Optimization Practice [83 marks]

Optimization Practice [83 marks] The following diagram shows a semicircle centre O, diameter [AB], with radius 2. Let P be a point on the circumfere...
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Optimization Practice

[83 marks]

The following diagram shows a semicircle centre O, diameter [AB], with radius 2. Let P be a point on the circumference, with ˆ B = θ radians. PO

1a. Find the area of the triangle OPB, in terms of

[2 marks]

θ.

Markscheme evidence of using area of a triangle

(M1)

e.g. A = 12 × 2 × 2 × sin θ A = 2 sin θ

A1

N2

[2 marks]

1b. Explain why the area of triangle OPA is the same as the area triangle OPB.

Markscheme METHOD 1 ˆ A =π − θ PO

(A1)

area ΔOPA = 12 2 × 2 × sin(π − θ) (= 2 sin(π − θ)) A1 since sin(π − θ) = sin θ

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 then both triangles have the same area [3 marks]

AG

N0

R3

[3 marks]

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

1c. Show that

[3 marks]

S = 2(π − 2 sin θ) .

Markscheme area semicircle = 12 × π(2)2 (= 2π) A1 area ΔAPB = 2 sin θ + 2 sin θ (= 4 sin θ) A1 S = area of semicircle − area ΔAPB (= 2π − 4 sin θ) M1 S = 2(π − 2 sin θ)

AG

N0

[3 marks]

1d.

Find the value of θ when S is a local minimum, justifying that it is a minimum.

[8 marks]

Markscheme METHOD 1 attempt to differentiate

(M1)

e.g. dS = −4 cos θ dθ setting derivative equal to 0 correct equation

(M1)

A1

e.g. −4 cos θ = 0 , cos θ = 0 , 4 cos θ = 0 θ=

π 2

A1

N3

EITHER evidence of using second derivative

(M1)

S ′′ (θ) = 4 sin θ A1 A1

S ′′ ( π2 ) = 4

it is a minimum because S ′′ ( π2 ) > 0 R1 N0 OR (M1)

evidence of using first derivative for θ < π2 , S ′ (θ) < 0 (may use diagram)

A1

for θ > π2 , S ′ (θ) > 0 (may use diagram) A1 it is a minimum since the derivative goes from negative to positive

R1

N0

METHOD 2 2π − 4 sin θ is minimum when 4 sin θ is a maximum R3 4 sin θ is a maximum when sin θ = 1 (A2) θ=

π 2

A3

N3

[8 marks]

1e. Find a value of

[2 marks]

θ for which S has its greatest value.

Markscheme S is greatest when 4 sin θ is smallest (or equivalent) θ = 0 (or π ) A1

(R1)

N2

[2 marks]

2

A farmer wishes to create a rectangular enclosure, ABCD, of area 525 m2, as shown below.

2.

The fencing used for side AB costs $11 per metre. The fencing for the other three sides costs $3 per metre. The farmer creates an enclosure so that the cost is a minimum. Find this minimum cost.

[7 marks]

Markscheme METHOD 1 correct expression for second side, using area = 525

(A1)

e.g. let AB = x , AD = 525 x attempt to set up cost function using $3 for three sides and $11 for one side e.g. 3(AD + BC + CD) + 11AB correct expression for cost

A2

e.g.

525 × 3 + 525 x x 525 525 × 3 + AB AB 3150 + 14x x

× 3 + 11x + 3x , × 3 + 11AB + 3AB ,

EITHER sketch of cost function

(M1) (A1)

identifying minimum point e.g. marking point on graph, x = 15

A1

minimum cost is 420 (dollars)

N4

OR correct derivative (may be seen in equation below) e.g. C ′ (x) =

−1575 x2

(A1)

+ −1575 + 14 2 x

setting their derivative equal to 0 (seen anywhere)

(M1)

e.g. −3150 x2

+ 14 = 0 A1

minimum cost is 420 (dollars)

N4

METHOD 2 correct expression for second side, using area = 525 e.g. let AD = x , AB = 525 x attempt to set up cost function using $3 for three sides and $11 for one side (M1) e.g. 3(AD + BC + CD) + 11AB correct expression for cost

A2

e.g. 3 (x + x + 525 ) + 525 × 11 , x x

3 (AD + AD + 525 ) + 525 × 11 , AD AD 6x + 7350 x

EITHER sketch of cost function

(M1)

identifying minimum point

(A1)

e.g. marking point on graph, x = 35 minimum cost is 420 (dollars)

A1

N4

(A1)

(M1)

OR correct derivative (may be seen in equation below)

(A1)

e.g. C ′ (x) = 6 − 7350 2 x

setting their derivative equal to 0 (seen anywhere)

(M1)

e.g. 6 − 7350 =0 2 x

minimum cost is 420 (dollars)

A1

N4

[7 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 θ radians, where 0 ≤ θ ≤ π2 . 3a.

Write down an expression in terms of θ for

[2 marks]

(i) x; (ii) y.

Markscheme (i) x = 3 cos θ

A1

N1

(ii) y = 3 sin θ

A1

N1

[2 marks]

3b.

Let the area of the rectangle be A. Show that A = 18 sin 2θ .

[3 marks]

Markscheme (M1)

finding area e.g. A = 2x × 2y , A = 8 × 12 bh substituting

A1

e.g. A = 4 × 3 sin θ × 3 cos θ , 8 × 12 × 3 cos θ × 3 sin θ A = 18(2 sin θ cos θ) A1 AG

A = 18 sin 2θ

N0

[3 marks]

3c.

(i)

Find

dA dθ

[8 marks]

.

(ii) Hence, find the exact value of θ which maximizes the area of the rectangle. (iii) Use the second derivative to justify that this value of θ does give a maximum.

Markscheme (i) dA dθ

A2

= 36 cos 2θ

N2

(ii) for setting derivative equal to 0

(M1)

e.g. 36 cos 2θ = 0 , dA =0 dθ 2θ = θ=

(A1)

π 2

π 4

A1

N2

(iii) valid reason (seen anywhere)

R1

e.g. at π , 4 d2 A

< 0 ; maximum when

dθ2 f ′′ (x)