| Exam Board | CAIE |
|---|---|
| Module | P1 (Pure Mathematics 1) |
| Year | 2003 |
| Session | November |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Chain Rule |
| Type | Optimization with constraint |
| Difficulty | Standard +0.3 This is a standard optimization problem requiring volume constraint manipulation, surface area formula derivation, and basic differentiation. While it involves multiple steps, each step uses routine techniques (expressing one variable in terms of another, differentiating a sum of power functions, solving for stationary points, and using second derivative test). The algebraic manipulation is straightforward and this type of question appears frequently in P1/C1 textbooks, making it slightly easier than average. |
| Spec | 1.07n Stationary points: find maxima, minima using derivatives |
| Answer | Marks | Guidance |
|---|---|---|
| \(\rightarrow A = 4x^2 + 216-x\) | B1, M1, A1 [3] | Co from volume \(=\) lbh. Attempts most of the faces(4 or more). Co – answer was given. |
| Answer | Marks | Guidance |
|---|---|---|
| \(\rightarrow x = 3\) | M1, DM1, A1 [3] | Reasonable attempt at differentiation. Sets his differential to 0 and uses. Co. (\(answer = \pm 3\) loses last A mark). |
| Answer | Marks | Guidance |
|---|---|---|
| \(\rightarrow\) Positive when \(x=3\) Minimum. | A1, ∇, M1, A1 [3] | For putting his \(x\) into his A. Allow in (ii). Correct method – could be signs of \(\frac{dA}{dx}\) A mark needs \(\frac{d^2A}{dx^2}\) correct algebraically, \(+ x=3 +\) minimum. It does not need "24". |
**(i)** $y = 72-(2x^2)$ or $36-x^2$
$A = 4x^2+6xy$
$\rightarrow A = 4x^2 + 216-x$ | B1, M1, A1 [3] | Co from volume $=$ lbh. Attempts most of the faces(4 or more). Co – answer was given.
**(ii)** $\frac{dA}{dx} = 8x - 216 \cdot x^2$
$= 0$ when $8x^3=216$
$\rightarrow x = 3$ | M1, DM1, A1 [3] | Reasonable attempt at differentiation. Sets his differential to 0 and uses. Co. ($answer = \pm 3$ loses last A mark).
**(iii)** Stationary value $= 108$ cm$^2$
$\frac{d^2A}{dx^2}=8+432 \cdot x^3$
$\rightarrow$ Positive when $x=3$ Minimum. | A1, ∇, M1, A1 [3] | For putting his $x$ into his A. Allow in (ii). Correct method – could be signs of $\frac{dA}{dx}$ A mark needs $\frac{d^2A}{dx^2}$ correct algebraically, $+ x=3 +$ minimum. It does not need "24".8 A solid rectangular block has a base which measures $2 x \mathrm {~cm}$ by $x \mathrm {~cm}$. The height of the block is $y \mathrm {~cm}$ and the volume of the block is $72 \mathrm {~cm} ^ { 3 }$.\\
(i) Express $y$ in terms of $x$ and show that the total surface area, $A \mathrm {~cm} ^ { 2 }$, of the block is given by
$$A = 4 x ^ { 2 } + \frac { 216 } { x }$$
Given that $x$ can vary,\\
(ii) find the value of $x$ for which $A$ has a stationary value,\\
(iii) find this stationary value and determine whether it is a maximum or a minimum.
\hfill \mbox{\textit{CAIE P1 2003 Q8 [9]}}