Differential Calculus 201-SN2-RE • CEGEP, Montreal

Revision sheet: applied optimization (201-SN2-RE)

Q: How do I recognize the constraint and the function to optimize?

The function to optimize is the one whose maximum or minimum the question asks for, often an area, a volume, a cost or a time. The constraint is the quantity fixed by the statement, the one that is available or imposed. The words given, available, fixed, imposed almost always signal the constraint.

Q: Why must a variable be eliminated before differentiating?

Because an ordinary derivative is computed with respect to a single variable only. As long as the objective function depends on two unknowns, there is nothing to solve. The constraint provides the relation that expresses one in terms of the other and brings the problem down to a single variable.

Q: How do I justify that a critical number really is a maximum?

Three possible arguments. On a closed bounded interval, compare the function at the critical numbers and at both endpoints. If there is only one critical number and the derivative changes sign there, the local maximum is absolute. Otherwise, use the sign of the second derivative, which must be strictly negative.

Q: Why is the domain so important in optimization?

Because the algebraic formula can still be computed outside its physical meaning and often produces a second, spurious root. For a box cut from a sheet of cardboard thirty by forty, the derivative vanishes twice, and only the domain rules out the value that would give negative dimensions.

Q: Is profit maximal when revenue is maximal?

No, almost never. Profit is revenue minus cost, so it is maximal where marginal revenue equals marginal cost. As long as the next unit sold brings in more than it costs, production should continue, and it stops before the maximum of revenue as soon as the cost is increasing.

Q: Why is an optimal can as tall as it is wide?

Because for a fixed volume, the total surface area is minimal when the height is twice the radius, that is, when the height equals the diameter. The computation shows it by setting to zero the derivative of the area expressed in terms of the radius alone. It is an excellent consistency check on this type of problem.

This sheet does not repeat the list of steps of an optimization problem: it is in your notes. It deals with what the marking scheme really penalizes, namely translating the statement, the domain, and the justification of the extremum, three steps where a numerically exact answer can earn only half the marks on a 201-SN2-RE (formerly 201-NYA) exam.

It is written for CEGEP students in the Science program (Sciences de la nature). Once you have read the sheet, the corrected exercise set on the same chapter puts every reflex to the test.

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The thread of the chapter

In an optimization problem, the differentiation is the easy part. Marks are lost in the three steps around it: translating the statement into a constraint and an objective function, establishing the domain, and JUSTIFYING that the critical number really is the extremum asked for.

This chapter is part of 201-SN2-RE, Differential Calculus

The essentials

The six-step method, and where each step costs marks

• 1. DIAGRAM and variable names. Without a diagram, the constraint is wrong one time out of two.
• 2. Write the function to OPTIMIZE, the one whose maximum or minimum the question asks for.
• 3. Write the CONSTRAINT, the relation imposed by the statement (given perimeter, fixed volume, available material).
• 4. Use the constraint to eliminate a variable: the objective function must depend on only ONE unknown.
• 5. Establish the physically admissible DOMAIN, by combining all the positivity conditions.
• 6. Differentiate, find the critical numbers, and JUSTIFY that the candidate kept really is the extremum asked for.

The objective function

V(x)=x(30-2x)(40-2x)

is zero at both ends of the domain and peaks inside: this shape is what makes the only critical number necessarily a maximum.

Steps 2 and 3 carry the most marks and are the most rushed. Swapping them brings down everything else, however flawless.

Justifying the extremum: three arguments, pick one

• CLOSED bounded interval: the Extreme Value Theorem guarantees existence; compare $f$ at the critical numbers AND at both endpoints.
• A SINGLE critical number on the interval, with a change of sign of $f'$ : the local extremum is then ABSOLUTE. It is the most economical argument.
• Second Derivative Test: $f''(c)<0$ gives a maximum, $f''(c)>0$ a minimum. Silent if $f''(c)=0$ .
• OPEN interval: the endpoints are not reached. You must show that $f$ tends to a smaller value there, or to infinity in the right direction.

An optimization answer without justification is worth half the marks, even when the number is exact.

Economics: the three functions and the optimality condition

• Cost $C(x)$ , MARGINAL cost $C'(x)$ : the cost of the next unit produced.
• Revenue $R(x)=x\,p(x)$ , where $p(x)$ is the unit price, which often depends on the quantity.
• Profit $P(x)=R(x)-C(x)$ , maximal where $P'(x)=0$ , that is, where $R'(x)=C'(x)$ .
• In plain words: keep producing as long as the next unit brings in more than it costs.

