| Exam Board | CAIE |
|---|---|
| Module | P1 (Pure Mathematics 1) |
| Year | 2011 |
| Session | November |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Solving quadratics and applications |
| Type | Optimisation via quadratic model |
| Difficulty | Moderate -0.3 This is a straightforward applied quadratic optimization problem requiring standard techniques: forming a perimeter equation, substituting to get area as a quadratic function, then finding the maximum using differentiation or completing the square. The L-shape adds minimal complexity as the perimeter/area relationships are direct. Slightly easier than average due to being a routine textbook-style optimization with clear steps and the quadratic already simplified in part (ii). |
| Spec | 1.02c Simultaneous equations: two variables by elimination and substitution1.02e Complete the square: quadratic polynomials and turning points1.07n Stationary points: find maxima, minima using derivatives1.07p Points of inflection: using second derivative |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(y = \frac{1}{6(48-8x)}\) oe | B1 | [1] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(A = 4xy + 2xy\) or \(3xy + 3xy = 6xy\) | M1 | |
| \(A = x(48 - 8x) = 48x - 8x^2\) | A1 | [2] AG |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(\frac{\delta A}{\delta x} = 48 - 16x\) | B1 | |
| \(A = 72\) cao | M1A1 | Attempt to solve derivative \(= 0\). Expect \(x = 3\) |
| \(\frac{\delta^2 A}{\delta x^2} = -16\ (< 0) \Rightarrow\) Maximum | B1 | [4] www Accept other complete methods |
## Question 7:
### Part (i):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $y = \frac{1}{6(48-8x)}$ oe | B1 | [1] |
### Part (ii):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $A = 4xy + 2xy$ or $3xy + 3xy = 6xy$ | M1 | |
| $A = x(48 - 8x) = 48x - 8x^2$ | A1 | [2] AG |
### Part (iii):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\frac{\delta A}{\delta x} = 48 - 16x$ | B1 | |
| $A = 72$ cao | M1A1 | Attempt to solve derivative $= 0$. Expect $x = 3$ |
| $\frac{\delta^2 A}{\delta x^2} = -16\ (< 0) \Rightarrow$ Maximum | B1 | [4] www Accept other complete methods |
---7\\
\includegraphics[max width=\textwidth, alt={}, center]{56d376c5-b91f-488d-89e2-18edcb14052d-3_534_895_255_625}
The diagram shows the dimensions in metres of an L-shaped garden. The perimeter of the garden is 48 m .\\
(i) Find an expression for $y$ in terms of $x$.\\
(ii) Given that the area of the garden is $A \mathrm {~m} ^ { 2 }$, show that $A = 48 x - 8 x ^ { 2 }$.\\
(iii) Given that $x$ can vary, find the maximum area of the garden, showing that this is a maximum value rather than a minimum value.
\hfill \mbox{\textit{CAIE P1 2011 Q7 [7]}}