MATH 203 Calculus I • Concordia University, Montreal

Revision sheet: Applied optimization (MATH 203)

This sheet is not a summary of section 4.6 of Thomas' Calculus: you have the textbook and the lecture notes. It answers one question only, what makes students lose marks on applied optimization in MATH 203 at Concordia University, and which precise gesture avoids each loss. It is the last chapter of the course, and the final almost always ends with one of these problems.

Almost every student knows that the maximum is where $f' = 0$ . The marks are lost elsewhere: in the algebra that builds $f$ from the constraint, in the algebra that solves $f'(x) = 0$ , and in the sentence that proves the extremum is absolute. The gesture that protects all three is the same: write $f'$ as ONE factored fraction. A scientific calculator is allowed, so decimals appear where the context asks for them, but never in place of the justification.

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The thread of the chapter

The derivative is one line and the marks are lost in the ALGEBRA around it: substitute the constraint and simplify before differentiating, write $f'$ as ONE FACTORED FRACTION, read the critical numbers from its numerator and the absolute extremum from the sign of its factors on the whole domain, then answer the quantity asked.

This chapter is part of MATH 203, Calculus I (Concordia)

What is on this sheet

1. The essentials
2. The rules in table form
3. The mistakes that cost marks
4. Which method to choose
5. How the answer is expected to be written
6. Check before you hand in
7. The typical problem, taken apart
8. Learn by heart
9. Frequently asked questions

Topics covered

Applied optimization
Objective function
Constraint equation
Closed Interval Method
First Derivative Test
Absolute extreme values
Critical points
Negative exponents
Radical equations
Inscribed solids
Similar triangles
Revenue and profit maximization

The essentials

The method, and where the algebra sits in it

• Read, draw, name the variables with units. Write the CONSTRAINT and solve it for the variable that is easy to isolate: $xy = 1800$ gives $y = \frac{1800}{x}$ .
• Write the OBJECTIVE in one variable and SIMPLIFY before differentiating: $x^2 \cdot \frac{2700 - x^2}{4x} = \frac{2700x - x^3}{4}$ , the $4x$ dividing the whole numerator.
• State the DOMAIN from the situation (every length $\ge 0$ , every cap), and say whether it is closed or open.
• Differentiate, then write $f'$ as ONE fraction and FACTOR it: $24 - \frac{21600}{x^2} = \frac{24(x - 30)(x + 30)}{x^2}$ . The numerator gives the critical numbers; the signs of the factors give the rest.
• Justify the absolute extremum, then answer every quantity asked, with units: dimensions, cost, price, distance.

On a MATH 203 final, the set-up and the justification together are usually worth more than the derivative itself. A correct derivative with no domain and no justification earns about half the question.

The two licences for an ABSOLUTE extremum

• Closed and bounded domain $[a, b]$ , $f$ continuous: Closed Interval Method, the value at each critical number compared with $f(a)$ AND $f(b)$ .
• Open or unbounded domain, such as $(0, \infty)$ : no endpoint to evaluate. The sign of $f'$ on the WHOLE domain decides: negative then positive around $c$ proves an absolute minimum.
• The factored fraction makes the sign readable: in $\frac{2(\pi r^3 - 1000)}{r^2}$ the denominator is positive, so only $\pi r^3 - 1000$ matters.
• A critical number outside the domain is struck off; if none is left, $f'$ has one sign and the optimum is at an ENDPOINT, where $f' \ne 0$ .
• A single critical number is not automatically a maximum: the open cup's only critical number is a minimum, and the cup has no maximum area.

The cup's area

A(r) = \pi r^2 + \frac{2000}{r}

blows up at both ends of

(0, \infty)

: no endpoint to test, but

A'

changes sign once, near

r = 6.83

Write the licence in one sentence with its hypothesis: “ $A$ is continuous on the closed interval $[0, 60]$ , so by the Closed Interval Method...”. That sentence is the method mark.

The rules in table form

Each row reads from left to right: the assumptions, then the result. A red cell is not an answer, it is the finding that the form settles nothing and the instruction to rewrite it. Every case is followed by a worked example.

The algebra before and after the derivative

Read a line as: faced with this form, make this gesture, and this is what you get. The red lines are rules that do not exist, written on real assignments.

