| Exam Board | CAIE |
|---|---|
| Module | P1 (Pure Mathematics 1) |
| Year | 2010 |
| Session | June |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Tangents, normals and gradients |
| Type | Optimization with constraints |
| Difficulty | Moderate -0.3 This is a standard optimization problem requiring surface area formula manipulation, differentiation of a polynomial, and second derivative test. While it involves multiple steps (3+3+2 marks), each step uses routine A-level techniques with no novel insight required. The algebra is straightforward and the problem type is commonly practiced, making it slightly easier than average. |
| Spec | 1.07n Stationary points: find maxima, minima using derivatives1.07p Points of inflection: using second derivative |
| Answer | Marks | Guidance |
|---|---|---|
| (i) \(4xh + 2x^2 = 96 \rightarrow h = \frac{24}{x} - \frac{x}{2}\) | M1, A1 | Needs to consider at least 5 areas. co. |
| \(V = x^2h \rightarrow V = 24x - \frac{x^3}{2}\). | M1 [3] | for \(V = x^2h\) with \(h\) as \(f(x)\). |
| (ii) \(\frac{dV}{dx} = 24 - \frac{3x^2}{2} = 0\) when \(x = 4 \rightarrow V = 64\). | B1, M1, A1 [3] | co. Sets differential to 0 and solves. co. |
| (iii) \(\frac{d^2V}{dx^2} = -3x \rightarrow\) Max. | M1 A1√ [2] | Any valid method. co. |
(i) $4xh + 2x^2 = 96 \rightarrow h = \frac{24}{x} - \frac{x}{2}$ | M1, A1 | Needs to consider at least 5 areas. co.
$V = x^2h \rightarrow V = 24x - \frac{x^3}{2}$. | M1 [3] | for $V = x^2h$ with $h$ as $f(x)$.
(ii) $\frac{dV}{dx} = 24 - \frac{3x^2}{2} = 0$ when $x = 4 \rightarrow V = 64$. | B1, M1, A1 [3] | co. Sets differential to 0 **and** solves. co.
(iii) $\frac{d^2V}{dx^2} = -3x \rightarrow$ Max. | M1 A1√ [2] | Any valid method. co.A solid rectangular block has a square base of side $x$ cm. The height of the block is $h$ cm and the total surface area of the block is $96$ cm$^2$.
\begin{enumerate}[label=(\roman*)]
\item Express $h$ in terms of $x$ and show that the volume, $V$ cm$^3$, of the block is given by
$$V = 24x - \frac{1}{2}x^3.$$ [3]
\end{enumerate}
Given that $x$ can vary,
\begin{enumerate}[label=(\roman*)]
\setcounter{enumi}{1}
\item find the stationary value of $V$, [3]
\item determine whether this stationary value is a maximum or a minimum. [2]
\end{enumerate}
\hfill \mbox{\textit{CAIE P1 2010 Q8 [8]}}