With

R(x)=x(80-x)

and

C(x)=20x+300

, the gap

R-C

(the profit, in green) is largest at

x=30

, where the tangent to

R

is parallel to

C

R'(30)=C'(30)=20

The mistakes that cost marks

These are the errors I correct most often in session. Each one costs marks on a paper, even when the reasoning behind it is right.

1. Confusing the constraint with the function to optimize

the whole question

What not to write

“We have $200$ m of fencing, so I maximize $L+2\ell=200$ .”

What to write

“The constraint is $L+2\ell=200$ ; the function to OPTIMIZE is the area $A=L\ell$ .”

The fence runs along three sides:

L+2\ell=200

is the CONSTRAINT. What we maximize is the green area

A=L\ell

Why: The constraint is an imposed EQUALITY; it is not optimized. The foolproof marker is the verb of the question: what you are asked to make maximal or minimal is the objective function; what is “available”, “given” or “imposed” is the constraint.

2. Differentiating a function of two variables

the whole question, there is nothing to solve

What not to write

“ $A=L\ell$ , so $A'=\ell+L$ .”

What to write

“The constraint gives $L=200-2\ell$ , so $A(\ell)=200\ell-2\ell^{2}$ and $A'(\ell)=200-4\ell$ .”

Why: A derivative only makes sense with respect to ONE variable. Step 4 exists precisely to eliminate the second one with the constraint, and a solution that skips it writes an expression with no status, which no computation can exploit.

3. Forgetting the physically admissible domain

2 marks, and the whole answer if the critical number found falls outside the domain

What not to write

“ $V(x)=x(30-2x)(40-2x)$ is a polynomial, so I study it on $\mathbb{R}$ .”

What to write

“We need $x>0$ , $30-2x>0$ and $40-2x>0$ , so $x\in\left(0, 15\right)$ : the most restrictive condition wins.”

V

can be computed beyond

15

, but it no longer describes any box there:

V(16)=-256

. Only the green part,

0<x<15

, makes sense, and that is where we look for the maximum.

Why: The polynomial can still be computed outside it and gives negative volumes: $V(16)=-256$ . Here the two critical numbers are $5.66$ and $17.68$ , and it is the domain, and the domain alone, that rules out the second. Without it, the solution has two candidates and no reason to choose.

4. Announcing the extremum without justifying it

half the marks for the question, although the number is exact

What not to write

“ $A'(\ell)=0$ gives $\ell=50$ , so the maximum area is $5000$ m $^{2}$ .”

What to write

“ $\ell=50$ is the only critical number and $A'$ changes from $+$ to $-$ : the local maximum is therefore ABSOLUTE on $\left(0, 100\right)$ , and it is $5000$ m $^{2}$ .”

A(\ell)=200\ell-2\ell^{2}

rises then falls:

A'

changes from

+

-

\ell=50

, the only critical number in

(0, 100)

. That sentence is what turns the local maximum into an ABSOLUTE maximum.

Why: The marking scheme of an optimization problem always sets aside a mark for the justification. A critical number is a candidate: it could be a minimum, or neither. A fifteen-word sentence is enough to win back that mark.

5. Not testing the endpoints of a closed interval

the whole question

What not to write

“The only critical number of $f(x)=x^{3}-3x$ on $\left[0, 3\right]$ is $x=1$ , so the maximum is $f(1)=-2$ .”

What to write

“Candidates: $f(0)=0$ , $f(1)=-2$ , $f(3)=18$ . The absolute maximum is $18$ , reached at the ENDPOINT $x=3$ ; $x=1$ gives the minimum.”

The critical number

x=1

gives the MINIMUM; the maximum of

f

\left[0, 3\right]

is reached at the endpoint

x=3

, where the derivative does not vanish at all.

Why: On a closed bounded interval, the Extreme Value Theorem guarantees that the extrema exist, but nothing says they are inside. Comparing the three values takes thirty seconds and it is the only way to conclude.

6. Treating an open interval as if it were closed

1 mark for rigour, more if the question asks for an explanation

What not to write

“The domain is $\left(0, 15\right)$ , so I compare $V$ at the endpoints $0$ and $15$ .”

What to write

“The endpoints are not reached. I use the single critical number argument instead, or I show that $V\to 0$ as $x\to 0^{+}$ and $x\to 15^{-}$ .”

(0, 15)

, the endpoints are not candidates:

V

tends to

0

near them, but they are never reached. Conclude with the single critical number, or with these two limits.

Why: The Extreme Value Theorem requires a CLOSED BOUNDED interval. On an open one, the extremum may not exist at all. Here it does exist, but the argument that proves it is not the comparison with the endpoints: it is the uniqueness of the critical number with a change of sign.

