CEGEP Calculus I Applications of the Derivative Free · no sign-up

CEGEP Calculus I — Optimization

By Marius — private math tutor in Montreal. Updated 30 August 2026

Optimization is where the differential calculus of a whole semester finally gets pointed at something: a yard to enclose, a box to build, a cost per unit to bring down. Almost none of the difficulty is in the differentiating. It is in turning a paragraph of English into one function of one variable, knowing what that variable is physically allowed to be, and then saying — with a reason a marker will accept — that the number you found really is the largest or smallest value there is. Have a look on this page, then print the free PDF when you want to write on it.

Practice worksheet — free PDF

4 pages 5 questions Letter size, print-ready

Download the worksheet Read the questions here

No email, no account, no watermark. Teachers: photocopy it for your classes freely. Worked solutions and 4 harder problems come with the CEGEP Calculus I bundle.

All 5 questions

Each question targets one named concept from the sheet. Read them here, or print the PDF — it has working space under each one.

Q1Expressing a Quantity as a Function of One Variable

A depot manager has $180 m$ of snow fencing to enclose a rectangular salt storage yard. One full side of the yard is the straight wall of an existing garage and needs no fencing; the fencing is used for the other three sides. Let $x$ be the length, in metres, of each of the two sides perpendicular to the garage wall.
1. Express the enclosed area $A$ as a function of $x$ alone.
2. State the domain that the physical situation allows, and say in one sentence what goes wrong at each excluded value.
Do not differentiate anything — this question asks only for the model.
Q2Optimization on a Closed Interval

During a $6$ -hour maintenance window the electrical power drawn by a data centre's cooling loop is modelled by $P (t) = t^{3} - 9 t^{2} + 24 t + 5 kilowatts,$ where $t$ is the number of hours since the window opened, $0 \leq t \leq 6$ . Find the absolute maximum and the absolute minimum power drawn during the window, and state the time at which each occurs.
Q3Optimization Without a Closed Interval

A rooftop garden needs a galvanised planter box with a square base and no lid, holding exactly $32 000 {cm}^{3}$ of soil. Let $x$ be the side of the square base, in centimetres.
1. Express the total area of sheet metal used as a function of $x$ , and state the domain.
2. Find the dimensions that use the least metal, and the amount used.
3. The domain is not a closed interval, so the closed-interval method does not apply. Justify that your answer really is a global minimum.
Q4Justifying and Interpreting an Optimum

A workshop that builds sea kayaks finds that when it makes $x$ kayaks per week its average cost per kayak is $A (x) = 0.5 x + \frac{800}{x} + 120 dollars per kayak, x > 0 .$ A student differentiates, solves $A^{'} (x) = 0$ , and writes down the single line “the minimum is $40$ ”.
1. Reproduce the calculation, then rewrite that conclusion as a full sentence naming what $40$ is, what the minimum value is, and the units of each.
2. Justify, using the second derivative, that this really is a minimum, and explain why the test settles the question over the whole domain $x > 0$ without any interval being tested.
3. The owner reads the note and concludes “so building $40$ kayaks a week minimises our costs”. Explain precisely what is wrong with that reading.
Q5Synthesis — drawing on several topics in this set

A park authority is building a rectangular observation deck against a straight cliff face. The cliff forms one side and needs no railing; $60 m$ of railing is available for the other three sides. A safety bylaw requires the deck to project at least $18 m$ out from the cliff and at most $25 m$ .
1. Let $x$ be the projection from the cliff, in metres. Express the deck area $A$ as a function of $x$ , give the domain the geometry allows, and give the interval the bylaw allows.
2. Using the closed-interval method on the bylaw's interval, find the largest and the smallest deck the authority can build, with dimensions.
3. Interpret the result: state the answer in a sentence with units, and say how much deck area the bylaw costs compared with the best deck the railing alone would allow.

The 4 challenge problems for this topic are a separate, paid sheet and are not reproduced here.

Which stream is this for? This set is built for CEGEP Calculus I — 201-NYA-05 under the legacy numbering, 201-SN2-RE under the current one — and sits in competency 0M02, whose fourth element asks for the correct solution of optimization problems together with a sound interpretation of the results. It stops where that competency stops: there are no antiderivatives here, no substitution rule and no differential equations, because the ministerial devis places all three in Calculus II.

