| Exam Board | Edexcel |
|---|---|
| Module | AS Paper 1 (AS Paper 1) |
| Session | Specimen |
| Marks | 10 |
| 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 AS-level optimization problem combining area/perimeter formulas with calculus. Part (a) requires algebraic manipulation of geometric formulas (routine), part (b) tests understanding of domain constraints (straightforward), and part (c) applies standard differentiation to find a minimum. The multi-step nature and 10 total marks make it slightly above average difficulty, but all techniques are standard AS curriculum with no novel insights required. |
| Spec | 1.02z Models in context: use functions in modelling1.07i Differentiate x^n: for rational n and sums1.07n Stationary points: find maxima, minima using derivatives1.07o Increasing/decreasing: functions using sign of dy/dx |
| Answer | Marks |
|---|---|
| 16(a) | (cid:83)x2 |
| Answer | Marks | Guidance |
|---|---|---|
| 2 | B1 | 2.1 |
| Answer | Marks | Guidance |
|---|---|---|
| 2x | M1 | 1.1b |
| Use P(cid:32)2x(cid:14)2y(cid:14)(cid:83)xwith their y substituted | M1 | 2.1 |
| Answer | Marks | Guidance |
|---|---|---|
| x 2x x 2 | A1* | 1.1b |
| Answer | Marks |
|---|---|
| (b) | (cid:83)x2 |
| Answer | Marks | Guidance |
|---|---|---|
| 2x 2 | M1 | 2.4 |
| As x and y are distances they are positive so 0(cid:31) x(cid:31) | 500 |
| Answer | Marks | Guidance |
|---|---|---|
| (cid:83) | A1* | 3.2a |
| Answer | Marks |
|---|---|
| (c) | dP |
| Answer | Marks | Guidance |
|---|---|---|
| dx | M1 | 3.4 |
| Answer | Marks | Guidance |
|---|---|---|
| dx x2 2 | A1 | 1.1b |
| Answer | Marks | Guidance |
|---|---|---|
| dx | M1 | 1.1b |
| Answer | Marks | Guidance |
|---|---|---|
| perimeter = 59.8 M | A1 | 1.1b |
| Answer | Marks | Guidance |
|---|---|---|
| Question | Scheme | Marks |
Question 16:
--- 16(a) ---
16(a) | (cid:83)x2
Sets 2xy(cid:14) (cid:32)250
2 | B1 | 2.1
(cid:83)x2
250(cid:16)
2
Obtain y (cid:32) and substitute into P
2x | M1 | 1.1b
Use P(cid:32)2x(cid:14)2y(cid:14)(cid:83)xwith their y substituted | M1 | 2.1
250 (cid:83)x2 250 (cid:83)x
P(cid:32)2x(cid:14) (cid:16) (cid:14)(cid:83)x (cid:32)2x(cid:14) (cid:14) *
x 2x x 2 | A1* | 1.1b
(4)
(b) | (cid:83)x2
250(cid:16)
(cid:83)x2
2
x > 0 and y > 0 (distance) (cid:159) (cid:33)0 or 250(cid:16) (cid:33)0 o.e.
2x 2 | M1 | 2.4
As x and y are distances they are positive so 0(cid:31) x(cid:31) | 500
*
(cid:83) | A1* | 3.2a
(2)
(c) | dP
Differentiates P with negative index correct in ;x(cid:16)1 (cid:111) x(cid:16)2
dx | M1 | 3.4
dP 250 (cid:83)
(cid:32)2(cid:16) (cid:14)
dx x2 2 | A1 | 1.1b
dP
Sets (cid:32)0 and proceeds to x =
dx | M1 | 1.1b
250 (cid:83)x
Substitutes their x into P(cid:32)2x(cid:14) (cid:14) to give
x 2
perimeter = 59.8 M | A1 | 1.1b
(4)
(10 marks)
Question 16 continued
Notes:
(a)
B1: Correct area equation
M1: Rearranges their area equation to make y the subject of the formula and attempt to use with
an expression for P
M1: Use correct equation for perimeter with their y substituted
A1*: Completely correct solution to obtain and state printed answer
(b)
M1: States x > 0 and y > 0 and uses their expression from (a) to form inequality
A1*: Explains that x and y are positive because they are distances, and uses correct expression
for y to give the printed answer correctly
(c)
M1: Attempt to differentiate P (deals with negative power of x correctly)
A1: Correct differentiation
M1: Sets derived function equal to zero and obtains x =
(cid:167) 500 (cid:183)
A1: The value of x may not be seen (it is 8.37 to 3sf or (cid:168) (cid:184))
(cid:169)4(cid:14)(cid:83)(cid:185)
Need to see awrt 59.8 M with units included for the perimeter
Question | Scheme | Marks | AOs\includegraphics{figure_4}
Figure 4 shows the plan view of the design for a swimming pool.
The shape of this pool $ABCDEA$ consists of a rectangular section $ABDE$ joined to a semicircular section $BCD$ as shown in Figure 4.
Given that $AE = 2x$ metres, $ED = y$ metres and the area of the pool is $250\text{m}^2$,
\begin{enumerate}[label=(\alph*)]
\item show that the perimeter, $P$ metres, of the pool is given by
$$P = 2x + \frac{250}{x} + \frac{\pi x}{2}$$
[4]
\item Explain why $0 < x < \sqrt{\frac{500}{\pi}}$
[2]
\item Find the minimum perimeter of the pool, giving your answer to $3$ significant figures.
[4]
\end{enumerate}
\hfill \mbox{\textit{Edexcel AS Paper 1 Q16 [10]}}