AQA AS Paper 2 2023 June — Question 10 11 marks

Exam BoardAQA
ModuleAS Paper 2 (AS Paper 2)
Year2023
SessionJune
Marks11
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicTangents, normals and gradients
TypeOptimization with constraints
DifficultyStandard +0.3 This is a standard AS-level optimization problem with straightforward calculus. Part (a) involves basic area formulas and algebraic manipulation using a perimeter constraint. Part (b) requires differentiation and solving a linear equation to find the maximum, with verification using the second derivative test. While it has multiple steps and requires careful algebra, all techniques are routine AS-level content with no novel insights required.
Spec1.05c Area of triangle: using 1/2 ab sin(C)1.07n Stationary points: find maxima, minima using derivatives1.07r Chain rule: dy/dx = dy/du * du/dx and connected rates
A piece of wire of length 66 cm is bent to form the five sides of a pentagon. The pentagon consists of three sides of a rectangle and two sides of an equilateral triangle. The sides of the rectangle measure \(x\) cm and \(y\) cm and the sides of the triangle measure \(x\) cm, as shown in the diagram below. \includegraphics{figure_10}
    1. You are given that \(\sin 60° = \frac{\sqrt{3}}{2}\) Explain why the area of the triangle is \(\frac{\sqrt{3}}{4}x^2\) [1 mark]
    2. Show that the area enclosed by the wire, \(A\) cm\(^2\), can be expressed by the formula $$A = 33x - \frac{1}{4}(6 - \sqrt{3})x^2$$ [3 marks]
  2. Use calculus to find the value of \(x\) for which the wire encloses the maximum area. Give your answer in the form \(p + q\sqrt{3}\), where \(p\) and \(q\) are integers. Fully justify your answer. [7 marks]
Question 10:
AnswerMarks
10(a)(i)1
Explains the result using ab
2
sinC or other trigonometry
AnswerMarks Guidance
AG2.4 E1
Area of a triangle is ab sinC
2
Here a and b are both x and C =
60º
1 3 1
A = x2 = √3x2
2 2 4
AnswerMarks Guidance
Subtotal1
QMarking instructions AO
AnswerMarks Guidance
10(a)(ii)Obtains 3x + 2y = 66
OE3.1a B1
3
y = 33 – x
2
3
xy = (33 – x)x
2
3 1
A = (33 – x)x + √3x2
2 4
1
A = 33x – (6 – √3) x2
4
Uses their expression for y to
give an expression for xy in
terms of x
AnswerMarks Guidance
ACF1.1a M1
Completes correct derivation of
given formula
AnswerMarks Guidance
A G2.1 R1
Subtotal3
QMarking instructions AO
AnswerMarks Guidance
10(b)Differentiates, at least one term
correct1.1a M1
= 33 – (6 – √3)x
dx 2
dA
At a stationary point = 0
dx
1
33 – (6 – √3)x = 0
2
(6 – √3)x = 66
66
x =
6− 3
x = 12 + 2√3
d2A
1
= – (6 – √3)
dx2 2
Negative so maximum
AnswerMarks Guidance
O btains correct derivative1.1b A1
States that for a stationary (OE)
point
dA
= 0
dx
dA dy
Must state = 0 not = 0
dx dx
dy
but allow use of elsewhere
AnswerMarks Guidance
dx2.4 E1
Solves their linear equation = 0
to obtain x
AnswerMarks Guidance
PI ACF1.1a M1
Obtains correct exact value of x
AnswerMarks Guidance
ACF1.1b A1
dA
Differentiates their to obtain
dx
an expression independent of x
Or
Tests x values either side of
dA
12 + 2√3 in
AnswerMarks Guidance
dx1.1a M1
Completes reasoned argument
to deduce the maximum.
Must state that x = 12 + 2√3
CSO
Can be obtained independently
AnswerMarks Guidance
of E12.2a R1
Subtotal7
Question 10 Total11
QMarking instructions AO 
Question 10:
--- 10(a)(i) ---
10(a)(i) | 1
Explains the result using ab
2
sinC or other trigonometry
AG | 2.4 | E1 | 1
Area of a triangle is ab sinC
2
Here a and b are both x and C =
60º
1 3 1
A = x2 = √3x2
2 2 4
Subtotal | 1
Q | Marking instructions | AO | Marks | Typical solution
--- 10(a)(ii) ---
10(a)(ii) | Obtains 3x + 2y = 66
OE | 3.1a | B1 | 3x + 2y = 66
3
y = 33 – x
2
3
xy = (33 – x)x
2
3 1
A = (33 – x)x + √3x2
2 4
1
A = 33x – (6 – √3) x2
4
Uses their expression for y to
give an expression for xy in
terms of x
ACF | 1.1a | M1
Completes correct derivation of
given formula
A G | 2.1 | R1
Subtotal | 3
Q | Marking instructions | AO | Marks | Typical solution
--- 10(b) ---
10(b) | Differentiates, at least one term
correct | 1.1a | M1 | dA 1
= 33 – (6 – √3)x
dx 2
dA
At a stationary point = 0
dx
1
33 – (6 – √3)x = 0
2
(6 – √3)x = 66
66
x =
6− 3
x = 12 + 2√3
d2A
1
= – (6 – √3)
dx2 2
Negative so maximum
O btains correct derivative | 1.1b | A1
States that for a stationary (OE)
point
dA
= 0
dx
dA dy
Must state = 0 not = 0
dx dx
dy
but allow use of elsewhere
dx | 2.4 | E1
Solves their linear equation = 0
to obtain x
PI ACF | 1.1a | M1
Obtains correct exact value of x
ACF | 1.1b | A1
dA
Differentiates their to obtain
dx
an expression independent of x
Or
Tests x values either side of
dA
12 + 2√3 in
dx | 1.1a | M1
Completes reasoned argument
to deduce the maximum.
Must state that x = 12 + 2√3
CSO
Can be obtained independently
of E1 | 2.2a | R1
Subtotal | 7
Question 10 Total | 11
Q | Marking instructions | AO | Marks | Typical solution
A piece of wire of length 66 cm is bent to form the five sides of a pentagon.

The pentagon consists of three sides of a rectangle and two sides of an equilateral triangle.

The sides of the rectangle measure $x$ cm and $y$ cm and the sides of the triangle measure $x$ cm, as shown in the diagram below.

\includegraphics{figure_10}

\begin{enumerate}[label=(\alph*)]
\item
\begin{enumerate}[label=(\roman*)]
\item You are given that $\sin 60° = \frac{\sqrt{3}}{2}$

Explain why the area of the triangle is $\frac{\sqrt{3}}{4}x^2$ [1 mark]

\item Show that the area enclosed by the wire, $A$ cm$^2$, can be expressed by the formula
$$A = 33x - \frac{1}{4}(6 - \sqrt{3})x^2$$
[3 marks]
\end{enumerate}

\item Use calculus to find the value of $x$ for which the wire encloses the maximum area.

Give your answer in the form $p + q\sqrt{3}$, where $p$ and $q$ are integers.

Fully justify your answer.
[7 marks]
\end{enumerate}

\hfill \mbox{\textit{AQA AS Paper 2 2023 Q10 [11]}}
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