Optimization HL questions [65

marks]

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.

1a. Show that the volume of the cone may be expressed by [4 marks]
V = π3 (2Rh2 − h3 ).

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
⇒ r2 = R2 − (h − R)2 A1
πr2h
substitution of their r2 into formula for volume of cone V = 3
M1

= πh
3
(R2 − (h − R)2 )

= πh
3
(R2 − (h2 + R2 − 2hR)) A1
Note: This A mark is independent and may be seen anywhere for the
2
correct expansion of (h − R) .

= πh
3
(2hR − h2 )
= π
3
(2Rh2 − h3 ) AG
[4 marks]
1b. Given that there is one inscribed cone having a maximum volume, show [4 marks]
32πR 3
that the volume of this cone is 81 .

Markscheme
at max, dV
dh
=0 R1
dV = π
(4Rh − 3h2 )
dh 3
⇒ 4Rh = 3h2
4R
⇒h= 3
(since h ≠ 0) A1
EITHER
Vmax = π
3
(2Rh2 − h3 ) from part (a)

(2R( 3 4R 2
))
3
= π
3
) − ( 4R
3
A1

(2R 16R 2
− ( 64R ))
3
π
= 3 9 27
A1

OR
2
r2 = R2 − ( 4R
3
− R)
R2 8R 2
r2 = R2 − 9
= 9
A1
πr2 4R
⇒ Vmax = 3
( 3 )

= 9
( 9 )
4πR 8R 2
A1

THEN
32πR 3
= 81
AG
[4 marks]
 Points A , B and T lie on a line on an indoor soccer field. The goal, [AB] , is 2
metres wide. A player situated at point P kicks a ball at the goal. [PT] is
perpendicular to (AB) and is 6 metres from a parallel line through the centre of
[AB] . Let PT be x metros and let α = AP ^ B measured in degrees. Assume that the
ball travels along the floor.

2a. Find the value of α when x = 10. [4 marks]

Markscheme
* This question is from an exam for a previous syllabus, and may contain
minor differences in marking or structure.
EITHER
7 5
α = arctan 10
− arctan 10
(= 34.992 …∘ − 26.5651 …∘ ) (M1)(A1)(A1)

Note: Award (M1) for ^ T − BP

α = AP ^ T, (A1) for a correct AP
^ T and (A1)
^ T.
for a correct BP
OR
10
α = arctan 2 − arctan 7
(= 63.434 …∘ − 55.008 …∘ ) (M1)(A1)(A1)

Note: Award (M1) for α = PB ^ , (A1) for a correct PB

^ T − PAT ^ T and (A1)
^ .
for a correct PAT
OR

α = arccos( 125+149−4
) (M1)(A1)(A1)
2×√125×√149
Note: Award (M1) for use of cosine rule, (A1) for a correct numerator and
(A1) for a correct denominator.
THEN
= 8.43∘ A1
[4 marks]
2b. Show that tan α 2x [4 marks]
= x2+35
.

Markscheme
EITHER
7
−5
tan α = x x
M1A1A1
1+( 7x )( 5x )

Note: Award M1 for use of tan(A − B), A1 for a correct numerator and
A1 for a correct denominator.
2
= x
M1
1+ 352
x

OR
x− x
tan α = 5 7
M1A1A1
1+( x5 )( x7 )

Note: Award M1 for use of xxx, A1 for a correct numerator and A1 for a
correct denominator.
2x

= 35
2 M1
1+ x35

OR
x2+35
cos α = M1A1
√(x2+25)(x2+49)

Note: Award M1 for either use of the cosine rule or use of cos(A − B).
2x
sin α A1
√(x2+25)(x2+49)
2x
√(x 2+25)(x 2+49)
tan α = x 2+35
M1
√(x 2+25)(x 2+49)

THEN
2x
tan α = x2+35
AG

[4 marks]

tan
 The maximum for tan α gives the maximum for α.

2c. (i) Find dd (tan α). [11 marks]

x

(ii) Hence or otherwise find the value of α such that dd (tan α) = 0.

x
d2
(iii) Find
d x2
(tan α) and hence show that the value of α never exceeds 10°.

