A packaging company makes boxes for chocolates. An example of a box is shown below. This box is closed and the top and bottom of the box are identical regular hexagons of side x cm.

(a)     Show that the area of each hexagon is $\frac{{3\sqrt 3 {x^2}}}{2}{\text{c}}{{\text{m}}^2}$ .

(b)     Given that the volume of the box is ${\text{90 c}}{{\text{m}}^2}$ , show that when $x = \sqrt[3]{{20}}$ the total surface area of the box is a minimum, justifying that this value gives a minimum.

(a)     Area of hexagon $= 6 \times \frac{1}{2} \times x \times x \times \sin 60^\circ$     M1

$= \frac{{3\sqrt 3 {x^2}}}{2}$     AG

(b)     Let the height of the box be h

Volume $= \frac{{3\sqrt 3 h{x^2}}}{2} = 90$     M1

Hence $h = \frac{{60}}{{\sqrt 3 {x^2}}}$     A1

Surface area, $A = 3\sqrt 3 {x^2} + 6hx$     M1

$= 3\sqrt 3 {x^2} + \frac{{360}}{{\sqrt 3 }}{x^{ - 1}}$     A1

$\frac{{{\text{d}}A}}{{{\text{d}}x}} = 6\sqrt 3 x - \frac{{360}}{{\sqrt 3 }}{x^{ - 2}}$     A1

$\left( {\frac{{{\text{d}}A}}{{{\text{d}}x}} = 0} \right)$

$6\sqrt 3 {x^3} = \frac{{360}}{{\sqrt 3 }}$     M1

${x^3} = 20$

$x = \sqrt[3]{{20}}$     AG

$\frac{{{{\text{d}}^2}A}}{{{\text{d}}{x^2}}} = 6\sqrt 3  + \frac{{720{x^{ - 3}}}}{{\sqrt 3 }}$

which is positive when $x = \sqrt[3]{{20}}$, and hence gives a minimum value.     R1

[8 marks]

There were a number of wholly correct answers seen and the best candidates tackled the question well. However, many candidates did not seem to understand what was expected in such a problem. It was disappointing that a significant number of candidates were unable to find the area of the hexagon.