A company sells fruit juices in cylindrical cans, each of which has a volume of $340\,{\text{c}}{{\text{m}}^3}$. The surface area of a can is $A\,{\text{c}}{{\text{m}}^2}$ and is given by the formula

$A = 2\pi {r^2} + \frac{{680}}{r}$ ,

where $r$ is the radius of the can, in ${\text{cm}}$.

To reduce the cost of a can, its surface area must be minimized.

Find $\frac{{{\text{d}}A}}{{{\text{d}}r}}$

Calculate the value of $r$ that minimizes the surface area of a can.

$\left( {\frac{{{\text{d}}A}}{{{\text{d}}r}}} \right) = 4\pi r - \frac{{680}}{{{r^2}}}$        (A1)(A1)(A1)      (C3)

Note: Award (A1) for $4\pi r$ (accept $12.6r$), (A1) for $- 680$, (A1) for $\frac{1}{{{r^2}}}$ or ${r^{ - 2}}$

Award at most (A1)(A1)(A0) if additional terms are seen.

$4\pi r - \frac{{680}}{{{r^2}}} = 0$          (M1)

Note: Award (M1) for equating their $\frac{{{\text{d}}A}}{{{\text{d}}r}}$ to zero.

$4\pi {r^3} - 680 = 0$          (M1)

Note: Award (M1) for initial correct rearrangement of the equation. This may be assumed if ${r^3} = \frac{{680}}{{4\pi }}$ or $r = \sqrt[3]{{\frac{{680}}{{4\pi }}}}$ seen.

OR

sketch of $A$ with some indication of minimum point         (M1)(M1)

Note: Award (M1) for sketch of $A$, (M1) for indication of minimum point.

OR

sketch of $\frac{{{\text{d}}A}}{{{\text{d}}r}}$ with some indication of zero      (M1)(M1)

Note: Award (M1) for sketch of $\frac{{{\text{d}}A}}{{{\text{d}}r}}$, (M1) for indication of zero.

$(r = )\,\,3.78\,({\text{cm}})\,\,\,\,\,(3.78239...)$                (A1)(ft) (C3)  

Note: Follow through from part (a).

Question 15: Optimization
Many candidates were able to differentiate in part (a), but then were unable to relate this to part (b). However, it seemed that many more had not studied the calculus at all.

Question 15: Optimization
Many candidates were able to differentiate in part (a), but then were unable to relate this to part (b). However, it seemed that many more had not studied the calculus at all.