Chapter 6 · Application of Derivatives

Exercise 6.3 · Q17: Open box from an 18 cm square of tin: cut squares of side $3$ cm

Exercise 6.3Optimisation · Section 6.4

The question

A square piece of tin of side 18 cm is to be made into a box without top, by cutting a square from each corner and folding up the flaps to form the box. What should be the side of the square to be cut off so that the volume of the box is the maximum possible.

Exam answer

Let the side of each square cut off be $x$ cm. The box is $x$ cm high and its base is $(18 - 2x)$ cm by $(18 - 2x)$ cm, so $0 < x < 9$ ⁠.

\begin{aligned} V(x) &= x(18 - 2x)^2 \\ &= 4x^3 - 72x^2 + 324x \end{aligned}

\begin{aligned} V'(x) &= 12x^2 - 144x + 324 \\ &= 12(x - 3)(x - 9) \end{aligned}

$V'(x) = 0 \Rightarrow x = 3$ or $x = 9$ ⁠. As $x < 9$ ⁠, $x = 3$ ⁠.

\begin{aligned} V''(x) &= 24x - 144 \\ V''(3) &= -72 < 0 \end{aligned}

So $V$ is maximum at $x = 3$ ⁠. For $0 < x < 3$ ⁠, $V'(x) > 0$ ⁠, and for $3 < x < 9$ ⁠, $V'(x) < 0$ ⁠, so $V(3)$ is the greatest volume.

$\therefore$ The side of the square to be cut off is 3 cm.

Want the reason behind every step, the basics and the checks?

The one idea this problem is built on

The whole box is decided by one number, the side $x$ of the corner squares. Write the volume as a function of $x$ ⁠, find the range of $x$ that gives a real box, and maximise over that range. Small cuts give a wide but shallow box, large cuts a deep but narrow one; the best box is in between.

Given, to find, and the plan

Given: A square sheet 18 cm by 18 cm; equal squares are cut from the corners and the flaps folded up to make an open box.
To find: The side of the square to cut off so that the volume is largest.
Plan: Let the cut be $x$ cm. Read off the dimensions; find the allowed range; build $V(x)$ ⁠; solve $V'(x) = 0$ in the range; use the second derivative test, and the sign of $V'$ to show the maximum is global; answer with units.

Basics you need

1Building the function from the words

Choose the one number that fixes the situation, express every dimension in it, and state its physically sensible range. Basics: building the function to optimise

2Volume of a box

A cuboid of length $\ell$ ⁠, breadth $b$ and height $h$ has volume $\ell b h$ ⁠. Basics: area and volume formulas

3Second derivative test (Theorem 4, Section 6.4)

$f'(c) = 0$ and $f''(c) < 0$ give a point of local maxima. Lesson: derivative tests

4Local versus largest

Increasing up to $c$ ⁠, decreasing after it, on the whole allowed range: then $f(c)$ is the largest value. Lesson: optimisation problems

Solution

Choose the variable and read off the dimensions

Let $x$ cm be the side of each square cut off. After folding up the flaps:
- height of the box $= x$ ⁠,
- length $=$ breadth $= 18 - 2x$ (a square of side $x$ is removed at each end of every side).
The 18 cm square sheet with the corner squares (side $x$ ⁠) marked, drawn to scale with $x = 3$ cm. The dashed square becomes the base; the strips of width $x$ fold up to form the walls.

Each side of the sheet loses one square at each end, so it loses $2x$ ⁠. The flaps have width $x$ ⁠, and folding them up makes walls of height $x$ ⁠.

The base is a square because the sheet is a square. Basics: area and volume formulas
The allowed range of $x$

A real box needs a positive height and a positive base side:
$x > 0 \quad\text{and}\quad 18 - 2x > 0,$
so
$0 < x < 9.$
At $x = 0$ nothing is folded (no walls), and at $x = 9$ the four squares use up the whole sheet (no base). Either way the volume is 0, so these ends are excluded.

$18 - 2x > 0 \iff 2x < 18 \iff x < 9$ ⁠. Basics: interval notation
The volume as a function of $x$
$\begin{aligned} V(x) &= x(18 - 2x)^2 \\ &= x(324 - 72x + 4x^2) \\ &= 4x^3 - 72x^2 + 324x. \end{aligned}$
Volume = length × breadth × height = $(18 - 2x)(18 - 2x)\,x$ ⁠. Expanding gives a polynomial that is easy to differentiate.

$(18 - 2x)^2$ $= 324 - 2 \cdot 18 \cdot 2x + 4x^2$ $= 324 - 72x + 4x^2$ ⁠. Basics: volume formulas
Derivatives and the critical point
$\begin{aligned} V'(x) &= 12x^2 - 144x + 324 \\ &= 12(x^2 - 12x + 27) \\ &= 12(x - 3)(x - 9), \end{aligned}$ $V''(x) = 24x - 144.$
$V'(x) = 0$ gives $x = 3$ or $x = 9$ ⁠. But $x = 9$ is outside $0 < x < 9$ (no base), so the only critical point is $x = 3$ ⁠.

