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

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

Question

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

Let P be a point on the circumference, with \({\rm{P}}\widehat {\rm{O}}{\rm{B}} = \theta \) radians.


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


Find the area of the triangle OPB, in terms of \(\theta \) .

[2]
a.

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

[3]
b.

Show that \(S = 2(\pi  – 2\sin \theta )\) .

[3]
c.

Find the value of \(\theta \) when S is a local minimum, justifying that it is a minimum.

[8]
d.

Find a value of \(\theta \) for which S has its greatest value.

[2]
e.
Answer/Explanation

Markscheme

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]

a.

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]

b.

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]

c.

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]

d.

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

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

[2 marks]

e.

Question

Fred makes an open metal container in the shape of a cuboid, as shown in the following diagram.

M16/5/MATME/SP1/ENG/TZ2/09

The container has height \(x{\text{ m}}\), width \(x{\text{ m}}\) and length \(y{\text{ m}}\). The volume is \(36{\text{ }}{{\text{m}}^3}\).

Let \(A(x)\) be the outside surface area of the container.

Show that \(A(x) = \frac{{108}}{x} + 2{x^2}\).

[4]
a.

Find \(A'(x)\).

[2]
b.

Given that the outside surface area is a minimum, find the height of the container.

[5]
c.

Fred paints the outside of the container. A tin of paint covers a surface area of \({\text{10 }}{{\text{m}}^{\text{2}}}\) and costs $20. Find the total cost of the tins needed to paint the container.

[5]
d.
Answer/Explanation

Markscheme

correct substitution into the formula for volume     A1

eg\(\,\,\,\,\,\)\(36 = y \times x \times x\)

valid approach to eliminate \(y\) (may be seen in formula/substitution)     M1

eg\(\,\,\,\,\,\)\(y = \frac{{36}}{{{x^2}}},{\text{ }}xy = \frac{{36}}{x}\)

correct expression for surface area     A1

eg\(\,\,\,\,\,\)\(xy + xy + xy + {x^2} + {x^2},{\text{ area}} = 3xy + 2{x^2}\)

correct expression in terms of \(x\) only     A1

eg\(\,\,\,\,\,\)\(3x\left( {\frac{{36}}{{{x^2}}}} \right) + 2{x^2},{\text{ }}{x^2} + {x^2} + \frac{{36}}{x} + \frac{{36}}{x} + \frac{{36}}{x},{\text{ }}2{x^2} + 3\left( {\frac{{36}}{x}} \right)\)

\(A(x) = \frac{{108}}{x} + 2{x^2}\)    AG     N0

[4 marks]

a.

\(A'(x) =  – \frac{{108}}{{{x^2}}} + 4x,{\text{ }}4x – 108{x^{ – 2}}\)    A1A1     N2

Note:     Award A1 for each term.

[2 marks]

b.

recognizing that minimum is when \(A'(x) = 0\)     (M1)

correct equation     (A1)

eg\(\,\,\,\,\,\)\( – \frac{{108}}{{{x^2}}} + 4x = 0,{\text{ }}4x = \frac{{108}}{{{x^2}}}\)

correct simplification     (A1)

eg\(\,\,\,\,\,\)\( – 108 + 4{x^3} = 0,{\text{ }}4{x^3} = 108\)

correct working     (A1)

eg\(\,\,\,\,\,\)\({x^3} = 27\)

\({\text{height}} = 3{\text{ (m) }}({\text{accept }}x = 3)\)    A1     N2

[5 marks]

c.

attempt to find area using their height     (M1)

eg\(\,\,\,\,\,\)\(\frac{{108}}{3} + 2{(3)^2},{\text{ }}9 + 9 + 12 + 12 + 12\)

minimum surface area \( = 54{\text{ }}{{\text{m}}^{\text{2}}}\) (may be seen in part (c))     A1

attempt to find the number of tins     (M1)

eg\(\,\,\,\,\,\)\(\frac{{54}}{{10}},{\text{ }}5.4\)

6 (tins)     (A1)

$120     A1     N3

[5 marks]

d.

Question

A rectangle has sides x and y and its perimeter is 40m.

(a) Express y in term of x.
(b) Hence, express the area A of the rectangle in terms of x.
(c) Use \(\frac{dA}{dx}\) to find the maximum area of the rectangle, justify your answer.
(d) What is the minimum value of A?

Answer/Explanation

Ans:

(a) \(2x+2y=40 \Leftrightarrow x+y = 20 \Leftrightarrow y = 20 – x\)
(b) \(A=x(20-x)=20x-x^2\)
(c) \(\frac{dA}{dx}=20-2x\)
\(20-2x=0 \Leftrightarrow x = 10\)
\(\frac{d^2A}{dx^2}=-2<0\) so x = 10 gives a maximum.
It is a square of side x = 10 and the maximum area is A = 100
(d) The domain of A = x(20-x) is \(0 \leq x \leq 20\). At the endpoints x = 0 and x = 20, the area A = 0

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