| Exam Board | Edexcel |
|---|---|
| Module | C2 (Core Mathematics 2) |
| Year | 2008 |
| Session | January |
| Marks | 12 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Stationary points and optimisation |
| Type | Optimise geometric shape surface area/volume |
| Difficulty | Standard +0.3 This is a standard C2 optimisation problem with clear scaffolding through four parts. Students must form an expression for surface area using the volume constraint, differentiate a simple polynomial/rational function, find stationary points, apply the second derivative test, and evaluate. All steps are routine techniques with no novel insight required, making it slightly easier than the average A-level question. |
| Spec | 1.07n Stationary points: find maxima, minima using derivatives1.07o Increasing/decreasing: functions using sign of dy/dx1.07p Points of inflection: using second derivative |
| Answer | Marks |
|---|---|
| (a) (Total area) = \(3xy + 2x^2\) | B1 |
| (Vol:) \(x^2 y = 100\) (i.e. \(y = \frac{100}{x^2}\), \(xy = \frac{100}{x}\)) | B1 |
| Deriving expression for area in terms of \(x\) only | M1 |
| Answer | Marks |
|---|---|
| \[({\text{Area}} = \frac{300}{x} + 2x^2\] | AG; A1 (cso) (4) |
| (b) \[\frac{dA}{dx} = -\frac{300}{x^2} + 4x\] | M1A1 |
| Setting \(\frac{dA}{dx} = 0\) and finding a value for correct power of \(x\), for cand. | M1 |
| Answer | Marks | Guidance |
|---|---|---|
| \(x = 4.2172\) (awrt 4.22) (allow exact \(\sqrt[3]{75}\)) | A1 (4) | |
| (c) \[\frac{d^2A}{dx^2} = \frac{600}{x^3} + 4 =\] positive | M1A1 (2) | therefore minimum |
| (d) Substituting found value of \(x\) into (a) | M1 |
| Answer | Marks |
|---|---|
| Area = 106.707 (awrt 107) | A1 (2) [12] |
**(a)** (Total area) = $3xy + 2x^2$ | B1 |
(Vol:) $x^2 y = 100$ (i.e. $y = \frac{100}{x^2}$, $xy = \frac{100}{x}$) | B1 |
Deriving expression for area in terms of $x$ only | M1 |
(Substitution, or clear use of $y$ or $xy$ into expression for area)
$$({\text{Area}} = \frac{300}{x} + 2x^2$$ | AG; A1 (cso) (4) |
**(b)** $$\frac{dA}{dx} = -\frac{300}{x^2} + 4x$$ | M1A1 |
Setting $\frac{dA}{dx} = 0$ and finding a value for correct power of $x$, for cand. | M1 |
[$x^3 = 75$]
$x = 4.2172$ (awrt 4.22) (allow exact $\sqrt[3]{75}$) | A1 (4) |
**(c)** $$\frac{d^2A}{dx^2} = \frac{600}{x^3} + 4 =$$ positive | M1A1 (2) | therefore minimum
**(d)** Substituting found value of $x$ into (a) | M1 |
(Or finding $y$ for found $x$ and substituting both in $3xy + 2x^2$)
$[y = \frac{100}{4.2172^2} = 5.6228]$
Area = 106.707 (awrt 107) | A1 (2) [12] |
**Notes:**
**(a)** First B1: Earned for correct unsimplified expression, isw.
**(c)** For M1: Find $\frac{d^2A}{dx^2}$ and explicitly consider its sign, state > 0 or "positive"
A1: Candidate's $\frac{d^2A}{dx^2}$ must be correct for their $\frac{dA}{dx}$, sign must be + ve and conclusion "so minimum", (allow QED, √).
(May be wrong $x$, or even no value of $x$ found)
**Alternative:** M1: Find value of $\frac{dA}{dx}$ on either side of "$x = \sqrt[3]{75}$" and consider sign
A1: Indicate sign change of negative to positive for $\frac{dA}{dx}$, and conclude minimum.
**OR** M1: Consider values of A on either side of "$x = \sqrt[3]{75}$" and compare with "107"
A1: Both values greater than "$x = 107$" and conclude minimum.
Allow marks for (c) and (d) where seen; even if part labelling confused.\includegraphics{figure_4}
Figure 4 shows an open-topped water tank, in the shape of a cuboid, which is made of sheet metal. The base of the tank is a rectangle $x$ metres by $y$ metres. The height of the tank is $x$ metres.
The capacity of the tank is 100 m³.
\begin{enumerate}[label=(\alph*)]
\item Show that the area $A$ m² of the sheet metal used to make the tank is given by
$$A = \frac{300}{x} + 2x^2.$$
[4]
\item Use calculus to find the value of $x$ for which $A$ is stationary. [4]
\item Prove that this value of $x$ gives a minimum value of $A$. [2]
\item Calculate the minimum area of sheet metal needed to make the tank. [2]
\end{enumerate}
\hfill \mbox{\textit{Edexcel C2 2008 Q9 [12]}}