| Exam Board | Edexcel |
|---|---|
| Module | C2 (Core Mathematics 2) |
| Year | 2007 |
| Session | June |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Stationary points and optimisation |
| Type | Show formula then optimise: cylinder/prism (single variable) |
| Difficulty | Standard +0.3 This is a standard C2 optimization problem with a straightforward constraint equation leading to a simple cubic function. Part (a) is algebraic manipulation, part (b) requires basic differentiation and solving a quadratic, and part (c) asks for second derivative test. While it involves multiple steps, each step follows a routine procedure with no novel insight required, making it slightly easier than average. |
| Spec | 1.07i Differentiate x^n: for rational n and sums1.07n Stationary points: find maxima, minima using derivatives1.07o Increasing/decreasing: functions using sign of dy/dx |
| Answer | Marks | Guidance |
|---|---|---|
| \(4x^2 + 6xy = 600\) | M1 A1 | Attempting expression in \(x\) and \(y\) for total surface area, equated to 600 |
| \(V = 2x^2 y = 2x^2\left(\frac{600-4x^2}{6x}\right)\), \(V = 200x - \frac{4x^3}{3}\) | M1 A1cso | Substituting \(y\) into \(2x^2y\) to form expression in \(x\) only |
| Answer | Marks | Guidance |
|---|---|---|
| \(\frac{dV}{dx} = 200 - 4x^2\) | B1 | |
| Equate \(\frac{dV}{dx}\) to 0 and solve: \(x^2 = 50\) or \(x = \sqrt{50}\) | M1 A1 | Ignore \(x = -\sqrt{50}\) |
| \(V = 200(\sqrt{50}) - \frac{4}{3}(50\sqrt{50}) = 943 \text{ cm}^3\) | M1 A1 | Allow awrt |
| Answer | Marks | Guidance |
|---|---|---|
| \(\frac{d^2V}{dx^2} = -8x\), Negative, \(\therefore\) Maximum | M1, A1ft | Find second derivative and consider its sign; must conclude maximum not minimum |
## Question 10:
### Part (a):
| $4x^2 + 6xy = 600$ | M1 A1 | Attempting expression in $x$ and $y$ for total surface area, equated to 600 |
| $V = 2x^2 y = 2x^2\left(\frac{600-4x^2}{6x}\right)$, $V = 200x - \frac{4x^3}{3}$ | M1 A1cso | Substituting $y$ into $2x^2y$ to form expression in $x$ only |
### Part (b):
| $\frac{dV}{dx} = 200 - 4x^2$ | B1 | |
| Equate $\frac{dV}{dx}$ to 0 and solve: $x^2 = 50$ or $x = \sqrt{50}$ | M1 A1 | Ignore $x = -\sqrt{50}$ |
| $V = 200(\sqrt{50}) - \frac{4}{3}(50\sqrt{50}) = 943 \text{ cm}^3$ | M1 A1 | Allow awrt |
### Part (c):
| $\frac{d^2V}{dx^2} = -8x$, Negative, $\therefore$ Maximum | M1, A1ft | Find second derivative and consider its sign; must conclude maximum not minimum |10.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{22ebc302-765c-4734-b312-b286ccb20be9-15_538_529_205_744}
\captionsetup{labelformat=empty}
\caption{Figure 4}
\end{center}
\end{figure}
Figure 4 shows a solid brick in the shape of a cuboid measuring $2 x \mathrm {~cm}$ by $x \mathrm {~cm}$ by $y \mathrm {~cm}$. The total surface area of the brick is $600 \mathrm {~cm} ^ { 2 }$.
\begin{enumerate}[label=(\alph*)]
\item Show that the volume, $V \mathrm {~cm} ^ { 3 }$, of the brick is given by
$$V = 200 x - \frac { 4 x ^ { 3 } } { 3 }$$
Given that $x$ can vary,
\item use calculus to find the maximum value of $V$, giving your answer to the nearest $\mathrm { cm } ^ { 3 }$.
\item Justify that the value of $V$ you have found is a maximum.
\end{enumerate}
\hfill \mbox{\textit{Edexcel C2 2007 Q10 [11]}}