| Exam Board | CAIE |
|---|---|
| Module | P1 (Pure Mathematics 1) |
| Year | 2006 |
| Session | November |
| Marks | 10 |
| 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 surface area constraint setup, volume formula derivation, and basic differentiation. The question guides students through each step explicitly (showing specific results), making it slightly easier than average. The chain rule application is straightforward, and the optimization is routine A-level fare with no novel insights required. |
| Spec | 1.02z Models in context: use functions in modelling1.07n Stationary points: find maxima, minima using derivatives1.07o Increasing/decreasing: functions using sign of dy/dx1.07p Points of inflection: using second derivative |
| Answer | Marks | Guidance |
|---|---|---|
| (i) Height \(= 3x\). \(10xy + \frac{1}{2} \cdot 8x \cdot 3x \cdot 2 = 200\) \(\rightarrow y = \frac{200 - 24x^2}{10x}\) | B1 M1 A1 [3] | Anywhere in the question. Linking 200 with 4/5 areas. Allow slight slip in formulae (particularly with \(\frac{1}{2}\)) co. Answer given. |
| (ii) \(V = \frac{1}{2} \cdot 8x \cdot 3xy = 240x - 28.8x^2\) | M1 A1 [2] | M1 for statement "\(V =\) area \(\times y\)". co ag. |
| (iii) \(\frac{dV}{dx} = 240 - 86.4x^2\) \(= 0\) when \(x = 1\frac{2}{3}\). | M1 DM1 A1 [3] | Attempt at differentiating. Sets to 0 and attempts to solve. co. |
| (iv) \(\frac{d^2V}{dx^2} = -172.8x\) \(\rightarrow -ve \rightarrow\) Maximum | M1 A1∨ [2] | Looks at sign of 2nd differential. Correct deduction from correct diff. Ignore inclusion of -5/3. |
(i) Height $= 3x$. $10xy + \frac{1}{2} \cdot 8x \cdot 3x \cdot 2 = 200$ $\rightarrow y = \frac{200 - 24x^2}{10x}$ | B1 M1 A1 [3] | Anywhere in the question. Linking 200 with 4/5 areas. Allow slight slip in formulae (particularly with $\frac{1}{2}$) co. Answer given.
(ii) $V = \frac{1}{2} \cdot 8x \cdot 3xy = 240x - 28.8x^2$ | M1 A1 [2] | M1 for statement "$V =$ area $\times y$". co ag.
(iii) $\frac{dV}{dx} = 240 - 86.4x^2$ $= 0$ when $x = 1\frac{2}{3}$. | M1 DM1 A1 [3] | Attempt at differentiating. Sets to 0 and attempts to solve. co.
(iv) $\frac{d^2V}{dx^2} = -172.8x$ $\rightarrow -ve \rightarrow$ Maximum | M1 A1∨ [2] | Looks at sign of 2nd differential. Correct deduction from correct diff. Ignore inclusion of -5/3.
---9\\
\includegraphics[max width=\textwidth, alt={}, center]{dd2cb0ec-5df9-4d99-9e15-5ae1f1c07b96-4_387_903_799_623}
The diagram shows an open container constructed out of $200 \mathrm {~cm} ^ { 2 }$ of cardboard. The two vertical end pieces are isosceles triangles with sides $5 x \mathrm {~cm} , 5 x \mathrm {~cm}$ and $8 x \mathrm {~cm}$, and the two side pieces are rectangles of length $y \mathrm {~cm}$ and width $5 x \mathrm {~cm}$, as shown. The open top is a horizontal rectangle.\\
(i) Show that $y = \frac { 200 - 24 x ^ { 2 } } { 10 x }$.\\
(ii) Show that the volume, $V \mathrm {~cm} ^ { 3 }$, of the container is given by $V = 240 x - 28.8 x ^ { 3 }$.
Given that $x$ can vary,\\
(iii) find the value of $x$ for which $V$ has a stationary value,\\
(iv) determine whether it is a maximum or a minimum stationary value.
\hfill \mbox{\textit{CAIE P1 2006 Q9 [10]}}