7. Answering with the maximum when the dimensions are asked for

1 to 2 marks, although the whole computation is right

What not to write

“Question: what are the dimensions of the pen? Answer: $5000$ m $^{2}$ .”

What to write

“The dimensions are $100$ m parallel to the wall and $50$ m perpendicular to it; the corresponding area is $5000$ m $^{2}$ .”

The question “what dimensions?” expects TWO lengths:

100

m along the wall,

50

m perpendicular to it. The area of

5000

^{2}

is a complement, not the answer.

Why: The variable being optimized and the quantity asked for are two different things. Rereading the question when you conclude, and giving BOTH pieces of information whenever in doubt, is the most profitable protection of the chapter.

8. Looking for maximum profit where revenue is maximal

the whole economics question

What not to write

“Revenue is maximal at $x=40$ , so profit is too.”

What to write

“Profit is maximal where $P'(x)=0$ , that is, where $R'(x)=C'(x)$ , which is almost never the maximum of the revenue.”

Profit

P(x)=x(80-x)-(20x+300)

: maximal at

x=30

(green), where

R'=C'

. At

x=40

, the maximum of revenue (red), the profit has already dropped.

Why: The maximum of revenue ignores the cost. As long as the next unit brings in more than it costs, production should continue; it stops when marginal revenue and marginal cost become equal, which happens BEFORE the maximum of revenue as soon as the cost is increasing.

Which method to choose

Which argument justifies the extremum

Once the critical number is found, the argument depends on the NATURE of the domain, not on the problem.

If the domain is a closed bounded interval $\left[a,b\right]$ → Extreme Value Theorem: compare $f$ at the critical numbers AND at both endpoints

Example: $f(0)=0$ , $f(1)=-2$ , $f(3)=18$ : the maximum is $18$
If there is ONLY ONE critical number on the domain and $f'$ changes sign there → the local extremum is ABSOLUTE on that domain, and it is the shortest argument

Example: $A'(\ell)=200-4\ell$ changes from $+$ to $-$ at $\ell=50$
If $f''$ is easy to compute and $f''(c)\neq 0$ → Second Derivative Test: $f''(c)<0$ gives a maximum, $f''(c)>0$ a minimum

Example: $V''(5.66)=-144<0$ , so a maximum
If the domain is open and there are several critical numbers → full sign chart of $f'$ , plus the limits at the open endpoints

Example: show that $V\to 0$ at both endpoints of $\left(0, 15\right)$

The second argument covers the vast majority of exam problems and fits in one sentence. It is the one to favour when time is short.

Which constraint, depending on the type of problem

The statement always gives a FIXED quantity: that quantity, and that quantity alone, provides the constraint.

If the statement fixes a length of material (fence, wire, edging) → the constraint is a PERIMETER, and the objective is almost always an area

Example: $L+2\ell=200$ for a pen against a wall
If the statement fixes an area of material (cardboard, sheet metal, glass) → the constraint is an AREA, and the objective is a volume

Example: $2\pi r^{2}+2\pi rh=S$ for a cylindrical box
If the statement fixes a volume (capacity of a can, of a tank) → the constraint is the VOLUME, and the objective is the surface area to minimize

Example: $\pi r^{2}h=355$ gives $h=\frac{355}{\pi r^{2}}$ , then $h=2r$ at the optimum
If the statement fixes a starting shape that is cut or folded → there is no separate constraint: the geometry of the folding directly gives a function of a single variable

Example: $V(x)=x(30-2x)(40-2x)$ for a box cut from a sheet of cardboard
If the statement is about a path, a time or a travel cost → the objective is a total time or cost, and the constraint is the fixed position of the endpoints

Example: the minimal travel time gives back Snell's law

A problem where the constraint does not appear is a problem of the fourth type: the folding has already done the elimination.

How the answer is expected to be written

A marker ticks steps. Here they are in order, with the concluding sentence expected word for word.

An optimization problem, written up for the marker

When to use it: The long problem of the final exam, usually worth ten to fifteen marks.

1 Draw the diagram, name the variables and SAY so: “Let $x$ be the side of the square cut out, in centimetres.”
2 Write the function to optimize and the constraint separately, labelling them as such.
3 Eliminate a variable and give the objective function in a single unknown.
4 Establish the domain by listing each positivity condition, then give the interval.
5 Differentiate, solve $f'(x)=0$ , and explicitly rule out the solutions outside the domain.
6 JUSTIFY the extremum with one of the three arguments, then answer the question asked, with units.

