Show that y =12 − 3x .

Show that the volume V m³ of the container is given by

V = 24x² − 6x³

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

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

Find the maximum volume of the container.

Find the length and height of the container for which the volume is a maximum.

The shipping container is to be painted. One litre of paint covers an area of 15 m² . Paint comes in tins containing four litres.

Calculate the number of tins required to paint the shipping container.

\(4(2x) + 4y + 4x = 48\)     (M1)

Note: Award (M1) for setting up the equation.

\(12x + 4y = 48\)     (M1)

Note: Award (M1) for simplifying (can be implied).

\(y = \frac{{48 - 12x}}{{4}}\)   OR   \(3x + y =12\)     (A1)

\(y =12 - 3x\)     (AG)

Note: The last line must be seen for the (A1) to be awarded.

[3 marks]

\(V = 2x \times x \times (12 - 3x)\)     (M1)(A1)

Note: Award (M1) for substitution into volume equation, (A1) for correct substitution.

\(= 24x^2 - 6x^3\)     (AG)

Note: The last line must be seen for the (A1) to be awarded.

[2 marks]

\(\frac{{\text{d}V}}{{\text{d}x}} = 48x - 18x^2\)     (A1)(A1)

Note: Award (A1) for each correct term.

[2 marks]

\(48x -18x^2 = 0\)     (M1)(M1)

Note: Award (M1) for using their derivative, (M1) for equating their answer to part (c) to 0.

OR

(M1) for sketch of \(V = 24x^2 - 6x^3\), (M1) for the maximum point indicated     (M1)(M1)

OR

(M1) for sketch of \(\frac{{\text{d}V}}{{\text{d}x}} = 48x - 18x^2\), (M1) for the positive root indicated     (M1)(M1)

\(2.67\left( {\frac{{24}}{9},{\text{ }}\frac{8}{3},{\text{ }}2.66666...} \right)\)     (A1)(ft)(G2)

Note: Follow through from their part (c).

[3 marks]

\(V = 24 \times {\left( {\frac{8}{3}} \right)^2} - 6 \times {\left( {\frac{8}{3}} \right)^3}\)     (M1)

Note: Award (M1) for substitution of their value from part (d) into volume equation.

\(56.9({{\text{m}}^3})\left( {\frac{{512}}{9},{\text{ }}56.8888...} \right)\)     (A1)(ft)(G2)

Note: Follow through from their answer to part (d).

[2 marks]

\(\text{length} = \frac{{16}}{{3}}\)     (A1)(ft)(G1)

Note: Follow through from their answer to part (d). Accept 5.34 from use of 2.67

\(\text{height} = 12 - 3 \times \left( {\frac{{8}}{{3}}} \right) = 4\)     (M1)(A1)(ft)(G2)

Notes: Award (M1) for substitution of their answer to part (d), (A1)(ft) for answer. Accept 3.99 from use of 2.67.

[3 marks]

\(\text{SA} = 2 \times \frac{{16}}{{3}} \times 4 + 2 \times \frac{{8}}{{3}} \times 4 + 2 \times \frac{{16}}{{3}} \times \frac{{8}}{{3}}\)     (M1)

OR

\(\text{SA} = 4 \left( {\frac{{8}}{{3}}}\right)^2 + 6 \times \frac{{8}}{{3}} \times 4\)     (M1)

Note: Award (M1) for substitution of their values from parts (d) and (f) into formula for surface area.

92.4 (m²) (92.4444...(m²))     (A1)

Note: Accept 92.5 (92.4622...) from use of 3 sf answers.

\(\text{Number of tins} = \frac{{92.4444...}}{{15 \times 4}}( = 1.54)\)     (M1)

[4 marks]

Note: Award (M1) for division of their surface area by 60.

2 tins required     (A1)(ft)

Note: Follow through from their answers to parts (d) and (f).

Many candidates did not answer this question at all and others did not get past part (c). It was unclear if this was because they could not do the question or they ran out of time.

(a) This was very poorly done. Most candidates had no idea what they were supposed to do here. Many tried to find values for x.

Many candidates did not answer this question at all and others did not get past part (c). It was unclear if this was because they could not do the question or they ran out of time.

(a) This was very poorly done. Most candidates had no idea what they were supposed to do here. Many tried to find values for x.

(b) Similar comment as for part (a) although more candidates made an attempt at finding the Volume.

Many candidates did not answer this question at all and others did not get past part (c). It was unclear if this was because they could not do the question or they ran out of time.

(c) This part was very well done.

Many candidates did not answer this question at all and others did not get past part (c). It was unclear if this was because they could not do the question or they ran out of time.

(d) Not many correct answers seen. Many candidates graphed the wrong equation and found 1.333 as their answer.

Many candidates did not answer this question at all and others did not get past part (c). It was unclear if this was because they could not do the question or they ran out of time.

(e) Some managed to gain follow through marks for this part.

Many candidates did not answer this question at all and others did not get past part (c). It was unclear if this was because they could not do the question or they ran out of time.

(f) Again here follow through marks were gained by those who attempted it.

Many candidates did not answer this question at all and others did not get past part (c). It was unclear if this was because they could not do the question or they ran out of time.

(g) Very few correct answers for the surface area were seen. Most candidates thought that there were 4 equal faces 2 xy and 2 faces xy. Some managed to get follow through marks for the last part if they divided by 60.