How to do every concept on this sheet

This is the part worksheet sites usually leave out. Below is the actual reasoning behind each group of questions — not a full solution set, the decisions that get you to one. Read it before you start, or after you get stuck.

The set is a progression, and it is worth doing in order

The questions here are not interchangeable word problems. They are one method taken apart and rebuilt a stage at a time, so that when something goes wrong you can see which stage it went wrong at. Working them out of order is possible, but you lose the point of the design.

What each stage adds

Every stage keeps everything the one before it asked for.

1
Modelling only — no calculus at all
Q1 asks you to express the quantity as a function of one variable and to state the domain the physical situation allows, and then to stop. There is nothing to differentiate. This is deliberate: the modelling is where most of the marks and nearly all of the mistakes live, and doing it with the derivative switched off is the fastest way to get good at it.
2
The closed-interval method, with the endpoints tested
Q2 hands you the function already built and a closed interval to work on. Now the Extreme Value Theorem is available, so you may assert that the extrema exist before you go looking for them — and the endpoints join the critical numbers on the list of candidates.
3
No closed interval — you have to argue
Q3 lives on an open, unbounded domain, so the EVT is gone and there are no endpoints to evaluate. Finding the critical number is now the easy half; showing that the value there beats every other value on the domain is the question.
4
Saying what the answer means
Q4 gives you a bare, badly written conclusion and asks you to repair it: name what each number is, attach the units, justify the claim, and then say what the result does and does not tell the person who has to act on it.
5
All four at once
Q5 closes the set: build the model, notice that the domain the geometry allows and the interval you are actually restricted to are two different intervals, optimise on the second, and interpret the answer against the first.

Stating the domain is part of the answer, not the preamble

Students treat “state the domain” as bookkeeping to be done if there is time. It is not. The domain is what decides which method you are entitled to use, and a model with the wrong domain can be differentiated perfectly all the way to a value the situation cannot physically produce.

Ask the question the formula cannot answer. An algebraic expression returns a number for any input you feed it. Q1 makes exactly that point, and asks you to say in one sentence what goes wrong at each excluded value: the formula is perfectly happy at inputs where the enclosed figure has collapsed to a line. Only the physical situation knows which inputs describe a real object.

Every length must be positive — all of them. The variable you named is usually the obvious constraint, and the one that gets forgotten is the other dimension, the one the constraint equation eliminated. In Q1 it is that second condition, not the first, that pins down the upper end of the domain.

Open or closed is a physical question, not a habit. Sometimes an endpoint is a degenerate figure and is genuinely excluded; sometimes it is a perfectly buildable configuration and belongs in the domain. Decide it by asking whether that value describes something that could exist, and write the reason down in a clause.

The closed-interval method: the endpoints are candidates too

When the domain is a closed, bounded interval and the function is continuous on it, the Extreme Value Theorem promises that an absolute maximum and an absolute minimum exist, and that each occurs either at a critical number or at an endpoint. That promise is what makes the method finite: there is a list, and the answer is on it.

f continuous on [a, b] → the absolute maximum and minimum exist, at a critical number or at a or b

The closed-interval method on [a, b]

Q2 is this procedure and nothing else — which is exactly why it is worth running the same way every time.

1
Say why the EVT applies
Continuous on a closed and bounded interval. For a polynomial one clause does it, and writing that clause is the difference between a method and a guess. It is also the sentence that earns the reasoning mark.
2
Find the critical numbers, then discard those outside
Where the derivative is zero, and where it fails to exist. Keep only the ones lying inside the interval — and write down, in words, that the others were rejected for falling outside it. It is entirely possible for none to be left, and that is a legitimate outcome rather than a sign you have gone wrong.
3
Evaluate at every survivor and at both endpoints
Both endpoints, every time, whatever the sign chart looked like. They are not critical numbers, so nothing in your derivative work will ever remind you of them — and an endpoint carrying the absolute extremum is common, not exotic.
4
Compare the values and name both parts
“The absolute maximum value is M, and it occurs at t = c.” A bare input value is not a maximum, and a bare number with no location is not an answer. Give both, with units.

