually wanted: the manufacturer needs to know the cut size to set the machine AND the volume to quote a capacity. Giving one without the other answers half the question, and giving the input alone answers none of it.
Fill the middle
The box problem, with the optimising cut found.
Fill in the blanks
x = 4 \;\Longrightarrow\; V = 4(24-8)^1024 = ___ \text___^___
Why: Four times 256 is 1024 cubic inches. The cut size 4 is the input; the volume 1024 is the answer to a question about volume.
Two truths and a lie
All three are about interpreting a result.
Eliminate the wrong options
Two of these are true. Knock those out and keep the false one.
Survives elimination: B
Why: The survivor is the false one. Real cans are taller and thinner than the minimal-surface shape because the model omits costs the mathematics was never told about — handling, stacking, thicker end-caps. The disagreement is informative rather than an error: it identifies what the model left out, which is often the most valuable thing an optimisation produces.
Prediction
Commit before reasoning.
Predict first
The optimal can has height equal to diameter, and real drink cans do not. What follows?
Correct: The model omits real costs the calculation was never given.
\[ \text{minimising area alone} \ne \text{minimising total cost} \]
Why: The optimisation minimised metal area alone, and that is genuinely minimised by the squat shape. Real cans must also be gripped, stacked and made with thicker metal on the ends, which favours a taller thinner form. The mathematics is correct for the problem it was given, and the disagreement identifies which costs were left out — which is exactly the kind of feedback a modelling exercise should produce.
Comparison
Fill the blanks. Getting these the wrong way round is the section's characteristic failure.
Comparison matrix
| Objective | Constraint | |
|---|---|---|
| Identified by | the verb: maximise, minimise, largest | words like available, given, exactly |
| What you do with it | differentiate it, after reducing | solve it for one variable and substitute |
| In the fencing problem | the area | the 100 m of fence |
| If you differentiate it | you get useful information | you get zero, since it is fixed |
The last row is a free diagnostic. A derivative that vanishes identically means the constraint was differentiated, and the setup must be revisited before any more work is done.
Pattern
Given a described optimisation problem.
If the domain is not a closed bounded interval, the closed-interval method does not apply and the first derivative test must establish that the extremum is absolute.
Stewart, Calculus: Early Transcendentals 8e, §4.7 Optimization Problems §4.7, pp. 330-344
Check
Objective and constraint.
Check your understanding
With 100 m of fence, what is the largest rectangular area?
Answer: A
Why: A(x) = x(50-x) is maximised at x = 25, giving a 25 by 25 square of area 625.
Check
The physical domain.
Check your understanding
For the open box cut from a 24-inch sheet, what is the domain of the cut size x?
Answer: A
Why: Two cuts are removed from each side, so 24 - 2x must stay positive.
Check
Infeasible candidates.
Check your understanding
Revenue is x(200-x) for 0 <= x <= 80. Where is the maximum?
Answer: A
Why: The critical point at 100 is outside the domain, so the endpoints decide and 80 wins.
Real world
A logistics firm ships goods in rectangular containers with a square base. The carrier charges by girth plus length, and the combined figure may not exceed 300 centimetres, where girth is the perimeter of the square cross-section.
Discussion prompt
Find the largest volume that can be shipped, identify the objective and constraint, and say what happens if the carrier raises the limit.
Hint: Girth is four times the base side; the constraint is an equation in the base and the length.
Answer:
Let the base side be x and the length L. Girth plus length gives the constraint, and volume is the objective:
\[ 4x + L = 300, \qquad V = x^{2}L \]
\[ L = 300 - 4x \;\Longrightarrow\; V(x) = x^{2}(300-4x) = 300x^{2}-4x^{3} \]
The physical domain is 0 to 75, since the length cannot be negative. Differentiating:
\[ V'(x) = 600x - 12x^{2} = 12x(50-x) \;\Longrightarrow\; x = 50 \]
Both endpoints give zero volume, so the interior critical point is the maximum: a base of 50 by 50 cm with a length of 100 cm, giving 250 000 cubic centimetres, or a quarter of a cubic metre.
