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

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 \(\theta \) radians, where \(0 \le \theta  \le \frac{\pi }{2}\) .

Write down an expression in terms of \(\theta \) for

(i)     \(x\) ;

(ii)    \(y\) .

[2]
a.

Let the area of the rectangle be A.

Show that \(A = 18\sin 2\theta \) .

[3]
b.

(i)     Find \(\frac{{{\rm{d}}A}}{{{\rm{d}}\theta }}\) .

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

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

[8]
c.
Answer/Explanation

Markscheme

(i) \(x = 3\cos \theta \)     A1     N1 

(ii) \(y = 3\sin \theta \)     A1     N1

[2 marks]

a.

finding area     (M1)

e.g. \(A = 2x \times 2y\) , \(A = 8 \times \frac{1}{2}bh\) 

substituting     A1

e.g. \(A = 4 \times 3\sin \theta  \times 3\cos \theta \) , \(8 \times \frac{1}{2} \times 3\cos \theta  \times 3\sin \theta \)

\(A = 18(2\sin \theta \cos \theta )\)    A1

\(A = 18\sin 2\theta \)     AG     N0

[3 marks]

b.

(i) \(\frac{{{\rm{d}}A}}{{{\rm{d}}\theta }} = 36\cos 2\theta \)     A2     N2 

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

e.g. \(36\cos 2\theta  = 0\) , \(\frac{{{\rm{d}}A}}{{{\rm{d}}\theta }} = 0\)

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

\(\theta  = \frac{\pi }{4}\)     A1     N2

(iii) valid reason (seen anywhere)     R1

e.g. at \(\frac{\pi }{4}\), \(\frac{{{{\rm{d}}^2}A}}{{{\rm{d}}{\theta ^2}}} < 0\) ; maximum when \(f”(x) < 0\)

finding second derivative \(\frac{{{{\rm{d}}^2}A}}{{{\rm{d}}{\theta ^2}}} =  – 72\sin 2\theta \)     A1

evidence of substituting \(\frac{\pi }{4}\)     M1

e.g. \( – 72\sin \left( {2 \times \frac{\pi }{4}} \right)\) , \( – 72\sin \left( {\frac{\pi }{2}} \right)\) , \( – 72\)

\(\theta  = \frac{\pi }{4}\) produces the maximum area     AG     N0

[8 marks]

c.

Question

Let \({L_x}\) be a family of lines with equation given by \(r\) \( = \left( {\begin{array}{*{20}{c}} x \\ {\frac{2}{x}} \end{array}} \right) + t\left( {\begin{array}{*{20}{c}} {{x^2}} \\ { – 2} \end{array}} \right)\), where \(x > 0\).

Write down the equation of \({L_1}\).

[2]
a.

A line \({L_a}\) crosses the \(y\)-axis at a point \(P\).

Show that \(P\) has coordinates \(\left( {0,{\text{ }}\frac{4}{a}} \right)\).

[6]
b.

The line \({L_a}\) crosses the \(x\)-axis at \({\text{Q}}(2a,{\text{ }}0)\). Let \(d = {\text{P}}{{\text{Q}}^2}\).

Show that \(d = 4{a^2} + \frac{{16}}{{{a^2}}}\).

[2]
c.

There is a minimum value for \(d\). Find the value of \(a\) that gives this minimum value.

[7]
d.
Answer/Explanation

Markscheme

attempt to substitute \(x = 1\)     (M1)

eg\(\;\;\;\)r \( = \left( {\begin{array}{*{20}{c}} 1 \\ {\frac{2}{1}} \end{array}} \right) + t\left( {\begin{array}{*{20}{c}} {{1^2}} \\ { – 2} \end{array}} \right),{\text{ }}{L_1} = \left( {\begin{array}{*{20}{c}} 1 \\ 2 \end{array}} \right) + t\left( {\begin{array}{*{20}{c}} 1 \\ { – 2} \end{array}} \right)\)

correct equation (vector or Cartesian, but do not accept “\({L_1}\)”)

eg\(\;\;\;\)r \( = \left( {\begin{array}{*{20}{c}} 1 \\ 2 \end{array}} \right) + t\left( {\begin{array}{*{20}{c}} 1 \\ { – 2} \end{array}} \right),{\text{ }}y =  – 2x + 4\;\;\;\)(must be an equation)     A1     N2