Markscheme
(= )
d (tan α) 2(x2+35)−(2x)(2x) 70−2x2
(i) dx
= M1A1A1
(x2+35)2 (x2+35)2

Note: Award M1 for attempting product or quotient rule differentiation, A1

for a correct numerator and A1 for a correct denominator.
(ii) METHOD 1
EITHER
d (tan α) = 0 ⇒ 70 − 2x2 = 0 (M1)
dx
x = √35 (m) (= 5.9161 … (m)) A1
1
tan α = (= 0.16903 …) (A1)
√35
OR
attempting to locate the stationary point on the graph of
2x
tan α = x2+35
(M1)

x = 5.9161 … (m) (= √35 (m)) A1

tan α = 0.16903 … (= 1
) (A1)
√35
THEN
α = 9.59∘ A1
METHOD 2
EITHER
x )⇒
α = arctan( x22+35 dα = 70−2x2
M1
dx (x2+35)2+4x2
dα = 0 ⇒ x = √35 (m) (= 5.9161 (m)) A1
dx
OR
attempting to locate the stationary point on the graph of

( )
 x )
α = arctan( x22+35 (M1)

x = 5.9161 … (m) (= √35 (m)) A1

THEN

α = 0.1674 … (= arctan 1
) (A1)
√35
= 9.59∘ A1
2

(= )
d2 (x2+25) (−4x)−(2)(2x)(x2+35)(70−2x2) 4x(x2−105)
(iii)
d x2
(tan α) =
(x2+35)4 (x2+35)3
M1A1
d2
substituting x = √35 (= 5.9161 …) into d x2
(tan α) M1
d2
d x2
(tan α) < 0 (= −0.004829 …) and so α = 9.59∘ is the maximum value of
α R1
α never exceeds 10° AG
[11 marks]

2d. Find the set of values of x for which α ⩾ 7∘ . [3 marks]

Markscheme
2x
attempting to solve
x2+35
⩾ tan 7∘ (M1)
2x
Note: Award (M1) for attempting to solve
x2+35
= tan 7∘ .
x = 2.55 and x = 13.7 (A1)
2.55 ⩽ x ⩽ 13.7 (m) A1
[3 marks]

In triangle ABC, BC = √3cm , AB ^ =

^ C = θ and BCA π
.
3

3a. Show that length AB 3 [4 marks]

= .
√3 cos θ +sin θ
 Markscheme
* This question is from an exam for a previous syllabus, and may contain
minor differences in marking or structure.
any attempt to use sine rule M1
AB √3
sin π3
= A1
sin( 23π −θ )

√3
= sin 2π
cos θ −cos 23π sin θ
A1
3

Note: Condone use of degrees.

√3
= √3 1
A1
2 cos θ + 2 sin θ

AB √3
√3
= √3 1
2 2 cos θ + 2 sin θ

3
∴ AB = AG
√3 cos θ +sin θ
[4 marks]

3b. Given that AB has a minimum value, determine the value of θ for which [4 marks]
this occurs.
 Markscheme
METHOD 1
−3(−√3 sin θ +cos θ )
(AB)′ = 2
M1A1
(√3 cos θ +sin θ )

setting (AB)′ =0 M1
1
tan θ =
√3
π
θ= 6
A1
METHOD 2
√3 sin π3
AB =
sin( 23π −θ )

AB minimum when sin( 23π − θ) is maximum M1

sin( 23π − θ) = 1 (A1)

2π π
3
−θ= 2 M1
π
θ= 6
A1
METHOD 3
shortest distance from B to AC is perpendicular to AC R1
π π π
θ= 2 − 3
= 6
M1A2
[4 marks]
Total [8 marks]

^R
Consider the triangle PQR where Q P = 30∘ , PQ = (x + 2)cm and
PR = (5 − x)2 cm , where −2 < x < 5.

cm2 , of the triangle is given by

4a. Show that the area, A [2 marks]
A = 14 (x3 − 8x2 + 5x + 50).
 Markscheme
* This question is from an exam for a previous syllabus, and may contain
minor differences in marking or structure.
use of A = 12 qr sin θ to obtain A = 12 (x + 2)(5 − x)2 sin 30∘ M1

= 14 (x + 2)(25 − 10x + x2 ) A1

A = 14 (x3 − 8x2 + 5x + 50) AG

[2 marks]

4b. (i) State ddA . [3 marks]

x

(ii) Verify that ddA

x
= 0 when x = 13 .
 Markscheme
(i) dA
dx
= 14 (3x2 − 16x + 5) = 14 (3x − 1)(x − 5) A1
(ii) METHOD 1
EITHER

