# 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.

## 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.

## 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.

## 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.

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.

[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 $$k$$.

## 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.

## 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.