| Exam Board | OCR |
|---|---|
| Module | H240/02 (Pure Mathematics and Statistics) |
| Year | 2018 |
| Session | March |
| Marks | 7 |
| Topic | Stationary points and optimisation |
| Type | Optimise cost or profit model |
| Difficulty | Standard +0.3 This is a straightforward optimization problem with clearly defined functions and a given hint that there's exactly one minimum. Part (i) is trivial algebra (time = distance/speed, cost = rate × time). Parts (ii)-(iii) require standard calculus: substitute R into T, differentiate, set to zero, and solve a simple cubic equation. The algebra is clean and the problem requires no novel insight—just methodical application of standard A-level techniques. |
| Spec | 1.07n Stationary points: find maxima, minima using derivatives1.08a Fundamental theorem of calculus: integration as reverse of differentiation |
| Answer | Marks | Guidance |
|---|---|---|
| Hence \(T = \frac{500R}{v}\) | B1 | AG Must see Time \(= \frac{500}{v}\) |
| [1] |
| Answer | Marks | Guidance |
|---|---|---|
| \(T = \frac{500}{v}\left(270 + \frac{v^3}{200}\right)\) \(\left(= \frac{135000}{v} + \frac{5v^2}{2}\right)\) | M1 | |
| \(\frac{dT}{dv} = -\frac{135000}{v^2} + 5v\) oe | M1 | Attempt diff their \(T\) |
| \(-\frac{135000}{v^2} + 5v = 0\) \([v^3 = 27000]\) | M1 | Their \(\frac{dT}{dv} = 0\) |
| Required speed is \(30\) km/h | A1 | Allow \(v = 30\) km/h; not just \(v = 30\) |
| [4] |
| Answer | Marks | Guidance |
|---|---|---|
| Min cost = £6750 | M1, A1 | Subst their '30' into their \(T\) |
| Answer | Marks |
|---|---|
| [2] |
## 3(i)
Time $= \frac{500}{v}$, $T = \frac{500}{v} \times R$
Hence $T = \frac{500R}{v}$ | B1 | AG Must see Time $= \frac{500}{v}$
| [1]
## 3(ii)
$T = \frac{500}{v}\left(270 + \frac{v^3}{200}\right)$ $\left(= \frac{135000}{v} + \frac{5v^2}{2}\right)$ | M1 |
$\frac{dT}{dv} = -\frac{135000}{v^2} + 5v$ oe | M1 | Attempt diff their $T$
$-\frac{135000}{v^2} + 5v = 0$ $[v^3 = 27000]$ | M1 | Their $\frac{dT}{dv} = 0$
Required speed is $30$ km/h | A1 | Allow $v = 30$ km/h; not just $v = 30$
| [4]
## 3(iii)
$T_{\min} = \frac{135000}{30} + \frac{5 \times 30^2}{2}$
Min cost = £6750 | M1, A1 | Subst their '30' into their $T$
£ necessary
| [2]
---3 On a particular voyage, a ship sails 500 km at a constant speed of $v \mathrm {~km} / \mathrm { h }$. The cost for the voyage is $\pounds R$ per hour. The total cost of the voyage is $\pounds T$.\\
(i) Show that $T = \frac { 500 R } { v }$.
The running cost is modelled by the following formula.
$$R = 270 + \frac { v ^ { 3 } } { 200 }$$
The ship's owner wishes to sail at a speed that will minimise the total cost for the voyage. It is given that the graph of $T$ against $v$ has exactly one stationary point, which is a minimum.\\
(ii) Find the speed that gives the minimum value of $T$.\\
(iii) Find the minimum value of the total cost.
\hfill \mbox{\textit{OCR H240/02 2018 Q3 [7]}}