One consequence is worth carrying into Q2 and Q5. A local extremum and a global one are different claims: an interior critical number can produce a local maximum that is comfortably beaten by a value at the edge of the interval. Stopping at the critical numbers is not a shortcut — it is a different, and wrong, answer to the question that was asked.

No closed interval: the mark is in the argument

This is the most expensive habit on the whole topic, and Q3 exists to break it. You differentiate, you find one critical number, and you write “so the minimum is there”. On an open or unbounded interval that sentence has nothing holding it up.

The Extreme Value Theorem does not apply on an open or unbounded interval. It requires an interval that is closed and bounded, and (0, ∞) is neither. So there are no endpoints to evaluate and — this is the part that gets skipped — an extremum is not even guaranteed to exist until you show that one does. A critical point is a place where the function has stopped changing for an instant. By itself it says nothing whatever about how the function behaves far away at either end of the domain.

Two arguments that finish the job. Either show the second derivative is positive on the whole domain, so the function is concave up everywhere and its single critical point carries the absolute minimum; or build a sign chart for the first derivative and show the function decreases on the entire stretch to the left of the critical number and increases on the entire stretch to the right. Either one is complete, and either one takes a line.

What does not finish the job: evaluating the second derivative at the critical point alone. That is the second derivative test, and it establishes a local extremum only. It says nothing about the tail of the domain, so on an unbounded interval it leaves the global claim unproved. The word “everywhere” is doing real work in the argument above, and dropping it costs the mark.

f’’(x) > 0 for every x in the domain → the one critical point is the global minimum

Q3 asks for that justification in a part of its own, separately from the answer, precisely so that the arithmetic alone cannot carry you. Write the argument out even when a question does not carve out a part for it: on this material an unjustified global claim is the most reliable way to lose marks you have already done the work to earn.

Interpreting the result — the part that is not calculus

Competency 0M02 asks for a sound interpretation of the results, not only a correct solution, and Q4 tests that directly. It shows you a conclusion of the form “the minimum is 40” and asks what is wrong with it. Several things are.

The input and the output are different quantities. The value of the variable at which the optimum occurs and the optimal value of the function are two different numbers in two different units — a production level and a cost, a length and an area. A sentence naming only one of them has answered half the question.

Units, every time. “Dollars per kayak” and “kayaks per week” are not decoration; they are what makes the sentence mean anything. Carry them through the working, not just into the final line.

Minimising an average is not minimising a total. The last part of Q4 turns on this, and it is far better to meet it here than in an exam. A rate — cost per unit, metres per second, area per dollar — behaves quite differently from the total it came from, and the optimum of one is not the optimum of the other. Read what the function actually measures before you announce what has been optimised.

Answer the question that was asked. If the problem wants dimensions, give the dimensions, not the value of the variable you happened to solve for. Then check the result back against the constraint you started from: it costs one line and catches a whole class of algebra slips.

Two intervals, not one

Q5 pulls the set together, and the trap it sets is a realistic one. There is the domain the geometry allows, and there is the narrower interval some external restriction allows — a bylaw, a machine setting, a schedule. They are different intervals and they do different jobs.

You optimise on the feasible one. The unrestricted optimum is still worth locating, because the interesting sentence in the interpretation is usually the comparison: how much the restriction costs you against the best you could otherwise have done. And if the unrestricted optimum falls outside the feasible interval then it is not an answer at all — it is a value you are not permitted to use, and the real optimum will sit at an endpoint, where setting the derivative to zero would never have found it.

Getting the most out of it

Draw it, and label the drawing

A sketch with every length marked on it does in ten seconds what a paragraph of naming conventions does badly. Mark the variable you chose, mark the dimension the constraint will eliminate, and write the constraint and the quantity to be optimised as two separate equations before you combine them. Most modelling errors are visible in the picture and invisible in the algebra.

Write the domain down before you differentiate

Not afterwards, and not while checking. The domain decides which method you are allowed to use, so it has to exist before you choose one. Ask it as a single question: which values of this variable describe something that could actually be built, or could actually happen?

