AQA Paper 1 Specimen — Question 12 8 marks

Exam BoardAQA
ModulePaper 1 (Paper 1)
SessionSpecimen
Marks8
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicImplicit equations and differentiation
TypeFind stationary points
DifficultyChallenging +1.8 This is a constrained optimization problem requiring implicit differentiation or Lagrange multipliers to find the maximum y-value on the ellipse x² + 2xy + 2y² = 10. It demands multi-step reasoning (finding dy/dx = 0, solving the resulting system) and goes beyond routine calculus exercises, though the algebraic manipulation is manageable for strong A-level students.
Spec1.07s Parametric and implicit differentiation

A sculpture formed from a prism is fixed on a horizontal platform, as shown in the diagram. The shape of the cross-section of the sculpture can be modelled by the equation \(x^2 + 2xy + 2y^2 = 10\), where \(x\) and \(y\) are measured in metres. The \(x\) and \(y\) axes are horizontal and vertical respectively. \includegraphics{figure_12} Find the maximum vertical height above the platform of the sculpture. [8 marks]

Question 12:
AnswerMarks Guidance
12Finds the difference between the
maximum and minimum values of yAO3.1b M1
dy dy
2x2y2x 4y 0
dx dx
Highest and lowest points
dy
occur when 0
dx
dy
0 xy
dx
y2 2y2 2y2 10
y  10
 
∴Height  10   10
= 2 10 = 6.32 m
AnswerMarks Guidance
Uses implicit differentiationAO1.1a M1
Differentiates correctlyAO1.1b A1
States stationary points occur when
dy
0
AnswerMarks Guidance
dxAO2.4 R1
dy
Uses 0 to find x in terms of y (or
dx
AnswerMarks Guidance
vice versa)AO1.1a M1
Finds x = –yAO1.1b A1
Deduces maximum and minimum
values of y
FT ‘their’ expression provided all M1
AnswerMarks Guidance
marks have been awardedAO2.2a A1F
States the height of the sculpture
above the platform
FT ‘their’ max and min values for y
provided all M1 marks have been
AnswerMarks Guidance
awardedAO2.2a A1F
Total8
QMarking Instructions AO
Question 12:
12 | Finds the difference between the
maximum and minimum values of y | AO3.1b | M1 | x22xy2y2 10
dy dy
2x2y2x 4y 0
dx dx
Highest and lowest points
dy
occur when 0
dx
dy
0 xy
dx
y2 2y2 2y2 10
y  10
 
∴Height  10   10
= 2 10 = 6.32 m
Uses implicit differentiation | AO1.1a | M1
Differentiates correctly | AO1.1b | A1
States stationary points occur when
dy
0
dx | AO2.4 | R1
dy
Uses 0 to find x in terms of y (or
dx
vice versa) | AO1.1a | M1
Finds x = –y | AO1.1b | A1
Deduces maximum and minimum
values of y
FT ‘their’ expression provided all M1
marks have been awarded | AO2.2a | A1F
States the height of the sculpture
above the platform
FT ‘their’ max and min values for y
provided all M1 marks have been
awarded | AO2.2a | A1F
Total | 8
Q | Marking Instructions | AO | Marks | Typical Solution
A sculpture formed from a prism is fixed on a horizontal platform, as shown in the diagram.

The shape of the cross-section of the sculpture can be modelled by the equation

$x^2 + 2xy + 2y^2 = 10$, where $x$ and $y$ are measured in metres.

The $x$ and $y$ axes are horizontal and vertical respectively.

\includegraphics{figure_12}

Find the maximum vertical height above the platform of the sculpture.
[8 marks]

\hfill \mbox{\textit{AQA Paper 1  Q12 [8]}}