Show that \(A(x) = \frac{{108}}{x} + 2{x^2}\).

Find \(A'(x)\).

Given that the outside surface area is a minimum, find the height of the container.

Fred paints the outside of the container. A tin of paint covers a surface area of \({\text{10 }}{{\text{m}}^{\text{2}}}\) and costs $20. Find the total cost of the tins needed to paint the container.

correct substitution into the formula for volume     A1

eg\(\,\,\,\,\,\)\(36 = y \times x \times x\)

valid approach to eliminate \(y\) (may be seen in formula/substitution)     M1

eg\(\,\,\,\,\,\)\(y = \frac{{36}}{{{x^2}}},{\text{ }}xy = \frac{{36}}{x}\)

correct expression for surface area     A1

eg\(\,\,\,\,\,\)\(xy + xy + xy + {x^2} + {x^2},{\text{ area}} = 3xy + 2{x^2}\)

correct expression in terms of \(x\) only     A1

eg\(\,\,\,\,\,\)\(3x\left( {\frac{{36}}{{{x^2}}}} \right) + 2{x^2},{\text{ }}{x^2} + {x^2} + \frac{{36}}{x} + \frac{{36}}{x} + \frac{{36}}{x},{\text{ }}2{x^2} + 3\left( {\frac{{36}}{x}} \right)\)

\(A(x) = \frac{{108}}{x} + 2{x^2}\)    AG     N0

[4 marks]

\(A'(x) =  - \frac{{108}}{{{x^2}}} + 4x,{\text{ }}4x - 108{x^{ - 2}}\)    A1A1     N2

Note:     Award A1 for each term.

[2 marks]

recognizing that minimum is when \(A'(x) = 0\)     (M1)

correct equation     (A1)

eg\(\,\,\,\,\,\)\( - \frac{{108}}{{{x^2}}} + 4x = 0,{\text{ }}4x = \frac{{108}}{{{x^2}}}\)

correct simplification     (A1)

eg\(\,\,\,\,\,\)\( - 108 + 4{x^3} = 0,{\text{ }}4{x^3} = 108\)

correct working     (A1)

eg\(\,\,\,\,\,\)\({x^3} = 27\)

\({\text{height}} = 3{\text{ (m) }}({\text{accept }}x = 3)\)    A1     N2

[5 marks]

attempt to find area using their height     (M1)

eg\(\,\,\,\,\,\)\(\frac{{108}}{3} + 2{(3)^2},{\text{ }}9 + 9 + 12 + 12 + 12\)

minimum surface area \( = 54{\text{ }}{{\text{m}}^{\text{2}}}\) (may be seen in part (c))     A1

attempt to find the number of tins     (M1)

eg\(\,\,\,\,\,\)\(\frac{{54}}{{10}},{\text{ }}5.4\)

6 (tins)     (A1)

$120     A1     N3

[5 marks]

Many candidates answered part (a) of this question correctly, though some seemed to be working backwards from the given expression for area, which is not the intention of a "show that" question.

In part (b), while many candidates found the correct derivative, some did so using cumbersome methods such as the quotient rule, rather than using the simpler power rule.

It was disappointing to see the number of candidates who did not recognize that the derivative they had just found in part (b) would have to be equal to zero in order for the surface area to be a minimum. For the candidates who did set their derivative equal to zero, most were able to find the correct height.

In part (d) of this question, there were some arithmetic errors which kept candidates from finding the correct area. The most common error here, by far, was not considering that the number of tins purchased must be an integer.