Ask “is my interval closed and bounded?” every single time

Closed and bounded: cite the EVT, add both endpoints to the candidate list, compare the values. Open or unbounded: no EVT, no endpoints, and you owe an argument — concavity across the whole domain, or the sign of the first derivative on either side of the critical number. Nearly every avoidable lost mark on this material is the wrong branch of that one question.

Finish with a sentence, not a number

The last line of a solution should name the quantity, its value, its units and where it occurs, and — where the question invites it — say what that means for whoever asked. Practise writing that sentence even when nothing explicitly demands it. It is what the competency is assessed on, and it takes about fifteen seconds.

Want the solutions, or something more challenging?

The worksheet, the questions above and every explanation on this page stay free permanently. Three more PDFs exist for this topic — the worked answer key, a harder problem set, and the answer key to that. They come with the CEGEP Calculus I Solutions Bundle, which is what keeps the rest of the series free.

What else exists for Optimization

Three PDFs · 10 pages · all three are in the bundle below.

Answer key — 3 pages. All 5 questions worked step by step, including the restrictions and the justifications. Not a list of final answers.
Challenge problems — 4 pages, 4 problems. A separate sheet at exam-plus difficulty covering the same 4 concepts. Harder than anything on the free sheet.
Challenge answer key — 3 pages. Every challenge problem worked to the same standard, with the checks shown.
PDF, letter size, print-ready.

In the bundle See what's in it Not sold separately

The one thing that's for sale

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Every CEGEP Calculus I topic — the complete Solutions Bundle

One download, one payment, the whole program. Every answer key and every challenge set for all 8 CEGEP Calculus I worksheet sets — including this one.

8 sets · 24 PDFs · 82 pages$19.99

Worked solutions, not answer lists — every step written out
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Taking Secondary 1 Math as well? The Secondary 1 Math bundle covers all 15 of its sets — 45 PDFs, 182 pages — on the same terms.

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Common questions

Is this worksheet really free?

Yes — the questions are on this page to read, and the PDF downloads directly, no email and no account. The one paid item is optional: the complete CEGEP Calculus I Solutions Bundle, which covers every set at this level.

Which course is this for?

CEGEP Calculus I — 201-NYA-05 under the legacy numbering and 201-SN2-RE under the current one. They are the same course; colleges are partway through renumbering, so students search for both. This set covers the optimization element of competency 0M02, which is the same wherever in Québec you are taking it.

Why is there no differentiation on the first sheet?

Because the modelling is the part that goes wrong. Expressing the quantity as a function of one variable, and working out the domain the situation allows, are separately assessed skills, and practising them with the calculus switched off is the quickest way to stop losing marks before the derivative is even taken. The differentiating starts on the next sheet.

Do I really have to check the endpoints?

On a closed interval, yes, every time. An absolute maximum has to beat every value on the interval, and the endpoints are not critical numbers, so no amount of derivative work will surface them. It is entirely normal for the extremum to sit at an endpoint, and it is possible for no critical number to lie inside the interval at all.

How do I show a critical point is a global optimum with no closed interval?

Argue it, because the Extreme Value Theorem cannot: it needs an interval that is closed and bounded. Either show the second derivative keeps one sign across the whole domain — concave up everywhere means the single critical point carries the absolute minimum — or use a sign chart for the first derivative and show the function falls on all of the domain to the left of the critical number and rises on all of it to the right. Evaluating the second derivative at the point alone is not enough; that settles a local extremum only.

What counts as interpreting the result?

Naming what each number is, in units, and saying what it means for the situation. The value of the variable at which the optimum occurs and the optimal value itself are different quantities in different units, and a conclusion offering one bare number has answered half the question. Where the function is a rate — a cost per unit, say — it also means being clear about what has been optimised, since the optimum of an average is not the optimum of a total.

Is any of this really Calculus II material?

No. Optimization closes competency 0M02, and everything here uses differentiation only. Antiderivatives, the substitution rule and separable differential equations belong to 0M03 and open Calculus II, so nothing on this sheet needs them.

Can teachers use this in class?

Yes. Print and photocopy it for your own classes freely — I just ask that the tutorinmontreal.ca footer stays on the page.

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Yes — through one-on-one tutoring, in Montreal or online. Get in touch to arrange a session, or see the current rates.

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