Which justification should I use, the first or second derivative test?

Either, if it covers the whole domain. The second derivative test at a critical point shows only a local maximum or minimum; it becomes the absolute one when that is the only critical point in the domain, or when the second derivative keeps its sign on the whole domain. A first derivative sign chart across the domain does this in one step, and is the safer choice when the second derivative is awkward or zero.

Do I have to check endpoints every time?

Whenever the domain is a closed interval, yes. When the domain is open (a length that must be strictly positive, for example), you instead check the behaviour as the variable approaches each end, which is often enough to rule those ends out.

Is it acceptable to minimise the squared distance instead of the distance?

Yes, and it is usually the better choice. Squaring is an increasing function on non-negative values, so the two are minimised at the same point, and the algebra is considerably lighter.

MATH 100 optimization practice

An optimization question hands you a situation and asks for the best possible version of it: the largest area, the cheapest can, the closest point. The calculus involved is usually just a derivative set to zero. The marks are in everything around that.

Every one of these problems has the same skeleton. Name the quantity being optimised, write it as a function, use the constraint to reduce it to a single variable, differentiate, and then justify that the critical point you found really is the maximum or minimum on the domain the situation allows.

That last step is the one most often skipped and most often worth marks. A critical point is a candidate, not an answer, and a closed interval has endpoints that are candidates too.

Start practising Browse all 18 questions

In this topic

Taught: Before the final
Questions: 18
Total marks: 53
Suggested time: 93 min
Difficulty: Introductory to Capstone
Show the 5 hardest

Key results

Where extremes can be: $f \text{ has a local max or min at } c \implies f'(c) = 0 \text{ or } f'(c) \text{ does not exist}$
When $c$ is not an endpoint of the domain; $f'(c) = 0$ makes $c$ a critical point, and a missing $f'(c)$ a singular point.
The closed interval method: $\begin{aligned} &\max_{[a, b]} f \text{ and } \min_{[a, b]} f \text{ occur at} \\ &\text{critical points, singular points} \\ &\text{or the endpoints } a, b \end{aligned}$
When $f$ is continuous on the closed, finite interval $[a, b]$ ; both extremes then exist.
The second derivative test: $\begin{aligned} f'(c) = 0,\ f''(c) < 0 &\implies \text{local max at } c \\ f'(c) = 0,\ f''(c) > 0 &\implies \text{local min at } c \end{aligned}$
When $f''(c)$ exists; if $f''(c) = 0$ , the test gives no conclusion.
One turning point on an interval: $f'(x) > 0 \text{ for } x < c,\ \ f'(x) < 0 \text{ for } x > c \implies f(c) = \max_{I} f$
When $f$ is continuous on an interval $I$ containing $c$ and the signs hold at every other point of $I$ ; reversed signs make $f(c)$ the minimum.
One critical point, one concavity: $f'(c) = 0 \text{ and } f'' < 0 \text{ on } I \implies f(c) = \max_{I} f$
When $I$ is an interval containing $c$ ; with $f'' > 0$ on $I$ instead, $f(c)$ is the minimum.

Read the theoryUBC Math 100 §8 OptimizationCLP-1 §3.5 Optimisation

Worked example

Standard3 marks · 5 min

A rectangular plot is fenced on three sides, with a straight river forming the fourth side. With 240 m of fencing available, what is the largest area that can be enclosed?

Give the area in square metres.

Strategy

Write the quantity to be maximised, use the constraint to reduce it to one variable, differentiate, and confirm the critical point really is the maximum.

Working

$2x + y = 240 \implies y = 240 - 2x$
The constraint. Only three sides need fencing because the river bounds the fourth.
$A(x) = x(240 - 2x) = 240x - 2x^{2}, \qquad 0 \le x \le 120$
Neither side can be negative, so x runs from 0 to 120.
$A'(x) = 240 - 4x = 0 \implies x = 60$
Differentiate and solve for the critical point.
$A''(x) = -4 < 0$
A is concave down on the whole domain, so its only critical point, x = 60, gives the absolute maximum.
$y = 240 - 120 = 120, \qquad A = 60 \times 120 = 7200 \ \text{m}^{2}$
Back-substitute for the other dimension and the area.

Answer

A_{\max} = 7200 \ \text{m}^{2}

Check

Sides of 50 m or 70 m give 50 × 140 = 7000 m² and 70 × 100 = 7000 m², both less than 7200 m².

