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Date	November 2018	Marks available	3	Reference code	18N.2.SL.TZ0.T_6
Level	Standard Level	Paper	Paper 2	Time zone	Time zone 0
Command term	Show that	Question number	T_6	Adapted from	N/A

Question

Haruka has an eco-friendly bag in the shape of a cuboid with width 12 cm, length 36 cm and height of 9 cm. The bag is made from five rectangular pieces of cloth and is open at the top.

Nanako decides to make her own eco-friendly bag in the shape of a cuboid such that the surface area is minimized.

The width of Nanako’s bag is x cm, its length is three times its width and its height is y cm.

The volume of Nanako’s bag is 3888 cm³.

Calculate the area of cloth, in cm², needed to make Haruka’s bag.

[2]

Calculate the volume, in cm³, of the bag.

[2]

Use this value to write down, and simplify, the equation in x and y for the volume of Nanako’s bag.

[2]

Write down and simplify an expression in x and y for the area of cloth, A, used to make Nanako’s bag.

[2]

Use your answers to parts (c) and (d) to show that

$A = 3{x^2} + \frac{{10368}}{x}$ .

[2]

Find $\frac{{{\text{d}}A}}{{{\text{d}}x}}$ .

[3]

Use your answer to part (f) to show that the width of Nanako’s bag is 12 cm.

[3]

Markscheme

* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.

36 × 12 + 2(9 ×12) + 2(9 × 36) (M1)

Note: Award (M1) for correct substitution into surface area of cuboid formula.

= 1300 (cm²) (1296 (cm²)) (A1)(G2)

[2 marks]

36 × 9 ×12 (M1)

Note: Award (M1) for correct substitution into volume of cuboid formula.

= 3890 (cm³) (3888 (cm³)) (A1)(G2)

[2 marks]

3 $x$ × $x$ × $y$ = 3888 (M1)

Note: Award (M1) for correct substitution into volume of cuboid formula and equated to 3888.

$x$ ² $y$ = 1296 (A1)(G2)

Note: Award (A1) for correct fully simplified volume of cuboid.

Accept $y = \frac{{1296}}{{{x^2}}}$ .

[2 marks]

(A =) 3x² + 2(xy) + 2(3xy) (M1)

Note: Award (M1) for correct substitution into surface area of cuboid formula.

(A =) 3x² + 8xy (A1)(G2)

Note: Award (A1) for correct simplified surface area of cuboid formula.

[2 marks]

$A = 3{x^2} + 8x\left( {\frac{{1296}}{{{x^2}}}} \right)$ (A1)(ft)(M1)

Note: Award (A1)(ft) for correct rearrangement of their part (c) seen (rearrangement may be seen in part(c)), award (M1) for substitution of their part (c) into their part (d) but only if this leads to the given answer, which must be shown.

$A = 3{x^2} + \frac{{10368}}{x}$ (AG)

[2 marks]

$\left( {\frac{{{\text{d}}A}}{{{\text{d}}x}}} \right) = 6x - \frac{{10368}}{{{x^2}}}$ (A1)(A1)(A1)

Note: Award (A1) for $6x$ , (A1) for −10368, (A1) for ${x^{ - 2}}$ . Award a maximum of (A1)(A1)(A0) if any extra terms seen.

[3 marks]

$6x - \frac{{10368}}{{{x^2}}} = 0$ (M1)

Note: Award (M1) for equating their ${\frac{{{\text{d}}A}}{{{\text{d}}x}}}$ to zero.

$6{x^3} = 10368$ OR $6{x^3} - 10368 = 0$ OR ${x^3} - 1728 = 0$ (M1)

Note: Award (M1) for correctly rearranging their equation so that fractions are removed.

$x = \sqrt[3]{{1728}}$ (A1)

$x = 12$ (cm) (AG)

Note: The (AG) line must be seen for the final (A1) to be awarded. Substituting $x = 12$ invalidates the method, award a maximum of (M1)(M0)(A0).

