21M.1.SL.TZ1.12

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Ellis designs a gift box. The top of the gift box is in the shape of a right-angled triangle $GIK$ .

A rectangular section $HIJL$ is inscribed inside this triangle. The lengths of $GH, JK, HL$ , and $LJ$ are $p cm, q cm, 8 cm$ and $6 cm$ respectively.

The area of the top of the gift box is $A {cm}^{2}$ .

Ellis wishes to find the value of $q$ that will minimize the area of the top of the gift box.

Find $A$ in terms of $p$ and $q$ .

[1]

a.i.

Show that $A = \frac{192}{q} + 3 q + 48$ .

[3]

a.ii.

Find $\frac{d A}{d q}$ .

[2]

Write down an equation Ellis could solve to find this value of $q$ .

[1]

c.i.

Hence, or otherwise, find this value of $q$ .

[1]

c.ii.

Markscheme / solution

$A = \frac{1}{2} \times 6 \times q + \frac{1}{2} \times 8 \times p + 48$ OR $A = \frac{1}{2} (p + 6) (q + 8)$ OR $A = 3 q + 4 p + 48$ A1

[1 mark]

a.i.

valid attempt to link $p$ and $q$ , using tangents, similar triangles or other method (M1)

eg. $\tan θ = \frac{8}{p}$ and $\tan θ = \frac{q}{6}$ OR $\tan θ = \frac{p}{8}$ and $\tan θ = \frac{6}{q}$ OR $\frac{8}{p} = \frac{q}{6}$

correct equation linking $p$ and $q$ A1

eg. $p q = 48$ OR $p = \frac{48}{q}$ OR $q = \frac{48}{p}$

substitute $p = \frac{48}{q}$ into a correct area expression M1

eg. $(A =) \frac{1}{2} \times 6 \times q + \frac{1}{2} \times 8 \times \frac{48}{q} + 48$ OR $(A =) \frac{1}{2} (\frac{48}{q} + 6) (q + 8)$

$A = 3 q + \frac{192}{q} + 48$ AG

Note: The AG line must be seen with no incorrect, intermediate working, for the final M1 to be awarded.

[3 marks]

a.ii.

$\frac{- 192}{q^{2}} + 3$ A1A1

Note: Award A1 for $\frac{- 192}{q^{2}}$ , A1 for $3$ . Award A1A0 if extra terms are seen.

[2 marks]

$\frac{- 192}{q^{2}} + 3 = 0$ A1

[1 mark]

c.i.

$q = 8 cm$ A1

[1 mark]

c.ii.

Examiners’ report

[N/A]

a.i.

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a.ii.

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c.i.

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c.ii.