Factorise \(3{x^2} + 13x - 10\).

Solve the equation \(3{x^2} + 13x - 10 = 0\).

Consider a function \(f(x) = 3{x^2} + 13x - 10\) .

Find the equation of the axis of symmetry on the graph of this function.

Consider a function \(f(x) = 3{x^2} + 13x - 10\) .

Calculate the minimum value of this function.

Show that \(4{x^2} + 6xy = 300\).

Find an expression for \(y\) in terms of \(x\).

Hence show that the volume \(V\) of the box is given by \(V = 100x - \frac{4}{3}{x^3}\).

Find \(\frac{{{\text{d}}V}}{{{\text{d}}x}}\).

(i)     Hence find the value of \(x\) and of \(y\) required to make the volume of the box a maximum.

(ii)    Calculate the maximum volume.

\((3x - 2)(x + 5)\)     (A1)(A1)

[2 marks]

\((3x - 2)(x + 5) = 0\)

\(x = \frac{2}{3}\) or \(x = - 5\)     (A1)(ft)(A1)(ft)(G2)

[2 marks]

\(x = \frac{{ - 13}}{6}{\text{ }}( - 2.17)\)     (A1)(A1)(ft)(G2)

Note: (A1) is for \(x = \), (A1) for value. (ft) if value is half way between roots in (b).

[2 marks]

Minimum \(y = 3{\left( {\frac{{ - 13}}{6}} \right)^2} + 13\left( {\frac{{ - 13}}{6}} \right) - 10\)     (M1)

Note: (M1) for substituting their value of \(x\) from (c) into \(f(x)\) .

\( = - 24.1\)     (A1)(ft)(G2)

[2 marks]

\({\text{Area}} = 2(2x)x + 2xy + 2(2x)y\)     (M1)(A1)

Note: (M1) for using the correct surface area formula (which can be implied if numbers in the correct place). (A1) for using correct numbers.

\(300 = 4{x^2} + 6xy\)     (AG)

Note: Final line must be seen or previous (A1) mark is lost.

[2 marks]

\(6xy = 300 - 4{x^2}\)     (M1)

\(y = \frac{{300 - 4{x^2}}}{{6x}}\) or \(\frac{{150 - 2{x^2}}}{{3x}}\)     (A1)

[2 marks]

\({\text{Volume}} = x(2x)y\)     (M1)

\(V = 2{x^2}\left( {\frac{{300 - 4{x^2}}}{{6x}}} \right)\)     (A1)(ft)

\( = 100x - \frac{4}{3}{x^3}\)     (AG)

Note: Final line must be seen or previous (A1) mark is lost.

[2 marks]

\(\frac{{{\text{d}}V}}{{{\text{d}}x}} = 100 - \frac{{12{x^2}}}{3}\)  or  \(100 - 4{x^2}\)     (A1)(A1)

 Note: (A1) for each term.

[2 marks]

Unit penalty (UP) is applicable where indicated in the left hand column

(i)     For maximum \(\frac{{{\text{d}}V}}{{{\text{d}}x}} = 0\)  or  \(100 - 4{x^2} = 0\)     (M1)

\(x = 5\)     (A1)(ft)

\(y = \frac{{300 - 4{{(5)}^2}}}{{6(5)}}\)  or  \(\left( {\frac{{150 - 2{{(5)}^2}}}{{3(5)}}} \right)\)     (M1)

\( = \frac{{20}}{3}\)     (A1)(ft)

(UP)     (ii)    \(333\frac{1}{3}{\text{ c}}{{\text{m}}^3}{\text{ }}(333{\text{ c}}{{\text{m}}^3})\)

Note: (ft) from their (e)(i) if working for volume is seen.

[5 marks]

Most candidates made a good attempt to factorise the expression.

Many gained both marks here from a correct answer or ft from the previous part.

Many used the formula correctly. Some forgot to put \(x = \) .

Most candidates found this value from their GDC.

A good attempt was made to show the correct surface area.

Many could rearrange the equation correctly.

Although this was not a difficult question it probably looked complicated for the candidates and it was often left out.

Those who reached this length could usually manage the differentiation.

(i)     Many found the correct value of \(x\) but not of \(y\).

(ii)    This was well done and again the units were included in most scripts.