Optimization Practice [83 marks]

1a. [2 marks]

The following diagram shows a semicircle centre O, diameter [AB], with radius 2.

Let P be a point on the circumference, with radians.

Find the area of the triangle OPB, in terms of .

Markschemeevidence of using area of a triangle (M1)

e.g.

A1 N2

[2 marks]

P B = θO

θ

A = × 2 × 2 × sin θ12

A = 2 sin θ

[3 marks]1b. Explain why the area of triangle OPA is the same as the area triangle OPB.

MarkschemeMETHOD 1

(A1)

A1

since R1

then both triangles have the same area AG N0

METHOD 2

triangle OPA has the same height and the same base as triangle OPB R3

then both triangles have the same area AG N0

[3 marks]

P A =π − θO

area ΔOPA = 2 × 2 × sin(π − θ)12

(= 2 sin(π − θ))

sin(π − θ) = sin θ

1c. [3 marks]

Let S be the total area of the two segments shaded in the diagram below.

Show that .

Markschemearea semicircle

A1

A1

M1

AG N0

[3 marks]

S = 2(π − 2 sin θ)

= × π(212

)2

(= 2π)

area ΔAPB = 2 sin θ + 2 sin θ

(= 4 sin θ)

S = area of semicircle − area ΔAPB(= 2π − 4 sin θ)

S = 2(π − 2 sin θ)

1d. [8 marks]Find the value of when S is a local minimum, justifying that it is a minimum.θ

MarkschemeMETHOD 1

attempt to differentiate (M1)

e.g.

setting derivative equal to 0 (M1)

correct equation A1

e.g. ,

,

A1 N3

EITHER

evidence of using second derivative (M1)

A1

A1

it is a minimum because R1 N0

OR

evidence of using first derivative (M1)

for (may use diagram) A1

for (may use diagram) A1

it is a minimum since the derivative goes from negative to positive R1 N0

METHOD 2

is minimum when is a maximum R3

is a maximum when (A2)

A3 N3

[8 marks]

= −4 cosθdS

dθ

−4 cosθ = 0cosθ = 04 cosθ = 0

θ = π

2

(θ) = 4 sin θS ′′

( ) = 4S ′′ π

2

( ) > 0S ′′ π

2

θ < , (θ) < 0π

2S ′

θ > , (θ) > 0π

2S ′

2π − 4 sin θ

4 sin θ

4 sin θ

sin θ = 1

θ = π

2

1e. [2 marks]Find a value of for which S has its greatest value.

MarkschemeS is greatest when

is smallest (or equivalent) (R1)

(or ) A1 N2

[2 marks]

θ

4 sin θ

θ = 0π

2

2. [7 marks]

A farmer wishes to create a rectangular enclosure, ABCD, of area 525 m , as shown below.

The fencing used for side AB costs per metre. The fencing for the other three sides costs

per metre. The farmer creates an enclosure so that the cost is a minimum. Find this minimum cost.

2

$11$3

MarkschemeMETHOD 1

correct expression for second side, using area = 525 (A1)

e.g. let ,

attempt to set up cost function using $3 for three sides and $11 for one side (M1)

e.g.

correct expression for cost A2

e.g. ,

,

EITHER

sketch of cost function (M1)

identifying minimum point (A1)

e.g. marking point on graph,

minimum cost is 420 (dollars) A1 N4

OR

correct derivative (may be seen in equation below) (A1)

e.g.

setting their derivative equal to 0 (seen anywhere) (M1)

e.g.

minimum cost is 420 (dollars) A1 N4

METHOD 2

correct expression for second side, using area = 525 (A1)

e.g. let ,

attempt to set up cost function using for three sides and

for one side (M1)

e.g.

correct expression for cost A2

e.g. ,

,

EITHER

sketch of cost function (M1)

identifying minimum point (A1)

e.g. marking point on graph,

minimum cost is 420 (dollars) A1 N4

AB = x

AD = 525x

3(AD + BC + CD) + 11AB

× 3 + × 3 + 11x + 3x525x

525x

× 3 + × 3 + 11AB + 3AB525AB

525AB

+ 14x3150x

x = 15

(x) = + + 14C ′ −1575x2

−1575x2

+ 14 = 0−3150x2

AD = x

AB = 525x

$3$11

3(AD + BC + CD) + 11AB

3 (x + x + ) + × 11525x

525x

3 (AD + AD + ) + × 11525AD

525AD

6x + 7350x

x = 35

OR

correct derivative (may be seen in equation below) (A1)

e.g.

setting their derivative equal to 0 (seen anywhere) (M1)

e.g.

minimum cost is 420 (dollars) A1 N4

[7 marks]

(x) = 6 −C ′ 7350x2

6 − = 07350x2

3a. [2 marks]

A rectangle is inscribed in a circle of radius 3 cm and centre O, as shown below.

