| Date | May 2010 | Marks available | 2 | Reference code | 10M.2.sl.TZ2.5 |
| Level | SL only | Paper | 2 | Time zone | TZ2 |
| Command term | Show that | Question number | 5 | Adapted from | N/A |
Question
A dog food manufacturer has to cut production costs. She wishes to use as little aluminium as possible in the construction of cylindrical cans. In the following diagram, h represents the height of the can in cm and x, the radius of the base of the can in cm.
The volume of the dog food cans is 600 cm3.
Show that \(h = \frac{{600}}{{\pi {x^2}}}\).
Find an expression for the curved surface area of the can, in terms of x. Simplify your answer.
Hence write down an expression for A, the total surface area of the can, in terms of x.
Differentiate A in terms of x.
Find the value of x that makes A a minimum.
Calculate the minimum total surface area of the dog food can.
Markscheme
\(600 = \pi x^2 h\) (M1)(A1)
\(\frac{600}{\pi x^2} = h\) (AG)
Note: Award (M1) for correct substituted formula, (A1) for correct substitution. If answer given not shown award at most (M1)(A0).
[2 marks]
\(C = 2 \pi x \frac{600}{\pi x^2}\) (M1)
\(C = \frac{1200}{x}\) (or 1200x–1) (A1)
Note: Award (M1) for correct substitution in formula, (A1) for correct simplification.
[??? marks]
\(A = 2 \pi x^2 + 1200x^{-1}\) (A1)(A1)(ft)
Note: Award (A1) for multiplying the area of the base by two, (A1) for adding on their answer to part (b) (i).
For both marks to be awarded answer must be in terms of x.
[??? marks]
\(\frac{{{\text{d}}A}}{{{\text{d}}x}} = 4\pi x - \frac{{1200}}{{{x^2}}}\) (A1)(ft)(A1)(ft)(A1)(ft)
Notes: Award (A1) for \(4 \pi x\), (A1) for \(-1200\), (A1) for \(x^{-2}\). Award at most (A2) if any extra term is written. Follow through from their part (b) (ii).
[??? marks]
\(4 \pi x - \frac{1200}{x^2} = 0\) (M1)(M1)
\(x^3 = \frac{1200}{4 \pi}\) (or equivalent)
\(x = 4.57\) (A1)(ft)(G2)
Note: Award (M1) for using their derivative, (M1) for setting the derivative to zero, (A1)(ft) for answer.
Follow through from their derivative.
Last mark is lost if value of x is zero or negative.
[3 marks]
\(A = 2 \pi (4.57)^2 + 1200(4.57)^{-1}\) (M1)
\(A = 394\) (A1)(ft)(G2)
Note: Follow through from their answers to parts (b) (ii) and (d).
[2 marks]
Examiners report
This was the most difficult question for the candidates. It was clear that the vast majority of them had not had exposure to this style of question. Part (a) was well answered by most of the students. In part (b) the correct expression “in terms of x ” for the curve surface area was not frequently seen. In many cases the impression was that they did not know what “in terms of x ” meant as correct equivalent expressions were seen but where the h was also involved. Those candidates that made progress in the question, even with the wrong expression for the total area of the can, A were able to earn follow through marks.
This was the most difficult question for the candidates. It was clear that the vast majority of them had not had exposure to this style of question. Part (a) was well answered by most of the students. In part (b) the correct expression “in terms of x ” for the curve surface area was not frequently seen. In many cases the impression was that they did not know what “in terms of x ” meant as correct equivalent expressions were seen but where the h was also involved. Those candidates that made progress in the question, even with the wrong expression for the total area of the can, A were able to earn follow through marks.
This was the most difficult question for the candidates. It was clear that the vast majority of them had not had exposure to this style of question. Part (a) was well answered by most of the students. In part (b) the correct expression “in terms of x ” for the curve surface area was not frequently seen. In many cases the impression was that they did not know what “in terms of x ” meant as correct equivalent expressions were seen but where the h was also involved. Those candidates that made progress in the question, even with the wrong expression for the total area of the can, A were able to earn follow through marks.
This was the most difficult question for the candidates. It was clear that the vast majority of them had not had exposure to this style of question. Part (a) was well answered by most of the students. In part (b) the correct expression “in terms of x ” for the curve surface area was not frequently seen. In many cases the impression was that they did not know what “in terms of x ” meant as correct equivalent expressions were seen but where the h was also involved. Those candidates that made progress in the question, even with the wrong expression for the total area of the can, A were able to earn follow through marks.
This was the most difficult question for the candidates. It was clear that the vast majority of them had not had exposure to this style of question. Part (a) was well answered by most of the students. In part (b) the correct expression “in terms of x ” for the curve surface area was not frequently seen. In many cases the impression was that they did not know what “in terms of x ” meant as correct equivalent expressions were seen but where the h was also involved. Those candidates that made progress in the question, even with the wrong expression for the total area of the can, A were able to earn follow through marks.
This was the most difficult question for the candidates. It was clear that the vast majority of them had not had exposure to this style of question. Part (a) was well answered by most of the students. In part (b) the correct expression “in terms of x ” for the curve surface area was not frequently seen. In many cases the impression was that they did not know what “in terms of x ” meant as correct equivalent expressions were seen but where the h was also involved. Those candidates that made progress in the question, even with the wrong expression for the total area of the can, A were able to earn follow through marks.
