Chapter 6 · Application of Derivatives

Exercise 6.3 · Q18: Open box from a 45 cm by 24 cm sheet: cut squares of side $5$ cm

Exercise 6.3Optimisation · Section 6.4

The question

A rectangular sheet of tin 45 cm by 24 cm is to be made into a box without top, by cutting off square from each corner and folding up the flaps. What should be the side of the square to be cut off so that the volume of the box is maximum ?

Exam answer

Let the side of each square cut off be $x$ cm. The box is $x$ cm high, $(45 - 2x)$ cm long and $(24 - 2x)$ cm wide, so $0 < x < 12$ ⁠.

\begin{aligned} V(x) &= x(45 - 2x)(24 - 2x) \\ &= 4x^3 - 138x^2 + 1080x \end{aligned}

\begin{aligned} V'(x) &= 12x^2 - 276x + 1080 \\ &= 12(x - 5)(x - 18) \end{aligned}

$V'(x) = 0 \Rightarrow x = 5$ or $x = 18$ ⁠. As $x < 12$ ⁠, $x = 5$ ⁠.

\begin{aligned} V''(x) &= 24x - 276 \\ V''(5) &= -156 < 0 \end{aligned}

So $V$ is maximum at $x = 5$ ⁠. For $0 < x < 5$ ⁠, $V'(x) > 0$ ⁠, and for $5 < x < 12$ ⁠, $V'(x) < 0$ ⁠, so $V(5)$ is the greatest volume.

$\therefore$ The side of the square to be cut off is 5 cm.

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

The one idea this problem is built on

The side $x$ of the corner squares decides the whole box. The shorter side of the sheet (24 cm) limits how large $x$ can be, and that limit rejects one of the two solutions of $V'(x) = 0$ ⁠. The remaining critical point is then shown to give the largest volume on the whole range.

Given, to find, and the plan

Given: A rectangular sheet 45 cm by 24 cm; equal squares are cut from the four 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 length, width and height; find the allowed range; build $V(x)$ ⁠; solve $V'(x) = 0$ and keep only the solution in the range; second derivative test; sign of $V'$ for the global maximum; answer with units.

Basics you need

1Building the function from the words

One variable, every dimension written in it, and its physical range. Basics: building the function to optimise

2Volume of a box

Length × width × height. Basics: area and volume formulas

3Factorising a quadratic

$x^2 - 23x + 90$ ⁠: find two numbers with product 90 and sum 23 (they are 5 and 18). Basics: factorising

4Second derivative test and the global argument

$V'(c) = 0$ ⁠, $V''(c) < 0$ gives a local maximum (Theorem 4, Section 6.4); the sign of $V'$ on the whole range makes it the largest. 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 $= x$ ⁠,
- length $= 45 - 2x$ ⁠,
- width $= 24 - 2x$ ⁠.
The 45 cm by 24 cm sheet with the corner squares (side $x$ ⁠) marked, drawn to scale with $x = 5$ cm. The dashed rectangle becomes the base; the strips of width $x$ fold up into the walls.

Each side of the sheet loses a square of side $x$ at both ends, so each base dimension is the sheet dimension minus $2x$ ⁠. The flaps of width $x$ become walls of height $x$ ⁠.

Folding a flap of width $x$ upward makes a wall of height $x$ ⁠. Basics: building the function
The allowed range of $x$
$x > 0,\qquad 24 - 2x > 0,\qquad 45 - 2x > 0.$
The second gives $x < 12$ ⁠; then $x < 22.5$ (the third) holds automatically. So
$0 < x < 12.$
The narrower side (24 cm) runs out first: two squares of side 12 cm would use it up completely. At $x = 0$ or $x = 12$ the box is flat and holds nothing, so both ends are excluded.

When several conditions must all hold, the strictest one decides. Basics: interval notation
The volume as a function of $x$
$\begin{aligned} V(x) &= x(45 - 2x)(24 - 2x) \\ &= x(1080 - 138x + 4x^2) \\ &= 4x^3 - 138x^2 + 1080x. \end{aligned}$
Volume = length × width × height. Expanding gives a cubic that is easy to differentiate.

