| Exam Board | CAIE |
|---|---|
| Module | P1 (Pure Mathematics 1) |
| Year | 2008 |
| Session | November |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Chain Rule |
| Type | Optimization with constraint |
| Difficulty | Standard +0.3 This is a standard optimization problem requiring setting up a constraint equation (perimeter = 80), expressing area in terms of one variable, then differentiating to find stationary points. The algebra is straightforward and the problem follows a familiar textbook pattern, making it slightly easier than average for A-level. |
| Spec | 1.02z Models in context: use functions in modelling1.07n Stationary points: find maxima, minima using derivatives1.07o Increasing/decreasing: functions using sign of dy/dx |
| Answer | Marks | Guidance |
|---|---|---|
| \(4x + 2\pi r = 80\) | B1 | Connection of lengths |
| \(A = x^2 + \pi r^2\) | B1 | Connection of areas |
| \(\rightarrow A = \frac{(\pi + 4)x^2 - 160x + 1600}{\pi}\) | M1, A1 | Eliminates \(r\). co but answer given. |
| Answer | Marks | Guidance |
|---|---|---|
| \(\frac{dA}{dx} = \frac{2(\pi + 4)x - 160}{\pi}\) | M1, A1 | Attempt at differentiation. co. Ignore omission of \(\pi\). |
| \(= 0 \text{ when } x = \frac{160}{2(\pi + 4)} \text{ or } 11.2\) | DM1, A1 | Sets to 0 and solves. co |
**(i)**
$4x + 2\pi r = 80$ | B1 | Connection of lengths
$A = x^2 + \pi r^2$ | B1 | Connection of areas
$\rightarrow A = \frac{(\pi + 4)x^2 - 160x + 1600}{\pi}$ | M1, A1 | Eliminates $r$. co but answer given.
**(ii)**
$\frac{dA}{dx} = \frac{2(\pi + 4)x - 160}{\pi}$ | M1, A1 | Attempt at differentiation. co. Ignore omission of $\pi$.
$= 0 \text{ when } x = \frac{160}{2(\pi + 4)} \text{ or } 11.2$ | DM1, A1 | Sets to 0 and solves. co
---7\\
\includegraphics[max width=\textwidth, alt={}, center]{08729aab-586b-4210-94c9-77b1f6b1d873-3_385_360_1379_561}\\
\includegraphics[max width=\textwidth, alt={}, center]{08729aab-586b-4210-94c9-77b1f6b1d873-3_364_369_1379_1219}
A wire, 80 cm long, is cut into two pieces. One piece is bent to form a square of side $x \mathrm {~cm}$ and the other piece is bent to form a circle of radius $r \mathrm {~cm}$ (see diagram). The total area of the square and the circle is $A \mathrm {~cm} ^ { 2 }$.\\
(i) Show that $A = \frac { ( \pi + 4 ) x ^ { 2 } - 160 x + 1600 } { \pi }$.\\
(ii) Given that $x$ and $r$ can vary, find the value of $x$ for which $A$ has a stationary value.
\hfill \mbox{\textit{CAIE P1 2008 Q7 [8]}}