| Exam Board | CAIE |
|---|---|
| Module | P1 (Pure Mathematics 1) |
| Year | 2016 |
| Session | June |
| Marks | 6 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Solving quadratics and applications |
| Type | Optimisation via quadratic model |
| Difficulty | Standard +0.3 This is a standard applied quadratic optimization problem requiring students to form an equation from a constraint (total fencing), substitute to get area as a function of one variable, then find the maximum using differentiation or completing the square. The setup is straightforward with clear guidance in part (i), and part (ii) is routine calculus. Slightly easier than average due to the scaffolding provided. |
| Spec | 1.02e Complete the square: quadratic polynomials and turning points1.07n Stationary points: find maxima, minima using derivatives |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(A = 2y \times 4x\ (= 8xy)\) | B1 | |
| \(10y + 12x = 480\) | B1 | |
| \(\rightarrow A = 384x - 9.6x^2\) | B1 | Answer given |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(\frac{dA}{dx} = 384 - 19.2x\) | B1 | |
| \(= 0\) when \(x = 20\) | M1 | Sets to 0 and attempts to solve; might see completion of square |
| \(\rightarrow x = 20, y = 24\) | A1 | Needs both \(x\) and \(y\) |
| Uses \(x = -\frac{b}{2a} = \frac{-384}{-19.2} = 20\); \(y = 24\) | M1, A1 | Trial and improvement B3 |
## Question 5(i):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $A = 2y \times 4x\ (= 8xy)$ | **B1** | |
| $10y + 12x = 480$ | **B1** | |
| $\rightarrow A = 384x - 9.6x^2$ | **B1** | Answer given |
## Question 5(ii):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\frac{dA}{dx} = 384 - 19.2x$ | **B1** | |
| $= 0$ when $x = 20$ | **M1** | Sets to 0 and attempts to solve; might see completion of square |
| $\rightarrow x = 20, y = 24$ | **A1** | Needs both $x$ and $y$ |
| Uses $x = -\frac{b}{2a} = \frac{-384}{-19.2} = 20$; $y = 24$ | **M1, A1** | Trial and improvement **B3** |5\\
\includegraphics[max width=\textwidth, alt={}, center]{b6ae63ce-a8a8-45ef-9c75-2fab30de8ad9-2_364_625_1873_762}
A farmer divides a rectangular piece of land into 8 equal-sized rectangular sheep pens as shown in the diagram. Each sheep pen measures $x \mathrm {~m}$ by $y \mathrm {~m}$ and is fully enclosed by metal fencing. The farmer uses 480 m of fencing.\\
(i) Show that the total area of land used for the sheep pens, $A \mathrm {~m} ^ { 2 }$, is given by
$$A = 384 x - 9.6 x ^ { 2 }$$
(ii) Given that $x$ and $y$ can vary, find the dimensions of each sheep pen for which the value of $A$ is a maximum. (There is no need to verify that the value of $A$ is a maximum.)
\hfill \mbox{\textit{CAIE P1 2016 Q5 [6]}}