= 14 (3( 13 ) − 16 ( 13 ) + 5) = 0
dA 2
dx
M1A1

OR
dA
dx
= 14 (3 ( 13 ) − 1) (( 13 ) − 5) = 0 M1A1
THEN
1
so ddA = 0 when x = 3
AG
x
METHOD 2
solving ddA = 0 for x M1
x
1
−2 < x < 5 ⇒ x = 3
A1
1
so ddA = 0 when x = 3
AG
x
METHOD 3
a correct graph of ddA versus x M1
x
1
the graph clearly showing that ddA = 0 when x = 3
A1
x
1
so ddA = 0 when x = 3
AG
x
[3 marks]

4c. (i) d 2A 1 [7 marks]

Find
d x2
and hence justify that x= 3
gives the maximum area of
triangle P QR.
(ii) State the maximum area of triangle P QR.
(iii) Find QR when the area of triangle P QR is a maximum.
Markscheme
d 2A
(i)
d x2
= 12 (3x − 8) A1
d 2A
for x = 13 , d x2
= −3.5 (< 0) R1
1
so x=
3
gives the maximum area of triangle P QR AG

(ii) Amax = 343

27
(= 12.7) (cm2 ) A1
2
(iii) PQ = 73 (cm) and PR = ( 14 3
) (cm) (A1)
2 4 14 2
QR2 = ( 73 ) + ( 14
3
) − 2 ( 7
3
) ( 3
) cos 30∘ (M1)(A1)
= 391.702 …
QR = 19.8 (cm) A1
[7 marks]
Total [12 marks]
5. Engineers need to lay pipes to connect two cities A and B that are [15 marks]
separated by a river of width 450 metres as shown in the following
diagram. They plan to lay the pipes under the river from A to X and then under the
ground from X to B. The cost of laying the pipes under the river is five times the
cost of laying the pipes under the ground.
Let EX = x.

Let k be the cost, in dollars per metre, of laying the pipes under the ground.
(a) Show that the total cost C, in dollars, of laying the pipes from A to B is given
by C = 5k√202 500 + x2 + (1000 − x)k.

(b) (i) Find dC

d
.
x
(ii) Hence find the value of x for which the total cost is a minimum,
justifying that this value is a minimum.
(c) Find the minimum total cost in terms of k.
^B
The angle at which the pipes are joined is AX = θ.
(d) θ for the value of x calculated in (b).
Find
For safety reasons θ must be at least 120°.
Given this new requirement,
(e) (i) find the new value of x which minimises the total cost;
(ii) find the percentage increase in the minimum total cost.

Markscheme
* This question is from an exam for a previous syllabus, and may contain
minor differences in marking or structure.
(a) C = AX × 5k + XB × k (M1)

Note: Award (M1) for attempting to express the cost in terms of AX, XB
and k.

= 5k√4502 + x2 + (1000 − x)k A1

= 5k√202 500 + x2 + (1000 − x)k AG
[2 marks]

(b) (i) dC
dx
= k[ 2 5×2x
− 1] = k ( 5x
− 1) M1A1
√ 202 500+x2 √ 202 500+x2

Note: Award M1 for an attempt to differentiate and A1 for the correct

derivative.

(ii) attempting to solve dC

dx
=0 M1
5
=1 (A1)
√ 202 500+x2

x = 91.9 (m) (=
75√6
2 (m)) A1

METHOD 1
for example,
at x = 91 dC
dx
= −0.00895k < 0 M1

at x = 92 dC
dx
= 0.001506k > 0 A1

Note: Award M1 for attempting to find the gradient either side of x = 91.9
and A1 for two correct values.

thus x = 91.9 gives a minimum AG

METHOD 2
d 2C 1 012 500k
d x2
= 3
(x2+202 500) 2
2
at x = 91.9 ddxC2 = 0.010451k > 0 (M1)A1

Note: Award M1 for attempting to find the second derivative and A1 for
the correct value.

d 2C
Note: If
d x2
is obtained and its value at x = 91.9 is not calculated, award
(M1)A1 for correct reasoning eg, both numerator and denominator are
positive at x = 91.9 .

thus x = 91.9 gives a minimum AG

METHOD 3
 Sketching the graph of either C versus x or dC
d
versus x. M1
x
Clearly indicating that x = 91.9 gives the minimum on their graph.
A1
[7 marks]

(c) Cmin = 3205k A1

Note: Accept 3200k.

Accept 3204k.

[1 mark]

450
(d) arctan( 91.855865K ) = 78.463K∘ M1
180 − 78.463K = 101.537K
= 102∘ A1
[2 marks]

(e) (i) when θ = 120∘ , x = 260 (m) ( 450 (m)) A1

√3
133.728K
(ii) 3204.5407685K
× 100% M1

= 4.17 (% ) A1
[3 marks]

Total [15 marks]

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