[3 marks]

Examiners report

[N/A]

Syllabus sections

Topic 5 —Calculus » SL 5.1—Introduction of differential calculus

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18N.2.SL.TZ0.T_6a:
Calculate the area of cloth, in cm², needed to make Haruka’s bag.
19M.2.SL.TZ2.T_5b:
Write down the $y$ -intercept of the graph of $y = f\left( x \right)$ .
17M.2.SL.TZ2.T_6f:
Write down the number of possible solutions to the equation $f(x) = 5$ .
19M.2.SL.TZ2.T_5c:
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19M.2.AHL.TZ1.H_1:
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Find the coordinates of the point where $l$ meets the $x$ -axis.
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Find $\frac{{{\text{d}}y}}{{{\text{d}}x}}$ in terms of $x$ and $y$ .
18N.2.AHL.TZ0.H_5:
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22M.2.AHL.TZ2.12a:
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17M.2.SL.TZ2.T_6d.i:
Write down the $x$ -coordinates of these two points;
18M.1.SL.TZ1.T_5c:
Find the equation of the line DC. Write your answer in the form ax + by + d = 0 where a , b and d are integers.
16N.2.SL.TZ0.T_6b:
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16N.2.SL.TZ0.T_6c:
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16N.2.SL.TZ0.T_6e:
Find $\frac{{{\text{d}}A}}{{{\text{d}}r}}$ .
16N.2.SL.TZ0.T_6f:
Using your answer to part (e), find the value of $r$ which minimizes $A$ .
17M.2.AHL.TZ1.H_2b:
Determine the equation of the tangent to $C$ at the point $\left( {\frac{2}{{\text{e}}},{\text{ e}}} \right)$
EXN.1.SL.TZ0.9e:
Find the rate of change of the ball’s height above the ground when $t = \frac{1}{3}$ . Give your answer in the form $p π \sqrt{q} {ms}^{- 1}$ where $p \in ℚ$ and $q \in ℤ^{+}$ .
18M.1.SL.TZ1.T_5a:
Write down the coordinates of C, the midpoint of line segment AB.
18N.2.SL.TZ0.T_6d:
Write down and simplify an expression in x and y for the area of cloth, A, used to make Nanako’s bag.
18M.1.SL.TZ1.T_5b:
Find the gradient of the line DC.
17M.2.SL.TZ2.T_6a:
Write down the $y$ -intercept of the graph.
17M.2.SL.TZ2.T_6c.i:
Show that $a = 8$ .
17M.2.SL.TZ2.T_6b:
Find $f'(x)$ .
19M.2.SL.TZ2.T_5h:
Determine the range of $f\left( x \right)$ for $p$ ≤ $x$ ≤ $q$ .
17M.2.SL.TZ2.T_6g:
The equation $f(x) = m$ , where $m \in \mathbb{R}$ , has four solutions. Find the possible values of $m$ .
16N.2.SL.TZ0.T_6d:
Show that $A = \pi {r^2} + \frac{{1\,000\,000}}{r}$ .
16N.2.SL.TZ0.T_6g:
Find the value of this minimum area.
18N.2.SL.TZ0.T_6e:
Use your answers to parts (c) and (d) to show that

$A = 3{x^2} + \frac{{10368}}{x}$ .
16N.2.SL.TZ0.T_6a:
Write down a formula for $A$ , the surface area to be coated.
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Find the least number of cans of water-resistant material that will coat the area in part (g).
17M.2.SL.TZ2.T_6c.ii:
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17M.2.SL.TZ2.T_6d.ii:
Write down the intervals where the gradient of the graph of $y = f(x)$ is positive.
17M.2.SL.TZ2.T_6e:
Write down the range of $f(x)$ .
18N.2.SL.TZ0.T_6b:
Calculate the volume, in cm³, of the bag.
18N.2.SL.TZ0.T_6c:
Use this value to write down, and simplify, the equation in x and y for the volume of Nanako’s bag.
18N.2.SL.TZ0.T_6f:
Find $\frac{{{\text{d}}A}}{{{\text{d}}x}}$ .

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Examiners report

Syllabus sections

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