## Question 14:

### Part (a):

| Answer/Working | Mark | Guidance |
|---|---|---|
| Area of triangle $= \frac{1}{2}ab\sin C = \frac{1}{2} \times 2x \times 2x \times \sin 60 = \sqrt{3}x^2$ | M1 | Acceptable method for area of triangle. Using $\frac{1}{2}bh$ with Pythagoras fine. Accept $\frac{1}{2} \times 2x \times \sqrt{3}x$ with $\sqrt{3}x$ marked as perpendicular height |
| $S = 2 \times \sqrt{3}x^2 + 3 \times 2xl = 2x^2\sqrt{3} + 6xl$ | dM1 | For adding area of 3 rectangles and two triangles. Dependent on previous M mark |
| $S = 2\sqrt{3}x^2 + 6xl$ | A1* | Completing proof with no errors. Accept different order |

### Part (b):

| Answer/Working | Mark | Guidance |
|---|---|---|
| $960 = 2x^2\sqrt{3} + 6xl \Rightarrow l = \frac{960 - 2x^2\sqrt{3}}{6x}$ | M1 | Substitutes $S = 960$ and attempts to make $l$ or $lx$ subject |
| $l = \frac{960 - 2x^2\sqrt{3}}{6x}$ oe | A1 | $l = \frac{160}{x} - \frac{x\sqrt{3}}{3}$ or $lx = \frac{960 - 2x^2\sqrt{3}}{6}$ oe |
| $V = x^2\sqrt{3}\,l$ | B1 | States volume $V = x^2\sqrt{3}\,l$. May be implied by $V = x^2\sqrt{3} \times \textit{their}\,l$ |
| $V = x^2\sqrt{3} \times \left(\frac{960 - 2x^2\sqrt{3}}{6x}\right) = 160x\sqrt{3} - x^3$ | dM1 | Substitute $l$ into $V$ to get expression in $x$ only. Dependent on previous M |
| $V = 160x\sqrt{3} - x^3$ | A1* | Given answer. All aspects of proof must be correct |

### Part (c):

| Answer/Working | Mark | Guidance |
|---|---|---|
| $\frac{dV}{dx} = 160\sqrt{3} - 3x^2 = 0$ | M1 | Differentiating with at least one term correct and setting $= 0$ |
| $160\sqrt{3} - 3x^2 = 0$ | A1 | Accept awrt $277 - 3x^2 = 0$ |
| $x =$ awrt $9.6$ | A1 | Accept $x = \sqrt{\frac{160\sqrt{3}}{3}}$. On its own scores A0 |
| $V = 160 \times 9.611 \times \sqrt{3} - 9.611^3 = 1776$ | dM1 | Substitute their answer to $\frac{dV}{dx}=0$ into $V = 160x\sqrt{3} - x^3$. Dependent on previous M |
| $V =$ awrt $1776$ | A1 | |

### Part (d):

| Answer/Working | Mark | Guidance |
|---|---|---|
| $\frac{d^2V}{dx^2} = -6x < 0 \Rightarrow$ Maximum | M1 | Either find $\frac{d^2V}{dx^2} = -6x$ and substitute $x$ from (c), **or** state $\frac{d^2V}{dx^2} = -6x$ and state it is less than 0. Alternatives: find gradient either side, or find $V$ either side |
| $\frac{d^2V}{dx^2} = -6x < 0$, hence maximum | A1 | Must be calculated at $x$ from (c), with reason $\frac{d^2V}{dx^2} < 0$ and conclusion. Or $-6 \times \textit{positive} = \textit{negative}$ with reason and conclusion |

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