Form	Gesture	Result
$\frac{a}{x}$	write $ax^{-1}$	$-\frac{a}{x^2}$
Example: $\frac{d}{dx}\frac{21600}{x} = -\frac{21600}{x^2}$ , so $C'(10) = 24 - 216 = -192$ .
$a - \frac{b}{x^2}$	one fraction	$\frac{ax^2 - b}{x^2}$
Example: $24 - \frac{21600}{x^2} = \frac{24(x^2 - 900)}{x^2}$ , zero at $x = 30$ only on $(0, \infty)$ .
$\frac{A - B}{C}$	split	$\frac{A}{C} - \frac{B}{C}$
Example: $\frac{2 - \pi r^2/2}{2r} = \frac{1}{r} - \frac{\pi r}{4}$ ; at $r = 0.5$ : $2 - \frac{\pi}{8} \approx 1.6073$ .
$\sqrt{a + b}$	split	$\sqrt a + \sqrt b$ rule that does not exist
Example: $\sqrt{1.44 + 0.25} = 1.3$ , but $\sqrt{1.44} + \sqrt{0.25} = 1.7$ . What to do: Keep the radical and differentiate it by the chain rule; remove it only by squaring an ISOLATED radical.
$x \cdot g(x) = 0$	divide by $x$	$g(x) = 0$ loses a root
Example: $4x^3 - 14x = 0$ divided by $x$ loses $x = 0$ , where $x^4 - 7x^2 + 16$ has its local maximum $16$ . Same form, other result: $e^{-p/8}(8 - p) = 0$ may be divided by $e^{-p/8}$ , which is never $0$ : only $p = 8$ . What to do: Factor: $2x(2x^2 - 7) = 0$ gives $x = 0$ or $x = \pm\sqrt{\frac{7}{2}}$ .

The last two lines are the same mistake in two costumes: an operation that is only legal under a condition (a positive radicand split, a nonzero divisor), used without it.

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. Losing the minus sign of a negative exponent

the whole question

What not to write

“ $C(x) = 24x + \frac{21600}{x}$ , so $C'(x) = 24 + \frac{21600}{x^2} > 0$ , and the cheapest garden has $x$ as small as possible.”

What to write

“ $C(x) = 24x + 21600x^{-1}$ , so $C'(x) = 24 - 21600x^{-2} = \frac{24(x - 30)(x + 30)}{x^2}$ , negative on $(0, 30)$ and positive after: the minimum is at $x = 30$ .”

Why: The exponent $-1$ comes down as a factor with its sign. A check at one point, $C'(10) = -192$ , catches it, and so does common sense: a garden $1$ cm wide needs kilometres of fence.

2. Dividing a fraction in one term only

the whole question, and an impossible box

What not to write

“ $V = x^2 \cdot \frac{2700 - x^2}{4x} = 675x - x^2$ , so $V' = 675 - 2x = 0$ and $x = 337.5$ .”

What to write

“ $V = \frac{x(2700 - x^2)}{4} = 675x - \frac{x^3}{4}$ , so $V'(x) = \frac{3(30 - x)(30 + x)}{4}$ and $x = 30$ .”

Why: The denominator $4x$ divides every term of the numerator. A candidate far outside the domain, $337.5$ against $30\sqrt 3 \approx 52$ , is the alarm bell.

3. Dividing by x and losing a critical number

2 to 3 marks, and the wrong answer

What not to write

“ $D'(x) = 4x^3 - 14x = 0$ , so $4x^2 = 14$ and $x = \pm\sqrt{\frac{7}{2}}$ . On the arch $[-2, 2]$ the farthest point from the origin is $(\pm 2, 0)$ .”

What to write

“ $D'(x) = 2x(2x^2 - 7) = 0$ gives $x = 0$ or $x = \pm\sqrt{\frac{7}{2}}$ . $D(0) = 16$ is the largest value: the farthest point is $(0, 4)$ .”

D(x) = x^4 - 7x^2 + 16

has THREE critical numbers: two minima at

x = \pm 1.87

and a local maximum at

x = 0

, the one that dividing by

x

deletes.

Why: Dividing by $x$ assumes $x \ne 0$ . The deleted root was exactly the answer to the second question: factor, never divide by something that can vanish.

4. Splitting a square root, or squaring without checking

2 marks

What not to write

“ $\sqrt{1.44 + x^2} = 1.2 + x$ ” or “squaring $1.3x = 0.5\sqrt{1.44 + x^2}$ gives $x = \pm 0.5$ , two candidates.”