[2 marks]

a.

appropriate approach     (M1)

eg\(\;\;\;\left( {\begin{array}{*{20}{c}} 0 \\ y \end{array}} \right) = \left( {\begin{array}{*{20}{c}} a \\ {\frac{2}{a}} \end{array}} \right) + t\left( {\begin{array}{*{20}{c}} {{a^2}} \\ { – 2} \end{array}} \right)\)

correct equation for \(x\)-coordinate     A1

eg\(\;\;\;0 = a + t{a^2}\)

\(t = \frac{{ – 1}}{a}\)     A1

substituting their parameter to find \(y\)     (M1)

eg\(\;\;\;y = \frac{2}{a} – 2\left( {\frac{{ – 1}}{a}} \right),{\text{ }}\left( {\begin{array}{*{20}{c}} a \\ {\frac{2}{a}} \end{array}} \right) – \frac{1}{a}\left( {\begin{array}{*{20}{c}} {{a^2}} \\ { – 2} \end{array}} \right)\)

correct working     A1

eg\(\;\;\;y = \frac{2}{a} + \frac{2}{a},{\text{ }}\left( {\begin{array}{*{20}{c}} a \\ {\frac{2}{a}} \end{array}} \right) – \left( {\begin{array}{*{20}{c}} a \\ { – \frac{2}{a}} \end{array}} \right)\)

finding correct expression for \(y\)     A1

eg\(\;\;\;y = \frac{4}{a},{\text{ }}\left( {\begin{array}{*{20}{c}} 0 \\ {\frac{4}{a}} \end{array}} \right)\) \({\text{P}}\left( {0,{\text{ }}\frac{4}{a}} \right)\)     AG     N0

[6 marks]

b.

valid approach M1

eg\(\;\;\;\)distance formula, Pythagorean Theorem, \(\overrightarrow {{\text{PQ}}}  = \left( {\begin{array}{*{20}{c}} {2a} \\ { – \frac{4}{a}} \end{array}} \right)\)

correct simplification     A1

eg\(\;\;\;{(2a)^2} + {\left( {\frac{4}{a}} \right)^2}\)

\(d = 4{a^2} + \frac{{16}}{{{a^2}}}\)     AG     N0

[2 marks]

c.

recognizing need to find derivative     (M1)

eg\(\;\;\;d’,{\text{ }}d'(a)\)

correct derivative     A2

eg\(\;\;\;8a – \frac{{32}}{{{a^3}}},{\text{ }}8x – \frac{{32}}{{{x^3}}}\)

setting their derivative equal to \(0\)     (M1)

eg\(\;\;\;8a – \frac{{32}}{{{a^3}}} = 0\)

correct working     (A1)

eg\(\;\;\;8a = \frac{{32}}{{{a^3}}},{\text{ }}8{a^4} – 32 = 0\)

working towards solution     (A1)

eg\(\;\;\;{a^4} = 4,{\text{ }}{a^2} = 2,{\text{ }}a =  \pm \sqrt 2 \)

\(a = \sqrt[4]{4}\;\;\;(a = \sqrt 2 )\;\;\;({\text{do not accept }} \pm \sqrt 2 )\)     A1     N3

[7 marks]

Total [17 marks]

d.

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

Consider \(f(x) = \log k(6x – 3{x^2})\), for \(0 < x < 2\), where \(k > 0\).

The equation \(f(x) = 2\) has exactly one solution. Find the value of \(k\).

