| Exam Board | OCR |
|---|---|
| Module | H240/01 (Pure Mathematics) |
| Year | 2018 |
| Session | September |
| Marks | 9 |
| Topic | Stationary points and optimisation |
| Type | Show formula then optimise: composite/irregular shape |
| Difficulty | Standard +0.3 This is a standard optimization problem with a constraint, requiring volume equation setup, substitution to eliminate a variable, differentiation, and finding a minimum. Part (i) is guided (show that), part (ii) is routine calculus application with second derivative test, and part (iii) requires minimal reasoning. The geometry is straightforward (cylinder + hemisphere) and all formulas are provided. Slightly easier than average due to the scaffolding and standard technique application. |
| Spec | 1.02z Models in context: use functions in modelling1.07n Stationary points: find maxima, minima using derivatives1.08d Evaluate definite integrals: between limits |
| Answer | Marks | Guidance |
|---|---|---|
| (i) | \(\frac{2}{3}\pi r^3 + \pi r^2 h = 45\pi\) | B1 |
| \(A = \pi r^2 + 2\pi r^2 + 2\pi rh\) | B1 | Correct expression for surface area |
| \(h = \frac{45 - \frac{2}{3}r^3}{r^2} = 45r^{-2} - \frac{2}{3}r\) | M1 | Attempt to make \(h\) the subject and hence eliminate \(h\) |
| \(A = 3\pi r^2 + 2\pi(45r^{-2} - \frac{2}{3}r) = 3\pi r^2 + 90\pi r^{-1} - \frac{4}{3}\pi r\) | ||
| \(A = \frac{4}{3}\pi r^2 + \frac{90\pi}{r}\) | A1 | Simplify to obtain given answer |
Total: [4]
| Answer | Marks | Guidance |
|---|---|---|
| (ii) | \(\frac{dA}{dr} = \frac{10}{3}\pi r - 90\pi r^{-2}\) | M1 |
| \(\frac{10}{3}\pi r - 90\pi r^{-2} = 0 \Rightarrow r = 3\) | M1 | Equate to 0 and solve for \(r\) |
| \(A = 45\pi\) cm\(^2\) or \(141\) cm\(^2\) | A1 | Correct surface area, including units (FT their first derivative, provided it gives a minimum) |
| \(\frac{d^2A}{dr^2} = \frac{10}{3}\pi + 180\pi r^{-3} > 0\) hence minimum | A1IFT | FT their first derivative (Or using the sign-change of first derivative) |
Total: [4]
| Answer | Marks | Guidance |
|---|---|---|
| (iii) | E.g. Cheaper to manufacture as uses less material | E1 |
Total: [1]
(i) | $\frac{2}{3}\pi r^3 + \pi r^2 h = 45\pi$ | B1 | Equate correct volume to $45\pi$
| $A = \pi r^2 + 2\pi r^2 + 2\pi rh$ | B1 | Correct expression for surface area
| $h = \frac{45 - \frac{2}{3}r^3}{r^2} = 45r^{-2} - \frac{2}{3}r$ | M1 | Attempt to make $h$ the subject and hence eliminate $h$
| $A = 3\pi r^2 + 2\pi(45r^{-2} - \frac{2}{3}r) = 3\pi r^2 + 90\pi r^{-1} - \frac{4}{3}\pi r$ | | |
| $A = \frac{4}{3}\pi r^2 + \frac{90\pi}{r}$ | A1 | Simplify to obtain given answer
Total: [4]
(ii) | $\frac{dA}{dr} = \frac{10}{3}\pi r - 90\pi r^{-2}$ | M1 | Attempt differentiation
| $\frac{10}{3}\pi r - 90\pi r^{-2} = 0 \Rightarrow r = 3$ | M1 | Equate to 0 and solve for $r$
| $A = 45\pi$ cm$^2$ or $141$ cm$^2$ | A1 | Correct surface area, including units (FT their first derivative, provided it gives a minimum)
| $\frac{d^2A}{dr^2} = \frac{10}{3}\pi + 180\pi r^{-3} > 0$ hence minimum | A1IFT | FT their first derivative (Or using the sign-change of first derivative)
Total: [4]
(iii) | E.g. Cheaper to manufacture as uses less material | E1 | Sensible reason based on surface area
Total: [1]8\\
\includegraphics[max width=\textwidth, alt={}, center]{e3942549-bfc0-432a-bf49-7d01d44af01a-6_533_524_246_772}
The diagram shows a container which consists of a cylinder with a solid base and a hemispherical top. The radius of the cylinder is $r \mathrm {~cm}$ and the height is $h \mathrm {~cm}$. The container is to be made of thin plastic. The volume of the container is $45 \pi \mathrm {~cm} ^ { 3 }$.\\
(i) Show that the surface area of the container, $A \mathrm {~cm} ^ { 2 }$, is given by
$$A = \frac { 5 } { 3 } \pi r ^ { 2 } + \frac { 90 \pi } { r } .$$
[The volume of a sphere is $V = \frac { 4 } { 3 } \pi r ^ { 3 }$ and the surface area of a sphere is $S = 4 \pi r ^ { 2 }$.]\\
(ii) Use calculus to find the minimum surface area of the container, justifying that it is a minimum.\\
(iii) Suggest a reason why the manufacturer would wish to minimise the surface area.
\hfill \mbox{\textit{OCR H240/01 2018 Q8 [9]}}