Where this goes wrong

Using the constraint 2x + 2y = 240 out of habit, as though all four sides were fenced. That gives a 60 × 60 square of area 3600 and misses half the available area.

On the paper

Name the variables and state the domain. Write the quantity in one variable, justify the maximum or minimum on the whole domain, and answer the question asked, with units.

Leans on

Writing a function from a word problem, from Functions, graphs and inverses

Your turn

Standard3 marks · 5 min

An open-topped box is made from a 30 cm by 30 cm square sheet of card by cutting a square of side x from each corner and folding up the sides. Which value of x gives the largest volume?

V(x) = x(30 - 2x)^{2}, \qquad 0 < x < 15

Answer this question

What you should be able to do

Separate the objective from the constraint
One quantity is being optimised and another is fixed. Identifying which is which before writing anything down prevents the most expensive error in the topic.
Reduce to a single variable
Solve the constraint for whichever variable is easiest to remove, substitute, and simplify. You cannot differentiate usefully until the objective depends on one thing.
Establish the domain from the situation
A length cannot be negative and a cut-out corner cannot exceed half the sheet. The physical setup, not the algebra, decides which critical points are admissible.
Justify the extreme value
Use a first derivative sign chart across the whole domain, the second derivative test together with the fact that this is the only critical point in the domain, or the closed-interval method. Stating which test you used and what it showed is part of a complete answer.
Answer the question that was asked
Some questions want the dimensions, some want the optimal value, some want both. Read the final line again before writing your answer down.

Where marks are usually lost

Optimising the constraint by accident
In a fixed-volume can problem the volume is given, so it cannot be what you maximise. Surface area is the objective. Write down which is which before differentiating.
Assuming all four sides need fencing
When a river, wall or existing fence bounds one side, the constraint has three terms, not four. Sketching the region first makes the correct constraint obvious.
Keeping a critical point outside the domain
Solving $V'(x) = 0$ for a box made from a 30 cm sheet gives $x = 5$ and $x = 15$ . At $x = 15$ the box has no base at all, so it is rejected on physical grounds rather than algebraic ones.
Ignoring the endpoints on a closed interval
On $[a, b]$ the absolute extremes can occur at either end. Evaluating only the critical points is the standard way to lose a closed-interval question.
Stopping at the critical point
Without a second derivative, a sign chart or an endpoint comparison, nothing so far distinguishes a maximum from a minimum, or from a point that is neither, as $x = 0$ is for $x^{3}$ .

Practice questions

5 of the 18, across every level. Open a solution to read the working, or answer the question yourself and it counts towards your progress.