What to write

“Isolate: $1.3x = 0.5\sqrt{1.44 + x^2}$ . Square: $x^2 = 0.25$ . Check: $x = -0.5$ makes the left side negative, so only $x = 0.5$ .”

Why: $\sqrt{1.69} = 1.3$ , not $1.7$ . Squaring forgets signs, so it can create roots: every root of the squared equation is checked in the original one.

5. Rounding too early with a calculator

1 mark on the final answer

What not to write

“ $r \approx 6.83$ cm, so the least area is $3\pi(6.83)^2 \approx 439.66$ cm $^2$ .”

What to write

“ $r = \frac{10}{\sqrt[3]{\pi}}$ , kept in memory; the least area is $3\pi r^2 \approx 439.38$ cm $^2$ .”

Why: Squaring and multiplying by $3\pi$ amplifies the rounding error. Round once, at the very end, to the precision the question announces.

6. Keeping a critical number that a cap has excluded

the whole question

What not to write

“ $\Pi'(p) = 75e^{-p/8}(11 - p) = 0$ , so with prices capped at $10$ dollars, the best price is $11$ dollars.”

What to write

“On $[0, 10]$ , $11 - p > 0$ , so $\Pi$ is increasing: the best price is the cap, $10$ dollars, for a profit of about $803.32$ dollars.”

Why: A constraint that binds moves the optimum to the boundary, where $f' \ne 0$ . The domain is read BEFORE the critical numbers are kept.

7. Expanding a binomial square as two squares

the whole question

What not to write

“For the cone in the sphere of radius $9$ : $r^2 = 81 - (h - 9)^2 = 81 - (h^2 - 81) = 162 - h^2$ .”

What to write

“ $(h - 9)^2 = h^2 - 18h + 81$ , so $r^2 = 18h - h^2$ and $V(h) = \frac{\pi}{3}(18h^2 - h^3)$ , largest at $h = 12$ : $288\pi$ cm $^3$ .”

Why: $(a - b)^2$ has a middle term $-2ab$ . The five-second check: at $h = 0$ the cone is a point, so $r^2$ must be $0$ ; the false formula gives $162$ , a base wider than the sphere.

8. Answering with the critical number

1 to 2 marks, the conclusion line

What not to write

“The cheapest garden of $1800$ m $^2$ is $x = 30$ .”

What to write

“The cheapest garden is $30$ m along the street and $60$ m deep, and it costs $1440$ dollars.”

Why: $x = 30$ is a length and a candidate. The question asked for the garden and its cost: compute them from the constraint, with units.

Which method to choose

Which gesture, by the FORM of the objective and of the domain

Once the objective is written in one variable, look at its form, then at the domain

If a term $\frac{a}{x}$ or $\frac{a}{x^2}$ → rewrite with a negative exponent, differentiate, then bring $f'$ back to ONE fraction over $x^2$ or $x^3$

Example: $\pi r^2 + \frac{2000}{r}$ gives $\frac{2(\pi r^3 - 1000)}{r^2}$
If a common factor $x$ , or an exponential factor, in $f'$ → factor it out; keep $x = 0$ as a root, divide only by $e^{u}$ , which is never $0$

Example: $2x(2x^2 - 7)$ ; $75e^{-p/8}(11 - p)$
If a square root in $f$ → chain rule; in $f' = 0$ isolate the radical, square, check each root in the original equation. For a distance, optimize $d^2$ instead

Example: $1.3x = 0.5\sqrt{1.44 + x^2}$ gives $x = 0.5$ only
If a length inside a circle or a sphere → Pythagoras, the binomial square expanded in full, and the variable that keeps the square root away

Example: Cone in a sphere: $r^2 = 81 - (h - 9)^2 = 18h - h^2$
If a closed domain $[a, b]$ → Closed Interval Method: every interior critical number and both endpoints

Example: Cone: $V(0) = 0$ , $V(12) = 288\pi$ , $V(18) = 0$
If an open or unbounded domain → sign of $f'$ on the WHOLE domain, read on the factored fraction

Example: Garden: $\frac{24(x - 30)(x + 30)}{x^2}$ changes sign at $30$ only

If $f'$ has no zero in the domain, it has one sign there: the optimum is at an endpoint, or does not exist on an open domain. Say which, with the reason.

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.