Answer/Explanation

Markscheme

METHOD 1 – using discriminant

correct equation without logs     (A1)

eg\(\,\,\,\,\,\)\(6x – 3{x^2} = {k^2}\)

valid approach     (M1)

eg\(\,\,\,\,\,\)\( – 3{x^2} + 6x – {k^2} = 0,{\text{ }}3{x^2} – 6x + {k^2} = 0\)

recognizing discriminant must be zero (seen anywhere)     M1

eg\(\,\,\,\,\,\)\(\Delta  = 0\)

correct discriminant     (A1)

eg\(\,\,\,\,\,\)\({6^2} – 4( – 3)( – {k^2}),{\text{ }}36 – 12{k^2} = 0\)

correct working     (A1)

eg\(\,\,\,\,\,\)\(12{k^2} = 36,{\text{ }}{k^2} = 3\)

\(k = \sqrt 3 \)     A2     N2

METHOD 2 – completing the square

correct equation without logs     (A1)

eg\(\,\,\,\,\,\)\(6x – 3{x^2} = {k^2}\)

valid approach to complete the square     (M1)

eg\(\,\,\,\,\,\)\(3({x^2} – 2x + 1) =  – {k^2} + 3,{\text{ }}{x^2} – 2x + 1 – 1 + \frac{{{k^2}}}{3} = 0\)

correct working     (A1)

eg\(\,\,\,\,\,\)\(3{(x – 1)^2} =  – {k^2} + 3,{\text{ }}{(x – 1)^2} – 1 + \frac{{{k^2}}}{3} = 0\)

recognizing conditions for one solution     M1

eg\(\,\,\,\,\,\)\({(x – 1)^2} = 0,{\text{ }} – 1 + \frac{{{k^2}}}{3} = 0\)

correct working     (A1)

eg\(\,\,\,\,\,\)\(\frac{{{k^2}}}{3} = 1,{\text{ }}{k^2} = 3\)

\(k = \sqrt 3 \)     A2     N2

[7 marks]

Question

A closed cylindrical can with radius r centimetres and height h centimetres has a volume of 20\(\pi \) cm3.

The material for the base and top of the can costs 10 cents per cm2 and the material for the curved side costs 8 cents per cm2. The total cost of the material, in cents, is C.

Express h in terms of r.

[2]
a.

Show that \(C = 20\pi {r^2} + \frac{{320\pi }}{r}\).

[4]
b.

Given that there is a minimum value for C, find this minimum value in terms of \(\pi \).

[9]
c.
Answer/Explanation

Markscheme

correct equation for volume      (A1)
eg  \(\pi {r^2}h = 20\pi \)

\(h = \frac{{20}}{{{r^2}}}\)     A1 N2

[2 marks]

a.

attempt to find formula for cost of parts      (M1)
eg  10 × two circles, 8 × curved side

correct expression for cost of two circles in terms of r (seen anywhere)      A1
eg  \(2\pi {r^2} \times 10\)

correct expression for cost of curved side (seen anywhere)      (A1)
eg  \(2\pi r \times h \times 8\)

correct expression for cost of curved side in terms of     A1
eg  \(8 \times 2\pi r \times \frac{{20}}{{{r^2}}},\,\,\frac{{320\pi }}{{{r^2}}}\)

\(C = 20\pi {r^2} + \frac{{320\pi }}{r}\)      AG N0

[4 marks]

b.

recognize \(C’ = 0\) at minimum       (R1)
eg  \(C’ = 0,\,\,\frac{{{\text{d}}C}}{{{\text{d}}r}} = 0\)

correct differentiation (may be seen in equation)

\(C’ = 40\pi r – \frac{{320\pi }}{{{r^2}}}\)        A1A1

correct equation      A1
eg  \(40\pi r – \frac{{320\pi }}{{{r^2}}} = 0,\,\,40\pi r\frac{{320\pi }}{{{r^2}}}\)

correct working     (A1)
eg  \(40{r^3} = 320,\,\,{r^3} = 8\)

r = 2 (m)     A1

attempt to substitute their value of r into C
eg  \(20\pi  \times 4 + 320 \times \frac{\pi }{2}\)     (M1)

correct working
eg  \(80\pi  + 160\pi \)        (A1)

\(240\pi \) (cents)      A1 N3

Note: Do not accept 753.6, 753.98 or 754, even if 240\(\pi \) is seen.

[9 marks]

c.

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