Concluding sentence

The optimal cut is $x=\frac{35-\sqrt{325}}{3}\approx 5.66$ cm, which gives a box $28.7$ cm by $18.7$ cm and $5.66$ cm high, with a maximum volume of $3032$ cm $^{3}$ .

The trap: Ruling out the second root without saying so. The marker wants to see the sentence “ $17.68\notin\left(0, 15\right)$ , this solution is rejected”: it is what proves that the domain served a purpose.

Marking: Typically 2 marks for setting up the equations, 1 mark for the domain, 2 marks for differentiating and solving, 1 mark for rejecting the root outside the domain, 2 marks for the justification, 1 mark for the answer with units.

Check before you hand in

Five minutes of checking recover more marks than one more problem started in a hurry.

Check that the answer makes physical sense

Substitute the value found into every dimension of the problem and check that none is negative or zero.

$x=5.66$ gives $30-2x=18.7>0$ and $40-2x=28.7>0$ ✓. A negative width signals a root outside the domain accepted by mistake.
Compare with two neighbouring values

Evaluate the objective function a little before and a little after the point found: both values must be smaller for a maximum, larger for a minimum.

$V(5)=3000$ and $V(6)=3024$ are indeed smaller than $V(5.66)=3032$ ✓.
Check the order of magnitude with a rough bound

Bound the optimized quantity with an immediate computation, and check that the answer stays below it.

The volume of the box is less than $15\times 30\times 40=18000$ cm $^{3}$ , and much less in practice: $3032$ is plausible, $30320$ would not be.
Reread the question before writing the conclusion

Underline in the statement what is asked, dimensions or optimal value, and check that the final sentence answers it.

“What dimensions?” calls for three lengths, not a volume. Giving both costs nothing and secures the mark.
Check the known special case

Many classic problems have a remarkable answer: the square for a pen fenced on four sides, $h=2r$ for a can of given volume, $h=r$ for the Norman window.

An optimal can must be as tall as it is wide: $r\approx 3.84$ cm and $h\approx 7.67$ cm ✓. A result far from this ratio signals an error in the constraint.

The typical problem, taken apart

The box cut from a sheet of cardboard, as on the final exam

A square of side $x$ is cut from each of the four corners of a sheet of cardboard $30$ cm by $40$ cm, then the edges are folded up to form a box without a lid.

Find $x$ so that the volume is maximal, and give this volume.

Each dimension loses

x

TWICE, once at each end: that is where the factors

30-2x

and

40-2x

come from, and forgetting the factor

2

is the most frequent error in setting up the problem.

Step 1

After folding, the height is $x$ , the length $40-2x$ and the width $30-2x$ . The function to optimize is therefore $V(x)=x(30-2x)(40-2x)$ .

Why

There is no separate constraint here: the geometry of the folding has already reduced the problem to a single variable. Recognizing this type saves you from looking for a constraint that does not exist.

Step 2

Domain: $x>0$ to have a height, $30-2x>0$ that is $x<15$ , and $40-2x>0$ that is $x<20$ . The most restrictive wins: $\text{dom}=\left(0, 15\right)$ .

Why

The three conditions are written down before being combined. It is the domain, and the domain alone, that will later rule out the second root of the derivative.

Step 3

Expanding, $V(x)=4x^{3}-140x^{2}+1200x$ , so $V'(x)=12x^{2}-280x+1200=4\left(3x^{2}-70x+300\right)$ .

Why

Expanding before differentiating is faster here than applying the product rule twice, and factoring out $4$ simplifies the solving.

Step 4

$3x^{2}-70x+300=0$ gives $x=\frac{70\pm\sqrt{4900-3600}}{6}=\frac{70\pm\sqrt{1300}}{6}$ , that is, $x\approx 5.66$ or $x\approx 17.68$ . Now $17.68\notin\left(0, 15\right)$ : this solution is REJECTED.

Why

The rejection sentence earns marks. It proves that the domain established in the previous step served a purpose, and it is exactly what the marker looks for at that point in the solution.

Step 5

Justification: $V'$ is a quadratic with a positive leading coefficient, so it is positive before $5.66$ and negative after, on the whole domain. $V'$ changes from $+$ to $-$ at the ONLY critical number in the domain: the local maximum there is therefore ABSOLUTE.

Why

It is the shortest of the three arguments. We could also have invoked $V''(x)=24x-280$ , which equals $-144$ at $5.66$ , so it is negative, so a maximum. One argument is enough, provided it is written down.