Writing up an applied optimization problem

When to use it: Any question asking for the largest, smallest, cheapest, closest or most profitable, from a situation to model

1 Draw the figure, name the variables with units, and say which quantity is optimized.
2 Write the constraint and solve it for the easy variable; substitute and SIMPLIFY the objective to one variable.
3 State the domain with the reason for each bound, and whether it is closed or open.
4 Differentiate, naming the rules; write $f'$ as one factored fraction and list the critical numbers IN the domain.
5 Name the licence and apply it: Closed Interval Method with every value, or the sign of $f'$ on the whole domain.
6 Answer every quantity asked, with units and the rounding requested.

Concluding sentence

“ $C$ is differentiable on the open interval $(0, \infty)$ and $C'(x) = \frac{24(x - 30)(x + 30)}{x^2}$ is negative on $(0, 30)$ and positive on $(30, \infty)$ . So $C$ decreases then increases, and $C(30) = 1440$ is its absolute minimum: the cheapest garden is $30$ m by $60$ m, for $1440$ dollars.”

The trap: Stopping at $x = 30$ : the candidate is not the answer, and without the sign sentence the minimum is not proved.

Marking: Typically 3 marks for the set-up (constraint, objective, domain), 2 for the derivative and critical numbers, 3 for the justification, 2 for the final answer with units.

Check before you hand in

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

The answer satisfies the constraint

Put the optimal values back into the constraint. It takes five seconds and catches most substitution slips.

Box: $30^2 + 4 \times 30 \times 15 = 900 + 1800 = 2700$ . A box $30$ by $30$ by $20$ would fail at once.
The sign of f' at one convenient point

Evaluate f' at an easy value on each side of the candidate. The signs must match the sign table, or the derivative is wrong.

Garden: $C'(10) = -192 < 0$ and $C'(60) = 24 - 6 = 18 > 0$ . A formula that gives $C'(10) > 0$ has lost a minus sign.
The candidate is inside the domain

A critical number beyond a length, a width or a cap is a symptom: either the algebra is wrong or the optimum is at an endpoint.

$x = 337.5$ for a box made of $2700$ cm $^2$ of cardboard is impossible: the simplification of $V$ was wrong.
The shape is a classic

Many standard problems have a clean optimal shape. If yours has the same set-up and another shape, look again.

Open cup $h = r$ ; open square-based box $h = \frac{x}{2}$ ; closed box a cube; Norman window $h = r$ ; rectangle in a triangle: half its area.

The typical problem, taken apart

The open tank of least sheet metal

An open-top tank has a rectangular base whose length is twice its width, and a volume of $36$ m $^3$ . Find the dimensions that use the least sheet metal, and that area.

Every step must be justified as on a MATH 203 final.

Width

x

, length

2x

, height

h

, no top: the metal is the base

2x^2

plus four walls, two of area

2xh

and two of area

xh

Step 1

Constraint: $2x \cdot x \cdot h = 36$ , so $h = \frac{18}{x^2}$ . Metal: $S = 2x^2 + 2(2xh) + 2(xh) = 2x^2 + 6xh = 2x^2 + 6x \cdot \frac{18}{x^2} = 2x^2 + \frac{108}{x}$ .

Why

The constraint is used BEFORE differentiating, and $6x \cdot \frac{18}{x^2}$ is simplified to $\frac{108}{x}$ at once: one power of $x$ cancels, not two.

Step 2

Domain: any width $x > 0$ gives a tank, so $x \in (0, \infty)$ . As $x \to 0^+$ , $\frac{108}{x} \to \infty$ ; as $x \to \infty$ , $2x^2 \to \infty$ .

Why

The domain is open: the Closed Interval Method is unavailable, and saying so tells the marker which licence comes next.

Step 3

$S'(x) = 4x - 108x^{-2} = \frac{4x^3 - 108}{x^2} = \frac{4(x^3 - 27)}{x^2}$ . On $(0, \infty)$ the only zero is $x = 3$ , and the sign of $S'$ is that of $x^3 - 27$ : negative on $(0, 3)$ , positive on $(3, \infty)$ .

Why

One factored fraction gives the critical number AND the sign on the whole domain: the derivative line already contains the justification.

Step 4

So $S$ decreases then increases: $S(3) = 18 + 36 = 54$ is the absolute minimum. Then $h = \frac{18}{9} = 2$ : the tank is $3$ m wide, $6$ m long and $2$ m high.

Why

The licence is named, then the question is answered with the tank's dimensions and the area, not with $x = 3$ .

Step 5

Check: volume $3 \times 6 \times 2 = 36$ ; nearby widths do worse, $S(2) = 8 + 54 = 62$ and $S(4) = 32 + 27 = 59$ .