Introductory2 marks · 3 min
Find the global minimum value of $f(x) = x^{3} - 3x + 1$ on the closed interval $[0, 2]$ .
Strategy
The closed-interval method: list the critical points inside the interval and the two endpoints, evaluate f at each, and pick the smallest.
Working
$f'(x) = 3x^{2} - 3 = 0 \quad\Longrightarrow\quad x = \pm 1;\ \text{only } x = 1 \in (0, 2)$
Critical points.
$f(0) = 1, \qquad f(1) = -1, \qquad f(2) = 3$
Candidates.
$\min = -1 \ \text{at } x = 1$
Smallest of the three.
Answer
$-1$
The answer is negative 1, attained at x = 1.
Check
f(x) + 1 = (x − 1)²(x + 2), which is never negative for x ≥ 0, so f never drops below −1 on [0, 2].
Where this goes wrong
Answering x = 1, the location, when the question asks for the minimum value f(1) = −1. Read whether a question wants where or what.
Leans on
Finding critical points, from Curve sketching
Try a similar question
Find the absolute maximum value of $f(x) = x^{3} - 6x^{2} + 9x + 2$ on the closed interval $[0, 5]$ .
Answer this question
Introductory2 marks · 3 min
Find the coordinates of the local minimum of $h(x) = x^{3} - 27x + 5$ .
Strategy
Find the critical points, classify them with the second derivative, and evaluate h at the minimum to get both coordinates.
Working
$h'(x) = 3x^{2} - 27 = 3(x - 3)(x + 3) = 0 \quad\Longrightarrow\quad x = \pm 3$
Two critical points.
$h''(x) = 6x:\quad h''(3) = 18 > 0 \ (\text{min}), \quad h''(-3) = -18 < 0 \ (\text{max})$
Second derivative test.
$h(3) = 27 - 81 + 5 = -49$
The local minimum is at (3, −49); the local maximum is at (−3, 59).
Answer
$(3, -49)$
The answer is the point 3, negative 49.
Check
h(2) = −41 and h(4) = −39 both lie above h(3) = −49, so the curve dips to a low point at (3, −49).
Where this goes wrong
Reporting (−3, 59), the local maximum, or (3, 5) by forgetting to evaluate h. A local minimum is a point on the curve, with the function value as its second coordinate.
Leans on
The second derivative test, from Curve sketching
Try a similar question
Find the local maximum value of $f(x) = x^{3} - 3x^{2} - 9x + 5$ .
Answer this question
Standard2 marks · 4 min
Find the absolute maximum value of $f(x) = x^{3} - 6x^{2} + 9x + 2$ on the closed interval $[0, 5]$ .
Strategy
Use the closed-interval method: find the critical points inside the interval, evaluate the function at those points and at both endpoints, and compare the list.
Working
$f'(x) = 3x^{2} - 12x + 9 = 3(x - 1)(x - 3)$
Differentiate and factor. Both critical points lie inside [0, 5].
$f(0) = 2, \qquad f(5) = 125 - 150 + 45 + 2 = 22$
Evaluate at the endpoints. Skipping these is the usual way to lose this question.
$f(1) = 1 - 6 + 9 + 2 = 6, \qquad f(3) = 27 - 54 + 27 + 2 = 2$
Evaluate at the critical points.
$\max\{2,\ 22,\ 6,\ 2\} = 22$
The largest value is at the endpoint x = 5, well above the local maximum f(1) = 6.
Answer
$\max_{[0,5]} f = 22$
The answer is 22, at the endpoint x = 5.
Check
f(x) − 22 = (x − 5)(x² − x + 4), and x² − x + 4 is always positive, so f(x) ≤ 22 for every x up to 5.
Where this goes wrong
Checking only the critical points and answering with the local maximum f(1) = 6, or answering f(3) = 2 by picking the wrong critical point. On a closed interval the endpoints are candidates too, and here x = 5 gives the largest value.
Leans on
Critical points, from Differentiation rules
Try a similar question
Find the local maximum value of $f(x) = x^{3} - 3x^{2} - 9x + 5$ .
Answer this question
Challenging4 marks · 7 min
Find the $x$ -coordinate of the point on the curve $y = \sqrt{x}$ closest to the point $(4, 0)$ .
Strategy
Write the squared distance from a general point on the curve to the fixed point. Because the square root is increasing, whatever minimises the square also minimises the distance, so the root can be dropped.
Working
$\bigl(x, \sqrt{x}\bigr) \text{ is a general point on the curve}, \quad x \geq 0$
Parametrise by x so there is only one variable from the start.
$D^{2}(x) = (x - 4)^{2} + \bigl(\sqrt{x}\bigr)^{2} = (x - 4)^{2} + x$
Squaring the root collapses the second term, which is what makes this approach worth taking.
$\frac{d}{dx}D^{2} = 2(x - 4) + 1 = 2x - 7$
Differentiate the squared distance.
$2x - 7 = 0 \implies x = \frac{7}{2} = 3.5$
Solve for the critical point.
$\frac{d^{2}}{dx^{2}}D^{2} = 2 > 0$
The second derivative is 2 for every x ≥ 0, so D² is concave up on the whole domain and x = 7/2 gives its absolute minimum. The closest point is (3.5, √3.5).
Answer
$x = \frac{7}{2}$
The answer is x = 3.5, that is, 7/2.
Check
The segment from (3.5, √3.5) to (4, 0) has slope −2√3.5 and the tangent there has slope 1/(2√3.5): perpendicular, as the closest point needs.
Where this goes wrong
Taking the point straight above (4, 0), at x = 4. It is 2 away, but the curve comes closer a little to the left: minimise the squared distance (x − 4)² + x rather than guessing.
Leans on
Distance between two points, from Functions, graphs and inverses
Try a similar question
An open-topped box is made from a 30 cm by 30 cm square sheet of card by cutting a square of side x from each corner and folding up the sides. Which value of x gives the largest volume? $V(x) = x(30 - 2x)^{2}, \qquad 0 < x < 15$
Answer this question
Capstone5 marks · 10 min
A rowing boat is 6 km offshore from the nearest point $A$ on a straight coastline. The rower wants to reach a camp on the coast 10 km from $A$ . She can row at 5 km/h and walk along the coast at 13 km/h. To minimise her total travel time, how far from $A$ should she land?
Give the distance from A in kilometres.
Strategy
Model the time as a function of the landing point, differentiate, solve for the critical point, and compare against the endpoints because the domain is a closed interval.
Working
$T(x) = \frac{\sqrt{36 + x^{2}}}{5} + \frac{10 - x}{13}, \qquad 0 \leq x \leq 10$
Time is distance over speed for each leg.
$T'(x) = \frac{x}{5\sqrt{36 + x^{2}}} - \frac{1}{13} = 0 \quad\Longrightarrow\quad 13x = 5\sqrt{36 + x^{2}}$
Chain rule on the square root.
$169x^{2} = 25(36 + x^{2}) \quad\Longrightarrow\quad 144x^{2} = 900 \quad\Longrightarrow\quad x = \frac{30}{12} = 2.5$
Square both sides (x > 0, so no extraneous root).
$T(0) = 1.2 + \tfrac{10}{13} \approx 1.969,\quad T(2.5) = 1.3 + \tfrac{7.5}{13} \approx 1.877,\quad T(10) = \tfrac{\sqrt{136}}{5} \approx 2.332$
The critical point beats both endpoints: land 2.5 km from A, in hours about 1.88.
Answer
$x = 2.5 \text{ km from } A$
The answer is 2.5 kilometres from A.
Check
Landing 2 km or 3 km from A takes about 112.8 minutes, against 112.6 minutes for 2.5 km.
Where this goes wrong
Squaring 13x = 5√(36 + x²) as 169x² = 25·36 + x², dropping the 25 on the x² term. Keep the 25 on the whole bracket; and always compare with the endpoints, since rowing straight to A or straight to the camp are real options.
Leans on
The chain rule on a square root, from The chain rule and logarithms
Try a similar question
Find the $x$ -coordinate of the point on the curve $y = \sqrt{x}$ closest to the point $(4, 0)$ .
Answer this question