The point P(x , y) is a vertex of the rectangle and also lies on the circle. The angle between (OP) and the x-axis is radians, where

.

Write down an expression in terms of for

(i) ;

(ii) .

Markscheme(i)

A1 N1

(ii) A1 N1

[2 marks]

θ

0 ≤ θ ≤ π

2

θ

x

y

x = 3 cosθ

y = 3 sin θ

3b. [3 marks]Let the area of the rectangle be A.

Show that .A = 18 sin 2θ

Markschemefinding area (M1)

e.g. ,

substituting A1

e.g. ,

A1

AG N0

[3 marks]

A = 2x × 2y

A = 8 × bh12

A = 4 × 3 sin θ × 3 cosθ

8 × × 3 cosθ × 3 sin θ12

A = 18(2 sin θ cosθ)

A = 18 sin 2θ

3c. [8 marks](i) Find .

(ii) Hence, find the exact value of which maximizes the area of the rectangle.

(iii) Use the second derivative to justify that this value of does give a maximum.

Markscheme(i)

A2 N2

(ii) for setting derivative equal to 0 (M1)

e.g. ,

(A1)

A1 N2

(iii) valid reason (seen anywhere) R1

e.g. at,

; maximum when

finding second derivative A1

evidence of substituting M1

e.g. ,

,

produces the maximum area AG N0

[8 marks]

dA

dθ

θ

θ

= 36 cos2θdA

dθ

36 cos2θ = 0= 0dA

dθ

2θ = π

2

θ = π

4

π

4

< 0Ad2

dθ2

(x) < 0f ′′

= −72 sin 2θAd2

dθ2

π

4

−72 sin(2 × )π

4−72 sin( )π

2−72

θ = π

4

4a. [5 marks]

The diagram below shows a plan for a window in the shape of a trapezium.

Three sides of the window are long. The angle between the sloping sides of the window and the base is

, where .

Show that the area of the window is given by .

Markschemeevidence of finding height, h (A1)

e.g. ,

evidence of finding base of triangle, b (A1)

e.g. ,

attempt to substitute valid values into a formula for the area of the window (M1)

e.g. two triangles plus rectangle, trapezium area formula

correct expression (must be in terms of ) A1

e.g. ,

attempt to replace by

M1

e.g.

AG N0

[5 marks]

2 mθ

0 < θ < π

2

y = 4 sin θ + 2 sin 2θ

sin θ = h

22 sin θ

cosθ = b

22 cosθ

θ

2 ( × 2 cosθ × 2 sin θ) + 2 × 2 sin θ12

(2 sin θ)(2 + 2 + 4 cosθ)12

2 sin θ cosθ

sin 2θ

4 sin θ + 2(2 sin θ cosθ)

y = 4 sin θ + 2 sin 2θ

4b. [4 marks]Zoe wants a window to have an area of. Find the two possible values of

.5 m2

θ

Markschemecorrect equation A1

e.g. ,

evidence of attempt to solve (M1)

e.g. a sketch,

,

A1A1 N3

[4 marks]

y = 54 sin θ + 2 sin 2θ = 5

4 sin θ + 2 sin θ − 5 = 0

θ = 0.856( )49.0∘

θ = 1.25( )71.4∘

4c. [7 marks]John wants two windows which have the same area A but different values of .

Find all possible values for A .

Markschemerecognition that lower area value occurs at

(M1)

finding value of area at (M1)

e.g. , draw square

(A1)

recognition that maximum value of y is needed (M1)

(A1)

(accept ) A2 N5

[7 marks]

θ

θ = π

2

θ = π

2

4 sin( ) + 2 sin(2 × )π

2π

2

A = 4

A = 5.19615…

4 < A < 5.204 < A < 5.19

5a. [1 mark]

Let . Part of the graph of

is shown below.

Write down .

f(x) = 100(1+50 )e−0.2x

f

f(0)

Markscheme (exact),

A1 N1

[1 mark]

f(0) = 10051

1.96

5b. [2 marks]Solve .

Markschemesetting up equation (M1)

eg , sketch of graph with horizontal line at

A1 N2

[2 marks]

f(x) = 95

95 = 1001+50e−0.2x

y = 95

x = 34.3

5c. [3 marks]Find the range of .

