Find the value of $a$.

Find the value of $x$ that makes the volume a maximum.

$72 = 12x + 4h\;\;\;$(or equivalent)     (M1)

Note: Award (M1) for a correct equation obtained from the total length of the edges.

$V = 2{x^2}(18 - 3x)$     (A1)

$(a = ){\text{ }}36$     (A1)     (C3)

$\frac{{{\text{d}}V}}{{{\text{d}}x}} = 72x - 18{x^2}$     (A1)

$72x - 18{x^2} = 0\;\;\;$OR$\;\;\;\frac{{{\text{d}}V}}{{{\text{d}}x}} = 0$     (M1)

Notes: Award (A1) for  $- 18{x^2}$  seen. Award (M1) for equating derivative to zero.

$(x = ){\text{ 4}}$     (A1)(ft)     (C3)

Note: Follow through from part (a).

OR

Sketch of $V$ with visible maximum     (M1)

Sketch with $x \geqslant 0,{\text{ }}V \geqslant 0$ and indication of maximum (e.g. coordinates)     (A1)(ft)

$(x = ){\text{ 4}}$     (A1)(ft)     (C3)

Notes: Follow through from part (a).

Award (M1)(A1)(A0) for $(4,{\text{ }}192)$.

Award (C3) for $x = 4,{\text{ }}y = 192$.

The model in this question seemed to be too difficult for the vast majority of the candidates, and therefore was a strong discriminator between grade 6 and grade 7 candidates. An attempt to find an equation for the volume of the cube often started with V = x x 2x x h . Many struggled to translate the total length of the edges into a correct equation, and consequently were unable to substitute h. Some tried to write x in terms of h and got lost, others tried to work backwards from the expression given in the question.

As very few found a value for a, often part (b) was not attempted. When a derivative was calculated this was usually done correctly.