Standard +0.8 This is a constrained optimization problem requiring students to express the perimeter constraint (2r + rθ = 10), derive the area formula A = ½r²θ, substitute to get A as a function of one variable, differentiate, solve for the critical point, and verify it's a maximum using the second derivative test. While the individual techniques are standard A-level calculus, the multi-step problem-solving and need to manipulate the constraint equation makes this moderately challenging, above average difficulty.
7 A wire, 10 cm long, is bent to form the perimeter of a sector of a circle, as shown in the diagram. The radius is \(r \mathrm {~cm}\) and the angle at the centre is \(\theta\) radians.
\includegraphics[max width=\textwidth, alt={}, center]{20639e13-01cc-4d96-b694-fb3cf1828f4d-07_323_204_342_242}
Determine the maximum possible area of the sector, showing that it is a maximum.
For information: \(\dfrac{d^2A}{d\theta^2} = -\dfrac{200}{256} = -0.78125\) at \(\theta = 2\)
\(\theta = 1.5 \Rightarrow \dfrac{dA}{d\theta} > 0\), \(\theta = 2.5 \Rightarrow \dfrac{dA}{d\theta} < 0\) so max
B1
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## Question 7:
[Perimeter =] $2r + r\theta = 10$ | M1 | AO 1.1
$\theta = \dfrac{10 - 2r}{r}$ | B1 | AO 3.1a; OR $r = \dfrac{10}{2+\theta}$; expression for one of $r$, $\theta$ in terms of the other
$A = \dfrac{1}{2}r^2\theta$ so $A = \dfrac{r(10-2r)}{2} = 5r - r^2$ | M1 | AO 3.1a; OR $A = \dfrac{100\theta}{2(2+\theta)^2} = \dfrac{50\theta}{(2+\theta)^2}$; area in terms of either their $r$ or their $\theta$
$A = 2.5^2 - (2.5 - r)^2$ | M1 | AO 3.1a; Completing the square
This has a max when $2.5 - r = 0$ | B1 | AO 2.4; Convincing explanation that there is a max
Max $= 6.25$ [cm²] | A1 [6] | AO 2.2a
**Alternative method 1:**
$\dfrac{dA}{dr} = 5 - 2r$ | M1 | —
$\dfrac{d^2A}{dr^2} = -2$ so max | B1 | —
$\dfrac{dA}{dr} = 0 \Rightarrow r = 2.5$; Max $= 6.25$ [cm²] | A1 | —
**Alternative method 2:**
$\dfrac{dA}{d\theta} = \dfrac{50(2+\theta)^2 - 100\theta(2+\theta)}{(2+\theta)^4} = \dfrac{100 - 50\theta}{(2+\theta)^3}$ | M1 | Reasonable attempt at quotient rule
$\dfrac{dA}{d\theta} = 0 \Rightarrow \theta = 2$; $A = 6.25$ [cm²] | A1 | For information: $\dfrac{d^2A}{d\theta^2} = -\dfrac{200}{256} = -0.78125$ at $\theta = 2$
$\theta = 1.5 \Rightarrow \dfrac{dA}{d\theta} > 0$, $\theta = 2.5 \Rightarrow \dfrac{dA}{d\theta} < 0$ so max | B1 | —
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7 A wire, 10 cm long, is bent to form the perimeter of a sector of a circle, as shown in the diagram. The radius is $r \mathrm {~cm}$ and the angle at the centre is $\theta$ radians.\\
\includegraphics[max width=\textwidth, alt={}, center]{20639e13-01cc-4d96-b694-fb3cf1828f4d-07_323_204_342_242}
Determine the maximum possible area of the sector, showing that it is a maximum.
\hfill \mbox{\textit{OCR MEI Paper 3 2023 Q7 [6]}}