| Exam Board | OCR |
|---|---|
| Module | Further Additional Pure AS (Further Additional Pure AS) |
| Year | 2020 |
| Session | November |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Stationary points and optimisation |
| Type | Stationary points of surface (multivariable) |
| Difficulty | Challenging +1.2 This is a standard constrained optimization problem using partial differentiation, which is a Further Maths topic making it inherently harder than typical A-level. However, the problem follows a routine template: derive the constraint equation, find partial derivatives, solve simultaneous equations, and evaluate. The algebra is straightforward with symmetric variables leading to x=y, and no second derivative test is required. While above average due to the Further Maths content, it's a textbook exercise without novel insight. |
| Spec | 8.05a 3D surfaces: z = f(x,y) and implicit form, partial derivatives8.05d Partial differentiation: first and second order, mixed derivatives8.05e Stationary points: where partial derivatives are zero |
| Answer | Marks | Guidance |
|---|---|---|
| 2 | (a) | 1000 |
| Answer | Marks |
|---|---|
| x y | B1 |
| Answer | Marks |
|---|---|
| [4] | 3.1b |
| Answer | Marks |
|---|---|
| 1.1 | soi |
| Answer | Marks | Guidance |
|---|---|---|
| (b) | (i) | ∂A −1 ∂A −1 |
| Answer | Marks |
|---|---|
| x = y = 10 × 23 | M1 A1 |
| Answer | Marks |
|---|---|
| [5] | 1.1 1.1 |
| Answer | Marks |
|---|---|
| 1.1 | Partially differentiating A w.r.t. x or y; either correct |
| Answer | Marks | Guidance |
|---|---|---|
| (ii) | 2 | |
| Substg. x, y back into formula for A; 300 × 23 | M1 A1 | |
| [2] | 1.1 1.1 | 5 8 |
| Answer | Marks | Guidance |
|---|---|---|
| 2 | 4 | 8 |
Question 2:
2 | (a) | 1000
xyh = 1000 ⇒ h =
xy
A = xy + 2xh + 2yh
1 1
= xy + 2000 +
x y | B1
B1
M1
A1
[4] | 3.1b
1.1
2.1
1.1 | soi
Substitution of h expression from (a) (i)
AG shown with supporting working
(b) | (i) | ∂A −1 ∂A −1
= y + 2000 and = x + 2000
∂x x2 ∂y y2
Both p.d.s set to zero and solving
1
x = y = 10 × 23 | M1 A1
B1
M1
A1
[5] | 1.1 1.1
1.1
2.1
1.1 | Partially differentiating A w.r.t. x or y; either correct
2nd correct: FT 1st, with x ↔ y
x2y = xy2 = 2000
Both correct
(ii) | 2
Substg. x, y back into formula for A; 300 × 23 | M1 A1
[2] | 1.1 1.1 | 5 8
Any exact equivalent e.g. 150 × 23, 75 × 23 or awrt 476 BC
2 | 4 | 8 | 12 | 2 | 6 | 102 An open-topped rectangular box is to be manufactured with a fixed volume of $1000 \mathrm {~cm} ^ { 3 }$. The dimensions of the base of the box are $x \mathrm {~cm}$ by $y \mathrm {~cm}$. The surface area of the box is $A \mathrm {~cm} ^ { 2 }$.
\begin{enumerate}[label=(\alph*)]
\item Show that $\mathrm { A } = \mathrm { xy } + 2000 \left( \frac { 1 } { \mathrm { x } } + \frac { 1 } { \mathrm { y } } \right)$.
\item \begin{enumerate}[label=(\roman*)]
\item Use partial differentiation to determine, in exact form, the values of $x$ and $y$ for which $A$ has a stationary value.
\item Find the stationary value of $A$.
\end{enumerate}\end{enumerate}
\hfill \mbox{\textit{OCR Further Additional Pure AS 2020 Q2 [11]}}