Why

The constraint and two evaluations on either side catch a slip in the simplification of $S$ .

The conclusion, written out

“ $S(x) = 2x^2 + \frac{108}{x}$ on $(0, \infty)$ has $S'(x) = \frac{4(x^3 - 27)}{x^2}$ , negative on $(0, 3)$ and positive on $(3, \infty)$ , so its absolute minimum is $S(3) = 54$ . The tank is $3$ m by $6$ m by $2$ m and uses $54$ m $^2$ of sheet metal.”

The classic mistake on this problem: Writing $6x \cdot \frac{18}{x^2} = \frac{108}{x^2}$ (cancelling nothing), which gives $S' = 4x - \frac{216}{x^3}$ and a wrong width.

Learn by heart

• Constraint, objective in ONE variable, simplified; domain; critical numbers; licence; answer with units.
• $\frac{d}{dx}\frac{a}{x} = -\frac{a}{x^2}$ : the minus sign comes down with the exponent.
• Write $f'$ as ONE factored fraction: its numerator gives the critical numbers, its factors give the sign.
• Factor, never divide by an expression that can be $0$ ; $e^{u}$ is never $0$ .
• $\sqrt{a + b} \ne \sqrt a + \sqrt b$ ; square only an isolated radical, then check every root.
• Closed $[a, b]$ : compare with BOTH endpoints. Open domain: sign of $f'$ on the WHOLE domain.
• A cap or a short distance can exclude the critical number: the optimum is then at an endpoint, where $f' \ne 0$ .
• Round once, at the end, to the announced precision.

Frequently asked questions

How do I find the critical numbers when the derivative has a fraction in it?

Put everything over a common denominator so the derivative is a single fraction, then factor the numerator. The critical numbers are where the numerator is zero, plus any point of the domain where the denominator is zero. The same factored fraction then tells you the sign of the derivative on each interval, which is the justification you need.

How do I justify that my answer is an absolute maximum or minimum in an optimization problem?

Look at the domain. If it is a closed interval and the function is continuous, compare the value at each critical number with the values at both endpoints. If the domain is open or unbounded, show that the derivative keeps one sign before the critical number and the opposite sign after it, on the whole domain. Then write that sentence on your copy.

Can I divide both sides of f'(x) = 0 by x?

No, not unless you have already shown that x cannot be zero. Dividing by x deletes the solution x equals zero, and that point can be the answer to the question. Bring everything to one side and factor out the common factor instead. Dividing by an exponential factor is safe, because an exponential is never zero.

Should I round my answer in a MATH 203 optimization problem?

Give the exact value when it is natural, such as a square root of three or a fraction, and round only when the question asks for a decimal or the context needs one, like a length in centimetres or an amount of money. Keep the unrounded value in your calculator until the last step, since rounding early changes the final digits.

Practise it

Corrected exercises: Applied optimization, MATH 203 at Concordia

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.

10 corrected exercises
100 points
150 minutes

Do the exercises

Previous sheet Concavity and curve sketching

© Ahmed Squalli Houssaini. Revision sheet published at www.letuteurscientifique.ca/en/fiches/math203-optimization. Free for personal and classroom use; republishing it elsewhere requires written permission (legal notice).

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Revision sheet: Applied optimization (MATH 203)

The essentials

The method, and where the algebra sits in it

The two licences for an ABSOLUTE extremum

The rules in table form

The algebra before and after the derivative

The mistakes that cost marks

1. Losing the minus sign of a negative exponent

2. Dividing a fraction in one term only

3. Dividing by x and losing a critical number

4. Splitting a square root, or squaring without checking

5. Rounding too early with a calculator

6. Keeping a critical number that a cap has excluded

7. Expanding a binomial square as two squares

8. Answering with the critical number

Which method to choose

Which gesture, by the FORM of the objective and of the domain

How the answer is expected to be written

Writing up an applied optimization problem

Check before you hand in

The typical problem, taken apart

The open tank of least sheet metal

Learn by heart

Frequently asked questions

How do I find the critical numbers when the derivative has a fraction in it?

How do I justify that my answer is an absolute maximum or minimum in an optimization problem?

Can I divide both sides of f'(x) = 0 by x?

Should I round my answer in a MATH 203 optimization problem?

Corrected exercises: Applied optimization, MATH 203 at Concordia

See also

Looking for a MATH 203 tutor in Montreal?