Step 6

$V(5.66)=5.66\times 18.68\times 28.68\approx 3032$ cm $^{3}$ . Check: $V(5)=3000$ and $V(6)=3024$ , both smaller ✓.

Why

Two ten-second evaluations that confirm the maximum without differentiating again. They also catch arithmetic errors in the expansion of the triple product.

The conclusion, written out

By cutting out squares of side $x=\frac{35-\sqrt{325}}{3}\approx 5.66$ cm, we get a box $28.7$ cm by $18.7$ cm and $5.66$ cm high, whose volume of $3032$ cm $^{3}$ is maximal over the whole admissible domain $\left(0, 15\right)$ .

The classic mistake on this problem: Writing $V(x)=x(30-x)(40-x)$ , forgetting that each dimension loses $x$ at EACH end. The critical number is then no longer $x\approx 5.66$ but another value, and the whole problem is wrong despite a flawless method.

Learn by heart

• Objective function: what the question asks you to optimize. Constraint: what is given, fixed, available.
• The constraint serves to ELIMINATE a variable. Never differentiate an expression in two unknowns.
• Physical domain: combine all the positivity conditions; the most restrictive wins.
• Explicitly rule out the roots outside the domain: that sentence earns marks.
• Shortest justification: the only critical number in the domain, with a change of sign of $f'$ .
• CLOSED interval: also compare at both endpoints. OPEN interval: the endpoints are not reached.
• Economics: maximum profit where marginal revenue equals marginal cost, never at the maximum of revenue.
• Remarkable answers to know: the square, $h=2r$ for a can, $h=r$ for the Norman window.

Frequently asked questions

How do I recognize the constraint and the function to optimize?

The function to optimize is the one whose maximum or minimum the question asks for, often an area, a volume, a cost or a time. The constraint is the quantity fixed by the statement, the one that is available or imposed. The words given, available, fixed, imposed almost always signal the constraint.

Why must a variable be eliminated before differentiating?

Because an ordinary derivative is computed with respect to a single variable only. As long as the objective function depends on two unknowns, there is nothing to solve. The constraint provides the relation that expresses one in terms of the other and brings the problem down to a single variable.

How do I justify that a critical number really is a maximum?

Three possible arguments. On a closed bounded interval, compare the function at the critical numbers and at both endpoints. If there is only one critical number and the derivative changes sign there, the local maximum is absolute. Otherwise, use the sign of the second derivative, which must be strictly negative.

Why is the domain so important in optimization?

Because the algebraic formula can still be computed outside its physical meaning and often produces a second, spurious root. For a box cut from a sheet of cardboard thirty by forty, the derivative vanishes twice, and only the domain rules out the value that would give negative dimensions.

Is profit maximal when revenue is maximal?

No, almost never. Profit is revenue minus cost, so it is maximal where marginal revenue equals marginal cost. As long as the next unit sold brings in more than it costs, production should continue, and it stops before the maximum of revenue as soon as the cost is increasing.

Why is an optimal can as tall as it is wide?

Because for a fixed volume, the total surface area is minimal when the height is twice the radius, that is, when the height equals the diameter. The computation shows it by setting to zero the derivative of the area expressed in terms of the radius alone. It is an excellent consistency check on this type of problem.

Practise it

Corrected exercises: Applied optimization, 201-SN2-RE

A method is proved on a paper, not on a sheet. The set for the same chapter takes each of these traps into a problem, with the solution written out step by step.

15 corrected exercises
150 points
225 minutes

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Revision sheet: applied optimization (201-SN2-RE)

The essentials

The six-step method, and where each step costs marks

Justifying the extremum: three arguments, pick one

Economics: the three functions and the optimality condition

The mistakes that cost marks

1. Confusing the constraint with the function to optimize

2. Differentiating a function of two variables

3. Forgetting the physically admissible domain

4. Announcing the extremum without justifying it

5. Not testing the endpoints of a closed interval

6. Treating an open interval as if it were closed

7. Answering with the maximum when the dimensions are asked for

8. Looking for maximum profit where revenue is maximal

Which method to choose

Which argument justifies the extremum

Which constraint, depending on the type of problem

How the answer is expected to be written

An optimization problem, written up for the marker

Check before you hand in

The typical problem, taken apart

The box cut from a sheet of cardboard, as on the final exam

Learn by heart

Frequently asked questions

How do I recognize the constraint and the function to optimize?

Why must a variable be eliminated before differentiating?

How do I justify that a critical number really is a maximum?

Why is the domain so important in optimization?

Is profit maximal when revenue is maximal?

Why is an optimal can as tall as it is wide?

Corrected exercises: Applied optimization, 201-SN2-RE

See also

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