19N.1.SL.TZ0.S_8

pestleMathematicsAASLPaper 119N· sl-5-5-integration-introduction-areas-between-curve-and-x-axissource ↗

A small cuboid box has a rectangular base of length $3x$ cm and width $x$ cm, where $x > 0$ . The height is $y$ cm, where $y > 0$ .

The sum of the length, width and height is $12$ cm.

The volume of the box is $V$ cm³.

Write down an expression for $y$ in terms of $x$ .

[1]

Find an expression for $V$ in terms of $x$ .

[2]

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

[2]

Find the value of $x$ for which $V$ is a maximum.

[4]

d.i.

Justify your answer.

[3]

d.ii.

Find the maximum volume.

[2]

Markscheme / solution

$y = 12 - 4x$ A1 N1

[1 mark]

correct substitution into volume formula (A1)

eg $3x \times x \times y{\text{, }}x \times 3x \times \left( {12 - x - 3x} \right){\text{, }}\left( {12 - 4x} \right)\left( x \right)\left( {3x} \right)$

$V = 3{x^2}\left( {12 - 4x} \right)\,\,\left( { = 36{x^2} - 12{x^3}} \right)$ A1 N2

Note: Award A0 for unfinished answers such as $3{x^2}\left( {12 - x - 3x} \right)$ .

[2 marks]

$\frac{{{\text{d}}V}}{{{\text{d}}x}} = 72x - 36{x^2}$ A1A1 N2

Note: Award A1 for $72x$ and A1 for $- 36{x^2}$ .

[2 marks]

valid approach to find maximum (M1)

eg $V' = 0{\text{, }}72x - 36{x^2} = 0$

correct working (A1)

eg $x\left( {72 - 36x} \right){\text{, }}\frac{{ - 72 \pm \sqrt {{{72}^2} - 4 \cdot \left( { - 36} \right) \cdot 0} }}{{2\left( { - 36} \right)}}{\text{, }}36x = 72{\text{, }}36x\left( {2 - x} \right) = 0$

$x = 2$ A2 N2

Note: Award A1 for $x = 2$ and $x = 0$ .

[4 marks]

d.i.

valid approach to explain that $V$ is maximum when $x = 2$ (M1)

eg attempt to find ${V''}$ , sign chart (must be labelled ${V'}$ )

correct value/s A1

eg $V''\left( 2 \right) = 72 - 72 \times 2$ , $V'\left( a \right)$ where $a < 2$ and $V'\left( b \right)$ where $b > 2$

correct reasoning R1

eg $V''\left( 2 \right) < 0$ , ${V'}$ is positive for $x < 2$ and negative for $x > 2$

Note: Do not award R1 unless A1 has been awarded.

$V$ is maximum when $x = 2$ AG N0

[3 marks]

d.ii.

correct substitution into their expression for volume A1

eg $3 \times {2^2}\left( {12 - 4 \times 2} \right)$ , $36\left( {{2^2}} \right) - 12\left( {{2^3}} \right)$

$V = 48$ (cm³) A1 N1

[2 marks]

Examiners’ report

[N/A]

d.i.

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d.ii.

[N/A]