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Date	November 2020	Marks available	1	Reference code	20N.2.SL.TZ0.T_4
Level	Standard Level	Paper	Paper 2	Time zone	Time zone 0
Command term	Show that	Question number	T_4	Adapted from	N/A

Question

Hyungmin designs a concrete bird bath. The bird bath is supported by a pedestal. This is shown in the diagram.

The interior of the bird bath is in the shape of a cone with radius $r$ , height $h$ and a constant slant height of $50 cm$ .

Let $V$ be the volume of the bird bath.

Hyungmin wants the bird bath to have maximum volume.

Write down an equation in $r$ and $h$ that shows this information.

[1]

Show that $V = \frac{2500 π h}{3} - \frac{π h^{3}}{3}$ .

[1]

Find $\frac{d V}{d h}$ .

[2]

Using your answer to part (c), find the value of $h$ for which $V$ is a maximum.

[2]

Find the maximum volume of the bird bath.

[2]

To prevent leaks, a sealant is applied to the interior surface of the bird bath.

Find the surface area to be covered by the sealant, given that the bird bath has maximum volume.

[3]

Markscheme

* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.

$h^{2} + r^{2} = 50^{2}$ (or equivalent) (A1)

Note: Accept equivalent expressions such as $r = \sqrt{2500 - h^{2}}$ or $h = \sqrt{2500 - r^{2}}$ . Award (A0) for a final answer of $\pm \sqrt{2500 - h^{2}}$ or $\pm \sqrt{2500 - r^{2}}$ , or any further incorrect working.

[1 mark]

$\frac{1}{3} \times π \times (2500 - h^{2}) \times h$ OR $\frac{1}{3} \times π \times {(\sqrt{2500 - h^{2}})}^{2} \times h$ (M1)

Note: Award (M1) for correct substitution in the volume of cone formula.

$V = \frac{2500 π h}{3} - \frac{π h^{3}}{3}$ (AG)

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

[1 mark]

$(\frac{d V}{d h} =) \frac{2500 π}{3} - π h^{2}$ (A1)(A1)

Note: Award (A1) for $\frac{2500 π}{3}$ , (A1) for $- π h^{2}$ . Award at most (A1)(A0) if extra terms are seen. Award (A0) for the term $- \frac{3 π h^{2}}{3}$ .

[2 marks]

$0 = \frac{2500 π}{3} - π h^{2}$ (M1)

Note: Award (M1) for equating their derivative to zero. Follow through from part (c).

sketch of $\frac{d V}{d h}$ (M1)

Note: Award (M1) for a labelled sketch of $\frac{d V}{d h}$ with the curve/axes correctly labelled or the $x$ -intercept explicitly indicated.

$(h =) 28.9 (cm) (\sqrt{\frac{2500}{3}}, \frac{50}{\sqrt{3}}, \frac{50 \sqrt{3}}{3}, 28.8675 \dots)$ (A1)(ft)

Note: An unsupported $28.9 cm$ is awarded no marks. Graphing the function $V (h)$ is not an acceptable method and (M0)(A0) should be awarded. Follow through from part (c). Given the restraints of the question, $h \geq 50$ is not possible.

[2 marks]

$(V =) \frac{2500 \times π \times 28.8675 \dots}{3} - \frac{π {(28.8675 \dots)}^{3}}{3}$ (M1)

$\frac{1}{3} π {(40.828 \dots)}^{2} \times 28.8675 \dots$ (M1)

Note: Award (M1) for substituting their $28.8675 \dots$ in the volume formula.

$(V =) 50400 ({cm}^{3}) (50383.3 \dots)$ (A1)(ft)(G2)

Note: Follow through from part (d).

[2 marks]

$(S =) π \times \sqrt{2500 - {(28.8675 \dots)}^{2}} \times 50$ (A1)(ft)(M1)

Note: Award (A1) for their correct radius seen $(40.8248 \dots, \sqrt{2500 - {(28.8675 \dots)}^{2}})$ .
Award (M1) for correctly substituted curved surface area formula for a cone.

$(S =) 6410 ({cm}^{2}) (6412.74 \dots)$ (A1)(ft)(G2)

Note: Follow through from parts (a) and (d).

[3 marks]

Examiners report

[N/A]

Syllabus sections

Topic 5—Calculus » SL 5.7—Optimisation

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Find the surface area to be covered by the sealant, given that the bird bath has maximum volume.
EXN.2.SL.TZ0.2e:
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Find $\frac{d L}{d x}$
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Solve $\frac{d L}{d x} = 0$
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Interpret your answer to (b)(i) in context.

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