$(45 - 2x)(24 - 2x)$ $= 1080 - 90x - 48x + 4x^2$ ⁠, and $90 + 48 = 138$ ⁠. Basics: volume formulas
Derivatives and critical points
$\begin{aligned} V'(x) &= 12x^2 - 276x + 1080 \\ &= 12(x^2 - 23x + 90) \\ &= 12(x - 5)(x - 18), \end{aligned}$ $V''(x) = 24x - 276.$
$V'(x) = 0$ gives $x = 5$ or $x = 18$ ⁠. Since $18 > 12$ ⁠, $x = 18$ is outside the range: it would need a width $24 - 36 = -12$ cm. The only critical point is $x = 5$ ⁠.

$12 \times 23 = 276$ and $12 \times 90 = 1080$ ⁠, so the factor 12 comes out cleanly. Then $5 \times 18 = 90$ and $5 + 18 = 23$ ⁠.

Critical point: Note after Theorem 2, Section 6.4. Lesson: maxima and minima
Second derivative test at $x = 5$
$V''(5) = 120 - 276 = -156 < 0.$
By the second derivative test (Theorem 4), $x = 5$ is a point of local maxima of $V$ ⁠.

$V'(5) = 0$ and $V''(5) < 0$ is case (i) of Theorem 4: a peak.

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

For $0 < x < 12$ ⁠, $x - 18 < 0$ ⁠, so the sign of $V'(x) = 12(x - 5)(x - 18)$ is opposite to the sign of $x - 5$ ⁠:
- for $0 < x < 5$ ⁠: $V'(x) > 0$ ⁠, $V$ increasing;
- for $5 < x < 12$ ⁠: $V'(x) < 0$ ⁠, $V$ decreasing.
Hence $V(x) < V(5)$ for every other allowed $x$ ⁠.

$V(x) = x(45 - 2x)(24 - 2x)$ on $0 < x < 12$ ⁠: a single peak at $x = 5$ ⁠, where $V = 2450$ cm³; the volume is 0 at both excluded ends.

The local test compares $x = 5$ only with nearby cuts. Rising all the way to 5 and falling all the way after it is what makes it the best cut overall (Theorem 1, Section 6.3).

Sign of a product: one negative factor makes the product negative. Basics: sign of a product
Answer the question

The side of the square to be cut off is 5 cm. The box is then 35 cm long, 14 cm wide and 5 cm high, with volume
$V(5) = 5 \times 35 \times 14 = 2450\ \text{cm}^3.$
The question asks for the side of the square; the volume is a check.

$45 - 10 = 35$ ⁠, $24 - 10 = 14$ ⁠, $35 \times 14 = 490$ ⁠, $490 \times 5 = 2450$ ⁠. Basics: volume formulas

Result

Cut off squares of side 5 cm. The box is 35 cm × 14 cm × 5 cm and its volume, 2450 cm³, is the largest possible. (Textbook answer: $x = 5$ cm.)

Check it with numbers

$V(4) = 4 \times 37 \times 16 = 2368$ ⁠, $V(6) = 6 \times 33 \times 12 = 2376$ ⁠, $V(4.5) = 4.5 \times 36 \times 15 = 2430$ and $V(5.5) = 5.5 \times 34 \times 13 = 2431$ ⁠: all below $V(5) = 2450$ ⁠. A computer search over $x = 0.001, 0.002, \dots, 11.999$ finds no volume above 2450 cm³.

Common mistakes

Keeping $x = 18$ ⁠. It solves $V' = 0$ but is impossible: the 24 cm side cannot lose 36 cm.
Using the longer side for the range. $45 - 2x > 0$ alone would allow $x$ up to 22.5; the 24 cm side is the real limit.
Expansion slips. $(45 - 2x)(24 - 2x)$ $= 4x^2 - 138x + 1080$ ⁠; check the middle term: $-90x - 48x$ ⁠.
Answering 2450. The question asks for the side of the square, 5 cm.

What to take away

The allowed range comes from the shortest dimension; it is what rejects the second critical point.
One variable, its range, the function, critical points in the range, a test, a global argument, then the answer in words with units.
The square-sheet version is Question 17; the book's worked version is Example 36 (3 m by 8 m sheet).

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 $x = 5$ cm	Result	Answers to Exercise 6.3, Question 18

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Exercise 6.3 · Q18: Open box from a 45 cm by 24 cm sheet: cut squares of side 555 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 critical points

Second derivative test at x=5x = 5x=5

It is the largest box of all

Answer the question

Every fact used, and where it comes from

Exercise 6.3 · Q18: Open box from a 45 cm by 24 cm sheet: cut squares of side $5$ cm

The allowed range of $x$

The volume as a function of $x$

Second derivative test at $x = 5$