EXN.2.SL.TZ0.2

pestleMathematicsAISLPaper 2EXN· sl-5-3-differentiating-polynomials-n-e-zsource ↗

A box of chocolates is to have a ribbon tied around it as shown in the diagram below.

The box is in the shape of a cuboid with a height of $3$ cm. The length and width of the box are $x$ and $y$ cm.

After going around the box an extra $10$ cm of ribbon is needed to form the bow.

The volume of the box is $450 {cm}^{3}$ .

Find an expression for the total length of the ribbon $L$ in terms of $x$ and $y$ .

[2]

Show that $L = 2 x + \frac{300}{x} + 22$

[3]

Find $\frac{d L}{d x}$

[3]

Solve $\frac{d L}{d x} = 0$

[2]

Hence or otherwise find the minimum length of ribbon required.

[2]

Markscheme / solution

* This sample question was produced by experienced DP mathematics senior examiners to aid teachers in preparing for external assessment in the new MAA course. There may be minor differences in formatting compared to formal exam papers.

$L = 2 x + 2 y + 12 + 10 = 2 x + 2 y + 22$ A1A1

Note: A1 for $2 x + 2 y$ and A1 for $12 + 10$ or $22$ .

[2 marks]

$V = 3 x y = 450$ A1

$y = \frac{150}{x}$ A1

$L = 2 x + 2 (\frac{150}{x}) + 22$ M1

$L = 2 x + \frac{300}{x} + 22$ AG

[3 marks]

$L = 2 x + 300 x^{- 1} + 22$ (M1)

$\frac{d L}{d x} = 2 - \frac{300}{x^{2}}$ A1A1

Note: A1 for $2$ (and $0$ ), A1 for $\frac{300}{x^{2}}$ .

[3 marks]

$\frac{300}{x^{2}} = 2$ (M1)

$x = \sqrt{150} = 12.2 (12.2474 \dots)$ A1

[2 marks]

$L = 2 \sqrt{150} + \frac{300}{\sqrt{150}} + 22 = 71.0 (70.9897 \dots) cm$ (M1)A1

[2 marks]

Examiners’ report

[N/A]