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)