| Exam Board | Edexcel |
|---|---|
| Module | P2 (Pure Mathematics 2) |
| Year | 2019 |
| Session | June |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Stationary points and optimisation |
| Type | Optimise cost or profit model |
| Difficulty | Moderate -0.3 This is a straightforward stationary point problem requiring differentiation of a simple power function, solving dP/dx = 0, and using the second derivative test. The algebraic manipulation is minimal (solving 12 - (3/2)x^(1/2) = 0 gives x = 64 directly), and both parts follow standard textbook procedures with no conceptual challenges beyond basic calculus technique. |
| Spec | 1.07n 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 |
|---|---|---|
| (a) \(\frac{dP}{dx} = 12 - \frac{3}{2}x^{\frac{1}{2}}\) | M1A1 | M1: Attempts to differentiate \(x^n \to x^{n-1}\) seen at least once. It must be an \(x\) term and not the \(120 \to 0\). A1: \(\frac{dP}{dx} = 12 - \frac{3}{2}x^{\frac{1}{2}}\) with no need to see the lhs. Condone \(\frac{dy}{dx}\) all of the way through part (a). |
| Sets \(\frac{dP}{dx} = 0 \to 12 - \frac{3}{2}x^{\frac{1}{2}} = 0 \to x^n = ...\) | dM1 | Sets their \(\frac{dP}{dx} = 0\) and proceeds to \(x^n = k, k > 0\). Dependent upon the previous M. Don't be too concerned with the mechanics of process. Condone an attempted solution of \(\frac{dP}{dx} = ... 0\) where ... could be an inequality |
| \(x = 64\) | A1 | Condone \(x = \pm 64\) here |
| When \(x = 64 \Rightarrow P = 12 \times 64 - 64^{\frac{3}{2}} - 120 = ...\) | M1 | Substitutes their solution for \(\frac{dP}{dx} = 0\) into \(P\) and attempts to find the value of \(P\). |
| Profit \(= (£) 136\,000\) | A1 | CSO. Profit \(= (£) 136\,000\) or 136 thousand but not 136 or \(P = 136\). This cannot follow two values for \(x\), eg \(x = \pm 64\) Condone a lack of units or incorrect units such as \$ |
| (6 marks) | ||
| (b) \(\left(\frac{d^2P}{dx^2}\right) = -\frac{3}{4}x^{-\frac{1}{2}}\) and substitutes in their \(x = 64\) to find its value or state its sign | M1 | Achieves \(\frac{d^2P}{dx^2} = kx^{-\frac{1}{2}}\) and attempts to find its value at \(x = "\)64\("\). Alternatively achieves \(\frac{d^2P}{dx^2} = kx^{-\frac{1}{2}}\) and attempts to state its sign. Eg \(\frac{d^2P}{dx^2} = -\frac{3}{4}x^{-\frac{1}{2}} < 0\) |
| At \(x = 64\): \(\frac{d^2P}{dx^2} = -0.09375 < 0 \Rightarrow\) maximum | A1 | Achieves \(x = 64\), \(\frac{d^2P}{dx^2} = -\frac{3}{4}x^{-\frac{1}{2}}\) and states \(\frac{d^2P}{dx^2} = -\frac{32} < 0\) (at \(x = 64\)) then the profit is maximised. |
| (2 marks, 8 marks total) |
**(a)** $\frac{dP}{dx} = 12 - \frac{3}{2}x^{\frac{1}{2}}$ | M1A1 | M1: Attempts to differentiate $x^n \to x^{n-1}$ seen at least once. It must be an $x$ term and not the $120 \to 0$. A1: $\frac{dP}{dx} = 12 - \frac{3}{2}x^{\frac{1}{2}}$ with no need to see the lhs. Condone $\frac{dy}{dx}$ all of the way through part (a).
Sets $\frac{dP}{dx} = 0 \to 12 - \frac{3}{2}x^{\frac{1}{2}} = 0 \to x^n = ...$ | dM1 | Sets their $\frac{dP}{dx} = 0$ and proceeds to $x^n = k, k > 0$. Dependent upon the previous M. Don't be too concerned with the mechanics of process. Condone an attempted solution of $\frac{dP}{dx} = ... 0$ where ... could be an inequality
$x = 64$ | A1 | Condone $x = \pm 64$ here
When $x = 64 \Rightarrow P = 12 \times 64 - 64^{\frac{3}{2}} - 120 = ...$ | M1 | Substitutes their solution for $\frac{dP}{dx} = 0$ into $P$ and attempts to find the value of $P$.
Profit $= (£) 136\,000$ | A1 | CSO. Profit $= (£) 136\,000$ or 136 thousand but not 136 or $P = 136$. This cannot follow two values for $x$, eg $x = \pm 64$ Condone a lack of units or incorrect units such as \$
| (6 marks)
**(b)** $\left(\frac{d^2P}{dx^2}\right) = -\frac{3}{4}x^{-\frac{1}{2}}$ and substitutes in their $x = 64$ to find its value or state its sign | M1 | Achieves $\frac{d^2P}{dx^2} = kx^{-\frac{1}{2}}$ and attempts to find its value at $x = "$64$"$. Alternatively achieves $\frac{d^2P}{dx^2} = kx^{-\frac{1}{2}}$ and attempts to state its sign. Eg $\frac{d^2P}{dx^2} = -\frac{3}{4}x^{-\frac{1}{2}} < 0$
At $x = 64$: $\frac{d^2P}{dx^2} = -0.09375 < 0 \Rightarrow$ maximum | A1 | Achieves $x = 64$, $\frac{d^2P}{dx^2} = -\frac{3}{4}x^{-\frac{1}{2}}$ and states $\frac{d^2P}{dx^2} = -\frac{32} < 0$ (at $x = 64$) then the profit is maximised.
| (2 marks, 8 marks total)
**Additional notes:**
This requires the correct value of $x$, the correct value of the second derivative (allowing for awrt -0.09) a reason + conclusion.
Alt: Achieves $x = 64$, $\frac{d^2P}{dx^2} = -\frac{3}{4}x^{-\frac{1}{2}}$ and states as $x > 0$ or $\sqrt{x} > 0$ means that $\frac{d^2P}{dx^2} < 0$ then the profit is maximised.
Part (b) merely requires the use of calculus so allow
M1: Attempting to find the value of $\frac{dP}{dx}$ at two values either side, but close to their 64. Eg. For 64, allow the lower value to be $63.5 \leq x < 64$ and the upper value to be $64 < x \leq 64.5$
A1: Requires correct values, correct calculations with reason and conclusion
---5. A company makes a particular type of watch.
The annual profit made by the company from sales of these watches is modelled by the equation
$$P = 12 x - x ^ { \frac { 3 } { 2 } } - 120$$
where $P$ is the annual profit measured in thousands of pounds and $\pounds x$ is the selling price of the watch.
According to this model,
\begin{enumerate}[label=(\alph*)]
\item find, using calculus, the maximum possible annual profit.
\item Justify, also using calculus, that the profit you have found is a maximum.
\end{enumerate}
\hfill \mbox{\textit{Edexcel P2 2019 Q5 [8]}}