| Exam Board | AQA |
|---|---|
| Module | Paper 2 (Paper 2) |
| Year | 2022 |
| Session | June |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Stationary points and optimisation |
| Type | Optimise perimeter or area of 2D region |
| Difficulty | Standard +0.3 This is a standard optimization problem requiring students to express area in terms of one variable using the curve equation, then differentiate and find the maximum. The geometry is straightforward (isosceles triangle with horizontal top), the algebra is routine (area = ½ × base × height), and the calculus is standard A-level fare. Slightly easier than average due to part (a) scaffolding the setup. |
| Spec | 1.07n Stationary points: find maxima, minima using derivatives1.08e Area between curve and x-axis: using definite integrals |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| Height of triangle \(= 15 - q^2\) | M1 | Identifies height of triangle or rectangle as \(15-q^2\); PI by \((q, 15-q^2)\), \(h=15-q^2\) or \(y=15-q^2\) indicated on diagram |
| \(A = \frac{1}{2}\times 2q(15-q^2) = 15q - q^3\); since \(A = q(15-q^2) > 0\) then \(q\)'s upper limit \(c = \sqrt{15}\) | R1 | Completes rigorous argument showing given result; must be clear how base and height defined with use of \(\frac{1}{2}\times 2q(15-q^2)\) for whole triangle, or \(\left[\frac{1}{2}q(15-q^2)\right]\times 2\) for two half triangles, or explains area via rectangle on either side of \(y\)-axis |
| \(c = \sqrt{15}\) | B1 | Deduces \(c = \sqrt{15}\); ACF |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\frac{dA}{dq} = 15 - 3q^2\), max occurs at \(\frac{dA}{dq} = 0\) | E1 (AO 2.4) | Explains that maximum occurs when derivative equals 0; condone incorrect variables in their derivative |
| \(15 - 3q^2 = 0\), \(q = \sqrt{5}\) | M1 (AO 3.1a) | Differentiates w.r.t. \(q\); at least one term correct |
| Obtains \(15 - 3q^2\) | A1 (AO 1.1b) | |
| Solves \(\frac{dA}{dq} = 0\) to find \(q\) and substitutes to find maximum area | M1 (AO 1.1a) | |
| \(\therefore\) Max area \(= 15\sqrt{5} - 5\sqrt{5} = 10\sqrt{5}\) | A1 (AO 1.1b) | Obtains correct maximum area; ACF |
| \(\frac{d^2A}{dq^2} = -6\sqrt{5} < 0\), so local maximum | E1 (AO 2.4) | Gives justification for maximum; could be evaluation of second derivative as \(-13.42\ldots < 0\), or test of gradient either side, or explanation that this must be a max value as only turning point in the interval \(0 < q < \sqrt{15}\) and the area is 0 at the endpoints |
| Subtotal: 6 marks | Question 7 Total: 9 marks |
## Question 7(a):
| Answer/Working | Marks | Guidance |
|---|---|---|
| Height of triangle $= 15 - q^2$ | M1 | Identifies height of triangle or rectangle as $15-q^2$; PI by $(q, 15-q^2)$, $h=15-q^2$ or $y=15-q^2$ indicated on diagram |
| $A = \frac{1}{2}\times 2q(15-q^2) = 15q - q^3$; since $A = q(15-q^2) > 0$ then $q$'s upper limit $c = \sqrt{15}$ | R1 | Completes rigorous argument showing given result; must be clear how base and height defined with use of $\frac{1}{2}\times 2q(15-q^2)$ for whole triangle, or $\left[\frac{1}{2}q(15-q^2)\right]\times 2$ for two half triangles, or explains area via rectangle on either side of $y$-axis |
| $c = \sqrt{15}$ | B1 | Deduces $c = \sqrt{15}$; ACF |
## Question 7(b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\frac{dA}{dq} = 15 - 3q^2$, max occurs at $\frac{dA}{dq} = 0$ | E1 (AO 2.4) | Explains that maximum occurs when derivative equals 0; condone incorrect variables in their derivative |
| $15 - 3q^2 = 0$, $q = \sqrt{5}$ | M1 (AO 3.1a) | Differentiates w.r.t. $q$; at least one term correct |
| Obtains $15 - 3q^2$ | A1 (AO 1.1b) | |
| Solves $\frac{dA}{dq} = 0$ to find $q$ and substitutes to find maximum area | M1 (AO 1.1a) | |
| $\therefore$ Max area $= 15\sqrt{5} - 5\sqrt{5} = 10\sqrt{5}$ | A1 (AO 1.1b) | Obtains correct maximum area; ACF |
| $\frac{d^2A}{dq^2} = -6\sqrt{5} < 0$, so local maximum | E1 (AO 2.4) | Gives justification for maximum; could be evaluation of second derivative as $-13.42\ldots < 0$, or test of gradient either side, or explanation that this must be a max value as only turning point in the interval $0 < q < \sqrt{15}$ and the area is 0 at the endpoints |
**Subtotal: 6 marks | Question 7 Total: 9 marks**
---7 The curve $y = 15 - x ^ { 2 }$ and the isosceles triangle $O P Q$ are shown on the diagram
The curve $y = 15 - x ^ { 2 }$ and the isosceles triangle $O P Q$ are shown on the diagram below.\\
\includegraphics[max width=\textwidth, alt={}, center]{ad6590e8-6673-45ca-bef3-a14716978827-10_759_810_388_614}
Vertices $P$ and $Q$ lie on the curve such that $Q$ lies vertically above some point ( $q , 0$ ) The line $P Q$ is parallel to the $x$-axis.
7
\begin{enumerate}[label=(\alph*)]
\item Show that the area, $A$, of the triangle $O P Q$ is given by
$$A = 15 q - q ^ { 3 } \quad \text { for } 0 < q < c$$
where $c$ is a constant to be found.\\
7
\item Find the exact maximum area of triangle $O P Q$.
Fully justify your answer.
\end{enumerate}
\hfill \mbox{\textit{AQA Paper 2 2022 Q7 [9]}}