The shape is worth noticing: the length comes out at exactly twice the base side, which holds for any limit. If the carrier raises the cap to 400 cm, the same algebra gives x equal to 200 over 3 and L equal to 400 over 3 — the same two-to-one proportion, with the volume rising to about 593 000 cubic centimetres. The optimal SHAPE is independent of the limit, and only the scale changes.
That is the practical payoff of solving symbolically: a firm can fix its container proportions once and rescale them whenever a carrier's rules change, rather than re-optimising from scratch.
Commit first
Answer, then rate your confidence honestly.
Predict first
You differentiate and the derivative is zero for every value of the variable. What has gone wrong?
Correct: The constraint was differentiated instead of the objective.
\[ 2x+2y=100 \text{ always} \;\Longrightarrow\; \frac{d}{dx}(100) = 0 \]
Why: A constraint asserts that some quantity is fixed, so its derivative is identically zero — that is what being fixed means. An identically vanishing derivative therefore carries no information and points straight back to the setup. The fix is to reread the question for its verb, identify the quantity being maximised or minimised, and differentiate that one after using the constraint to reduce it to a single variable.
Explain it
They set up a fencing problem, differentiated the perimeter, got zero, and concluded the problem was broken.
Discussion prompt
In four sentences or fewer, explain what they did.
Hint: Ask them which quantity the problem said was fixed.
Answer:
Ask them what the 100 metres of fence is. It is fixed by the problem — the perimeter never changes — so of course its derivative is zero, and differentiating it can never tell them anything.
The quantity that varies is the AREA, which changes as they choose different shapes with that same fence. So the perimeter equation is used to write the area in terms of one variable, and it is the area that gets differentiated. The verb in the question — 'largest area' — names the objective every time.
Exit ticket
Name the weakest spot before you close the deck.
Predict first
Which of these would you least want handed to you cold?
Correct: Whichever you picked is tonight's ten minutes, and each has a one-line fix.
Why: For objective and constraint, find the question's verb — it names the objective, and anything stated as given is the constraint. For the domain, ask what each variable physically cannot be. For endpoints, remember constraints often make them binding. For conversion, reread the question and name what it actually asked for. Do five of your chosen kind rather than twenty mixed ones.
Connect it up
Paper. Fifteen minutes.
Draw it
At the top, write the five steps of the procedure, boxing steps two and three together and noting that these are where problems are lost. Below, work the fencing problem completely: draw the rectangle, label both quantities, mark which is objective and which constraint, reduce to one variable, state the domain, and optimise — then sketch the parabola with the physical domain shaded and both endpoints marked. In the middle of the page, draw the open box from a 24-inch sheet with the corner cuts shaded, derive the volume formula and the domain, and solve, noting which critical point had to be discarded and why. Beside it, set up the minimal-surface can and derive h equals 2r symbolically, then write one sentence on why real cans differ. At the bottom, take the nearest point on the square root curve to the point (4,0), minimise the SQUARE of the distance, and write the three different quantities the computation produces — input, squared distance and distance — with what each is. In a margin, write the diagnostic for having differentiated the constraint.
If your can derivation produced numbers before it produced h equals 2r, redo it symbolically — the general relation is the more valuable result and it disappears the moment a volume is substituted.
Recap
Five things, and only the last is calculus you had not already met.
| If you see | Then |
|---|---|
| 'Largest', 'minimise', 'cheapest' | That names the objective |
| 'Available', 'given', 'exactly' | That is the constraint |
| A derivative that vanishes identically | You differentiated the constraint |
| A distance to minimise | Minimise its square instead |
| A critical point outside the domain | Discard it and check the endpoints |
| An optimum at a boundary | The constraint is binding |
| An answer that is just a variable's value | Convert it into what was asked |
Section 4.8 returns to limits with a tool this chapter has repeatedly gestured at. L'Hopital's rule evaluates indeterminate forms by differentiating, and it finally proves the growth ranking Section 4.6 relied on.
OpenStax Calculus Volume 1, §4.7 Applied Optimization Problems §4.7, pp. 381-393 — everything on these slides traces back here
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