IB Math Analysis & Approaches Question bank-Topic: SL 5.8 Optimization SL Paper 1

evidence of using area of a triangle     (M1)

e.g. \(A = \frac{1}{2} \times 2 \times 2 \times \sin \theta \)

\(A = 2\sin \theta \)     A1     N2

[2 marks]

METHOD 1

\({\rm{P}}\widehat {\rm{O}}{\rm{A = }}\pi  – \theta \)     (A1)

\({\text{area }}\Delta {\rm{OPA}} = \frac{1}{2}2 \times 2 \times \sin (\pi  – \theta )\) \(( = 2\sin (\pi  – \theta ))\)     A1

since \(\sin (\pi  – \theta ) = \sin \theta \)     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]

area semicircle \( = \frac{1}{2} \times \pi {(2)^2}\) \(( = 2\pi )\)     A1

\({\text{area }}\Delta {\rm{APB}} = 2\sin \theta + 2\sin \theta \) \(( = 4\sin \theta )\)     A1

\(S{\text{ = area of semicircle}} – {\text{area }}\Delta {\rm{APB}}\) \(( = 2\pi – 4\sin \theta )\)     M1

\(S = 2(\pi – 2\sin \theta )\)     AG     N0

[3 marks]

METHOD 1

attempt to differentiate     (M1)

e.g. \(\frac{{{\rm{d}}S}}{{{\rm{d}}\theta }} = – 4\cos \theta \)

setting derivative equal to 0     (M1)

correct equation     A1

e.g. \( – 4\cos \theta = 0\) , \(\cos \theta = 0\) , \(4\cos \theta = 0\)

\(\theta = \frac{\pi }{2}\)     A1     N3

EITHER

evidence of using second derivative     (M1)

\(S”(\theta ) = 4\sin \theta \)   A1

\(S”\left( {\frac{\pi }{2}} \right) = 4\)     A1

it is a minimum because \(S”\left( {\frac{\pi }{2}} \right) > 0\)     R1     N0

OR

evidence of using first derivative      (M1)

for \(\theta < \frac{\pi }{2},S'(\theta ) < 0\) (may use diagram)     A1

for \(\theta > \frac{\pi }{2},S'(\theta ) > 0\) (may use diagram)    A1

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

METHOD 2

\(2\pi – 4\sin \theta \) is minimum when \(4\sin \theta \) is a maximum     R3

\(4\sin \theta \) is a maximum when \(\sin \theta = 1\)     (A2)

\(\theta = \frac{\pi }{2}\)     A3     N3

[8 marks]

S is greatest when \(4\sin \theta \) is smallest (or equivalent)     (R1)

\(\theta = 0\) (or \(\pi \) )     A1     N2

[2 marks]