$x^2 - 12x + 27 = (x - 3)(x - 9)$ because $3 \times 9 = 27$ and $3 + 9 = 12$ ⁠. The algebra produces $x = 9$ as well, but the physical range rejects it.

Factorising a quadratic: find two numbers with the right product and sum. Basics: factorising
Second derivative test at $x = 3$
$V''(3) = 72 - 144 = -72 < 0.$
By the second derivative test (Theorem 4), $x = 3$ is a point of local maxima of $V$ ⁠.

$V'(3) = 0$ and $V''(3) < 0$ ⁠: flat and bending downward, a peak.

Theorem 4(i), Section 6.4. Lesson: derivative tests
It is the largest box of all

For $0 < x < 9$ ⁠, the factor $x - 9$ is negative, so $V'(x) = 12(x - 3)(x - 9)$ has the opposite sign to $x - 3$ ⁠:
- for $0 < x < 3$ ⁠: $V'(x) > 0$ ⁠, $V$ increasing;
- for $3 < x < 9$ ⁠: $V'(x) < 0$ ⁠, $V$ decreasing.
So $V(x) < V(3)$ for every other allowed $x$ ⁠.

$V(x) = x(18 - 2x)^2$ on $0 < x < 9$ ⁠: one peak, at $x = 3$ ⁠, where $V = 432$ cm³. At both excluded ends the volume is 0.

The second derivative test only compares $x = 3$ with nearby cuts; the sign of $V'$ on the whole range compares it with every possible cut (Theorem 1, Section 6.3).

A product of one negative and one positive factor is negative. Basics: sign of a product
Answer the question

The side of the square to be cut off is 3 cm. The box is then $12$ cm by $12$ cm by $3$ cm, with volume
$V(3) = 3 \times 12^2 = 432\ \text{cm}^3.$
The question asks for the side of the square, so the answer is 3 cm; the volume is a useful check.

$18 - 2 \times 3 = 12$ ⁠. Basics: volume formulas

Result

Cut off squares of side 3 cm. The box is 12 cm × 12 cm × 3 cm and its volume, 432 cm³, is the largest possible. (Textbook answer: 3 cm.)

Check it with numbers

$V(2) = 2 \times 14^2 = 392$ ⁠, $V(4) = 4 \times 10^2 = 400$ ⁠, $V(2.5) = 2.5 \times 13^2 = 422.5$ and $V(3.5) = 3.5 \times 11^2 = 423.5$ ⁠: all below $V(3) = 432$ ⁠. A computer search over $x = 0.001, 0.002, \dots, 8.999$ finds no volume larger than 432 cm³.

Common mistakes

Using $18 - x$ for the base side. Each side loses a square at both ends: $18 - 2x$ ⁠.
Keeping $x = 9$ ⁠. It solves $V'(x) = 0$ but leaves no base (and $V(9) = 0$ ⁠, a minimum, not a maximum).
Stopping at the local test. $V''(3) < 0$ shows a local maximum; the sign of $V'$ on $(0, 9)$ shows it is the largest box.
Answering 432. The question asks for the side of the square, 3 cm.

What to take away

For a square sheet of side $a$ ⁠, the best cut is $x = \frac{a}{6}$ ⁠: here $\frac{18}{6} = 3$ ⁠.
Write the range first; it decides which critical points are real.
Same set-up with a rectangle: Question 18 and the book's Example 36.

Every fact used, and where it comes from

Fact	Used in	Source
Volume of a cuboid	Step 3	Mensuration (Basics: volume formulas)
Critical point	Step 4	Note after Theorem 2, Section 6.4 (Lesson: maxima and minima)
Second derivative test	Step 5	Theorem 4, Section 6.4 (Lesson: derivative tests)
Sign of $f'$ decides increasing or decreasing	Step 6	Theorem 1 and the Remark after it, Section 6.3 (Lesson: increasing and decreasing)
Optimisation set-up from a word problem	Steps 1–3	Section 6.4; Example 36 (Lesson: optimisation problems)
Answer 3 cm	Result	Answers to Exercise 6.3, Question 17

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Exercise 6.3 · Q17: Open box from an 18 cm square of tin: cut squares of side 333 cm

Basics you need

Solution

Choose the variable and read off the dimensions

The allowed range of xxx

The volume as a function of xxx

Derivatives and the critical point

Second derivative test at x=3x = 3x=3

It is the largest box of all

Answer the question

Every fact used, and where it comes from

Exercise 6.3 · Q17: Open box from an 18 cm square of tin: cut squares of side $3$ cm

The allowed range of $x$

The volume as a function of $x$

Second derivative test at $x = 3$