Example 36: Open box from a 3 m by 8 m sheet: largest volume $\frac{200}{27}$ m³ at $x=\frac{2}{3}$ m

Basics you need

Solution

1. Choose the variable and read off the dimensions

   Let $x$ metres be the side of each square removed. After folding up the flaps:

   • height of the box $= x$⁠,
   • length $= 8 - 2x$ (a square of side $x$ is removed at each end of the 8 m side),
   • breadth $= 3 - 2x$ (likewise from the 3 m side).
   Once $x$ is chosen, the box is completely determined, so the volume is a function of $x$ alone. Each long side loses two squares, one at each end, which is where the "$2x$" comes from.
   Folding a flap of width $x$ upwards makes a wall of height $x$⁠. Basics: building the function

2. The allowed range of $x$

   A real box needs all three dimensions positive:

   $$x > 0, \qquad 3 - 2x > 0, \qquad 8 - 2x > 0.$$

   The second gives $x < \frac{3}{2}$⁠, and then the third ($x < 4$⁠) holds automatically. So

   $$0 < x < \frac{3}{2}.$$
   The narrow side of the sheet is the real limit: we cannot cut two squares from a 3 m width if together they use up all of it. At $x = 0$ or $x = \frac{3}{2}$ the "box" is flat and holds nothing, so these ends are excluded.
   Solving $3 - 2x > 0$⁠: add $2x$ to both sides and divide by 2. Basics: interval notation

3. The volume as a function of $x$

   $$\begin{aligned} V(x) &= x(3 - 2x)(8 - 2x) \\ &= x(24 - 22x + 4x^2) \\ &= 4x^3 - 22x^2 + 24x. \end{aligned}$$
   Volume = length × breadth × height. Expanding gives a polynomial, which is easy to differentiate term by term.
   $(3 - 2x)(8 - 2x)$ $= 24 - 6x - 16x + 4x^2$ $= 4x^2 - 22x + 24$⁠. Basics: area and volume formulas

4. First and second derivatives

   $$\begin{aligned} V'(x) &= 12x^2 - 44x + 24 \\ &= 4(3x^2 - 11x + 6) \\ &= 4(x - 3)(3x - 2), \end{aligned}$$ $$V''(x) = 24x - 44.$$
   Factorising $V'$ shows at once where it is zero and what sign it has. Check: $(x - 3)(3x - 2)$ $= 3x^2 - 2x - 9x + 6$ $= 3x^2 - 11x + 6$⁠.
   Split $-11x$ as $-9x - 2x$ because $(-9)(-2) = 18 = 3 \times 6$⁠: $3x^2 - 9x - 2x + 6 = 3x(x - 3) - 2(x - 3)$⁠. Basics: factorising

5. Critical points, and why $x \ne 3$

   $$V'(x) = 0 \iff x = 3 \ \text{ or } \ x = \frac{2}{3}.$$

   But $x = 3$ is not in the allowed range $0 < x < \frac{3}{2}$⁠: squares of side 3 m cannot be cut from a sheet only 3 m wide, and the formula would give a breadth $3 - 2(3) = -3$ m. So the only critical point that describes a box is

   $$x = \frac{2}{3}.$$
   This answers the textbook's "(Why?)". The function $4x^3 - 22x^2 + 24x$ is a polynomial defined for every $x$⁠, so algebra alone happily produces $x = 3$⁠; only the physical range rules it out.
   A critical point is where $f' = 0$ or $f$ is not differentiable (Note after Theorem 2, Section 6.4). Lesson: maxima and minima

6. Second derivative test at $x = \frac{2}{3}$

   $$V''\left(\frac{2}{3}\right) = 24\cdot\frac{2}{3} - 44 = 16 - 44 = -28 < 0.$$

   By the second derivative test (Theorem 4), $x = \frac{2}{3}$ is a point of local maxima of $V$⁠.

   $V'\left(\frac{2}{3}\right) = 0$ and $V''\left(\frac{2}{3}\right) < 0$ is exactly case (i) of Theorem 4: the graph is flat there and bending downwards, so it is a peak.
   Theorem 4 (Second Derivative Test), Section 6.4. Lesson: derivative tests

7. It is the largest value on the whole range

   For $0 < x < \frac{3}{2}$ the factor $x - 3$ is always negative. So the sign of $V'(x) = 4(x - 3)(3x - 2)$ is opposite to the sign of $3x - 2$⁠:

   • for $0 < x < \frac{2}{3}$⁠: $3x - 2 < 0$⁠, so $V'(x) > 0$ and $V$ is increasing;
   • for $\frac{2}{3} < x < \frac{3}{2}$⁠: $3x - 2 > 0$⁠, so $V'(x) < 0$ and $V$ is decreasing.

   Hence $V(x) < V\left(\frac{2}{3}\right)$ for every other $x$ in the range: $x = \frac{2}{3}$ gives the largest box, not just a locally largest one.

   The second derivative test only speaks about points near $\frac{2}{3}$⁠. The question asks for the largest box of all, so we also need this global argument: $V$ climbs all the way up to $\frac{2}{3}$ and descends all the way after it (Theorem 1, Section 6.3).
   Increasing then decreasing on an interval means the turning point is the overall maximum on that interval. Basics: sign of a product

8. The largest volume

   $$\begin{aligned} V\left(\frac{2}{3}\right) &= \frac{2}{3}\left(3 - \frac{4}{3}\right)\left(8 - \frac{4}{3}\right) \\ &= \frac{2}{3}\cdot\frac{5}{3}\cdot\frac{20}{3} = \frac{200}{27}\ \text{m}^3. \end{aligned}$$

   The box is $\frac{20}{3}$ m long, $\frac{5}{3}$ m wide and $\frac{2}{3}$ m high, and the largest volume is $\frac{200}{27}$ m³ $\approx 7.407$ m³.

   Using the factored form $x(3 - 2x)(8 - 2x)$ avoids heavy fractions. The textbook uses the expanded form: $4\cdot\frac{8}{27} - 22\cdot\frac{4}{9} + 24\cdot\frac{2}{3}$ $= \frac{32 - 264 + 432}{27} = \frac{200}{27}$⁠, the same value.
   $3 - \frac{4}{3} = \frac{5}{3}$ and $8 - \frac{4}{3} = \frac{20}{3}$⁠.

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