Markschemeupper bound of is

(A1)

lower bound of is (A1)

range is A1 N3

[3 marks]

f

y

100

y

0

0 < y < 100

5d. [5 marks]Show that .(x) =f ′ 1000e−0.2x

(1+50 )e−0.2x 2

MarkschemeMETHOD 1

setting function ready to apply the chain rule (M1)

eg

evidence of correct differentiation (must be substituted into chain rule) (A1)(A1)

eg ,

correct chain rule derivative A1

eg

correct working clearly leading to the required answer A1

eg

AG N0

METHOD 2

attempt to apply the quotient rule (accept reversed numerator terms) (M1)

eg ,

evidence of correct differentiation inside the quotient rule (A1)(A1)

eg

,

any correct expression for derivative ( may not be explicitly seen) (A1)

eg

correct working clearly leading to the required answer A1

eg

,

AG N0

[5 marks]

100(1 + 50e−0.2x )−1

= −100(1 + 50u′ e−0.2x )−2

= (50 )(−0.2)v′ e−0.2x

(x) = −100(1 + 50 (50 )(−0.2)f ′ e−0.2x )−2 e−0.2x

(x) = 1000 (1 + 50f ′ e−0.2x e−0.2x )−2

(x) =f ′ 1000e−0.2x

(1+50 )e−0.2x 2

v −uu′ v′

v2

u −vv′ u′

v2

(x) =f ′ (1+50 )(0)−100(50 ×−0.2)e−0.2x e−0.2x

(1+50 )e−0.2x 2

100(−10) −0e−0.2x

(1+50 )e−0.2x 2

0

−100(50 ×−0.2)e−0.2x

(1+50 )e−0.2x 2

(x) =f ′ 0−100(−10)e−0.2x

(1+50 )e−0.2x 2

−100(−10)e−0.2x

(1+50 )e−0.2x 2

(x) =f ′ 1000e−0.2x

(1+50 )e−0.2x 2

5e. [4 marks]Find the maximum rate of change of .f

MarkschemeMETHOD 1

sketch of (A1)

eg

recognizing maximum on (M1)

eg dot on max of sketch

finding maximum on graph of A1

eg (,

) ,

maximum rate of increase is A1 N2

METHOD 2

recognizing (M1)

finding any correct expression for (A1)

eg

finding A1

maximum rate of increase is A1 N2

[4 marks]

(x)f ′

(x)f ′

(x)f ′

19.65x = 19.560…

5

(x) = 0f ′′

(x) = 0f ′′

(−200 )−(1000 )(2(1+50 )(−10 ))(1+50 )e−0.2x 2 e−0.2x e−0.2x e−0.2x e−0.2x

(1+50 )e−0.2x 4

x = 19.560…

5

6a. [2 marks]

Let

, where

.

Write down the equations of the vertical and horizontal asymptotes of the graph of

.

Markscheme (must be equations) A1A1 N2

[2 marks]

f(x) = 3x

x−q

x ≠ q

f

x = q, y = 3

6b. [2 marks]The vertical and horizontal asymptotes to the graph of

intersect at the point

.

Find the value of

.

Markschemerecognizing connection between point of intersection and asymptote (R1)eg

A1 N2[2 marks]

f

Q(1,3)

q

x = 1q = 1

6c. [4 marks]The vertical and horizontal asymptotes to the graph of

intersect at the point

.

The point

lies on the graph of

. Show that

.

Markschemecorrect substitution into distance formula A1eg

attempt to substitute

(M1)

eg

correct simplification of

(A1)

eg

correct expression clearly leading to the required answer A1eg

AG N0

[4 marks]

f

Q(1,3)

P(x, y)

f

PQ = +(x − 1) 2 ( )3x−1

2− −−−−−−−−−−−−−−√

+(x − 1) 2 (y − 3) 2− −−−−−−−−−−−−−−√

y = 3x

x−1

+(x − 1) 2 ( − 3)3x

x−1

2− −−−−−−−−−−−−−−−−−√

( − 3)3x

x−1

3x−3x(x−1)x−1

, 3x−3x+3x−1

+(x − 1) 2 ( )3x−3x+3x−1

2− −−−−−−−−−−−−−−−−−√PQ = +(x − 1) 2 ( )3

x−1

2− −−−−−−−−−−−−−−√

6d. [6 marks]The vertical and horizontal asymptotes to the graph of

intersect at the point

.

Hence find the coordinates of the points on the graph of

that are closest to

.

f

Q(1,3)

f

(1,3)

Printed for North Hills Preparatory

© International Baccalaureate Organization 2016 International Baccalaureate® - Baccalauréat International® - Bachillerato Internacional®

Markschemerecognizing that closest is when

is a minimum (R1)eg sketch of

,

(seen anywhere) A1A1attempt to find y-coordinates (M1)eg

A1A1 N4[6 marks]

PQ

PQ

(PQ (x) = 0)′

x = −0.73205 x = 2.73205

f(−0.73205)

(−0.73205,1.267949),(2.73205,4.73205)

(−0.732,1.27),(2.73,4.73)