Browse all 18 questions

Common questions

Which justification should I use, the first or second derivative test?: Either, if it covers the whole domain. The second derivative test at a critical point shows only a local maximum or minimum; it becomes the absolute one when that is the only critical point in the domain, or when the second derivative keeps its sign on the whole domain. A first derivative sign chart across the domain does this in one step, and is the safer choice when the second derivative is awkward or zero.
Do I have to check endpoints every time?: Whenever the domain is a closed interval, yes. When the domain is open (a length that must be strictly positive, for example), you instead check the behaviour as the variable approaches each end, which is often enough to rule those ends out.
Is it acceptable to minimise the squared distance instead of the distance?: Yes, and it is usually the better choice. Squaring is an increasing function on non-negative values, so the two are minimised at the same point, and the algebra is considerably lighter.

Differentiation rules
Power, sum, product and quotient rules, tangent lines and higher-order derivatives.
23 questions
Curve sketching
Increase, concavity, inflection points, asymptotes and full sketches.
29 questions
Related rates
Translating a moving situation into an equation, then differentiating.
26 questions

Key results

Worked example

Strategy

Working

Answer

Check

Where this goes wrong

On the paper

Leans on

Your turn

What you should be able to do

Separate the objective from the constraint

Reduce to a single variable

Establish the domain from the situation

Justify the extreme value

Answer the question that was asked

Where marks are usually lost

Optimising the constraint by accident

Assuming all four sides need fencing

Keeping a critical point outside the domain

Ignoring the endpoints on a closed interval

Stopping at the critical point

Practice questions

Strategy

Working

Answer

Check

Where this goes wrong

Leans on

Try a similar question

Strategy

Working

Answer

Check

Where this goes wrong

Leans on

Try a similar question

Strategy

Working

Answer

Check

Where this goes wrong

Leans on

Try a similar question

Strategy

Working

Answer

Check

Where this goes wrong

Leans on

Try a similar question

Strategy

Working

Answer

Check

Where this goes wrong

Leans on

Try a similar question

Common questions

Related topics

Differentiation rules

Curve sketching

Related rates