User interface language: English | Español

Date May 2019 Marks available 4 Reference code 19M.1.AHL.TZ2.H_8
Level Additional Higher Level Paper Paper 1 Time zone Time zone 2
Command term Show that Question number H_8 Adapted from N/A

Question

A right circular cone of radius r is inscribed in a sphere with centre O and radius R as shown in the following diagram. The perpendicular height of the cone is h , X denotes the centre of its base and B a point where the cone touches the sphere.

Show that the volume of the cone may be expressed by  V = π 3 ( 2 R h 2 h 3 ) .

[4]
a.

Given that there is one inscribed cone having a maximum volume, show that the volume of this cone is 32 π R 3 81 .

[4]
b.

Markscheme

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

attempt to use Pythagoras in triangle OXB       M1

r 2 = R 2 ( h R ) 2      A1

substitution of their r 2 into formula for volume of cone  V = π r 2 h 3        M1

= π h 3 ( R 2 ( h R ) 2 )

= π h 3 ( R 2 ( h 2 + R 2 2 h R ) )        A1

Note: This A mark is independent and may be seen anywhere for the correct expansion of  ( h R ) 2 .

= π h 3 ( 2 h R h 2 )

= π 3 ( 2 R h 2 h 3 )        AG

[4 marks]

a.

at max,  d V d h = 0        R1

d V d h = π 3 ( 4 R h 3 h 2 )

4 R h = 3 h 2

h = 4 R 3 (since  h 0 )     A1

EITHER

V max = π 3 ( 2 R h 2 h 3 )  from part (a)

= π 3 ( 2 R ( 4 R 3 ) 2 ( 4 R 3 ) 3 )      A1

= π 3 ( 2 R 16 R 2 9 ( 64 R 3 27 ) )      A1

OR

r 2 = R 2 ( 4 R 3 R ) 2

r 2 = R 2 R 2 9 = 8 R 2 9      A1

V max = π r 2 3 ( 4 R 3 )

= 4 π R 9 ( 8 R 2 9 )      A1

THEN

= 32 π R 3 81        AG

[4 marks]

b.

Examiners report

[N/A]
a.
[N/A]
b.

Syllabus sections

Topic 5—Calculus » SL 5.6—Stationary points, local max and min
Show 140 related questions
Topic 5—Calculus » SL 5.7—Optimisation
Topic 5—Calculus

View options