smc-641-10-Area

Calculus, MET2 2024 VCAA 18 MC

Find the value of $x$ which maximises the area of the trapezium below.

10
$5 \sqrt{2}$
7
$\sqrt{10}$

Show Answers Only

$B$

Show Worked Solution

$A$	$=\dfrac{h}{2}(a+b)$
	$=\dfrac{4x}{2}\sqrt{100-x^2}\quad\text{(Using Pythagoras)}$
	$=2x\sqrt{100-x^2}$
$\dfrac{dA}{dx}$	$=\dfrac{4(50-x^2)}{\sqrt{100-x^2}}\quad\text{(Using product rule)}$

$\text{For maximum }\ \dfrac{dA}{dx}=0$

$\dfrac{4(50-x^2)}{\sqrt{100-x^2}}$	$=0\quad(x \neq 10)$
$50-x^2$	$=0$
($\sqrt{50}-x)(\sqrt{50}+x)$	$=0$
$\therefore\ x$	$=\sqrt{50}=5\sqrt{2}\quad (0<x<10)$

$\text{Check gradient for max using table}$

$x$	$7$	$\sqrt{50}$	$7.2$
$A^{\prime}$	0.56	0	-1.06
$\text{Gradient}$	+	0	–

$\therefore\ x=5\sqrt{2}\ \text{maximises the area}$

$\Rightarrow B$

Calculus, MET2 2022 VCAA 1

The diagram below shows part of the graph of `y=f(x)`, where `f(x)=\frac{x^2}{12}`.

State the equation of the axis of symmetry of the graph of `f`. (1 mark)

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State the derivative of `f` with respect to `x`. (1 mark)

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The tangent to `f` at point `M` has gradient `-2` .

Find the equation of the tangent to `f` at point `M`. (2 marks)

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The diagram below shows part of the graph of `y=f(x)`, the tangent to `f` at point `M` and the line perpendicular to the tangent at point `M`.

i. Find the equation of the line perpendicular to the tangent passing through point `M`. (1 mark)

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ii. The line perpendicular to the tangent at point `M` also cuts `f` at point `N`, as shown in the diagram above.
Find the area enclosed by this line and the curve `y=f(x)`. (2 marks)

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Another parabola is defined by the rule `g(x)=\frac{x^2}{4 a^2}`, where `a>0`.
A tangent to `g` and the line perpendicular to the tangent at `x=-b`, where `b>0`, are shown below.

Find the value of `b`, in terms of `a`, such that the shaded area is a minimum. (4 marks)

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Show Answers Only

a. `x=0`

b. ` f^{\prime}(x)=1/6x`

c. `x=-12`

di. `y=1/2x + 18`

dii. Area`= 375` units²

e. `b = 2a^2`

Show Worked Solution

a. Axis of symmetry: `x=0`

b.	`f(x)`	`=\frac{x^2}{12}`
	` f^{\prime}(x)`	`= 1/6x`

c. At `M` gradient `= -2`

`1/6x`	`= -2`
`x`	`= -12`

When `x = -12`

`f(x) = (-12)^2/12 = 12`

Equation of tangent at `(-12 , 12)`:

`y – y_1`	`=m(x – x_1)`
`y – 12`	`= -2(x + 12)`
`y`	`= -2x -12`

d.i Gradient of tangent `= -2`

`:.` gradient of normal `= 1/2`

Equation at `M(- 12 , 12)`

`y – y_1`	`=m(x – x_1)`
`y – 12`	`= 1/2(x + 12)`
`y`	`=1/2x + 18`

d.ii Points of intersection of `f(x)` and normal are at `M` and `N`.

So equate ` y = x^2/12` and `y = 1/2x + 18` to find `N`

`x^2/12`	`=1/2x + 18`
`x^2 – 6x – 216`	`=0`
`(x + 12)(x – 18)`	`=0`

`:.\ x = -12` or `x = 18`

Area	`= \int_{-12}^{18}\left(\frac{1}{2} x+18-\frac{x^2}{12}\right) d x`
	`= [x^2/4 + 18x – x^3/36]_(-12)^18`
	`= [18^2/4 +18^2 – 18^3/36] – [12^2/4 + 18 xx (-12) – (-12)^3/36]`
	`= 375` units²

e. `g(x) = x^2/(4a^2)` `a > 0`

At `x = – b` `y = (-b)^2/(4a^2) = b^2/4a^2`

`g^{\prime}(x) = (2x)/(4a^2) = x/(2a^2)`

Gradient of tangent `= (-b)/(2a^2)`

Gradient of normal `= (2a^2)/b`

Equation of normal at `(- b , b^2/(4a^2))`

`y – y_1`	`= m(x – x_1)`
`y – b^2/(4a^2)`	`= (2a^2)/b(x – (-b))`
`y`	`= (2a^2x)/b + 2a^2 + b^2/(4a^2)`
`y`	`= (2a^2x)/b +(8a^4 + b^2)/(4a^2)`

Points of intersection of normal and parabola (Using CAS)

solve `((2a^2x)/b +(8a^4 + b^2)/(4a^2) = x^2/(4a^2),x)`

`x = – b` or `x = (8a^4+b^2)/b`

Calculate area using CAS

`A = \int_{-b}^{(8a^4+b^2)/b}\left(\frac{2a^2x}{b} +\frac{8a^4 + b^2}{4a^2} – frac{x^2}{4a^2} \right) dx`

`A = frac{64a^12 + 48a^8b^2 + 12a^4b^4 + b^6}{3a^2b^3}`

Using CAS Solve derivative of `A = 0` with respect to `b` to find `b`

solve`(d/(db)(frac{64a^12 + 48a^8b^2 + 12a^4b^4 + b^6}{3a^2b^3})=0,b)`

`b = – 2a^2` and `b = 2a^2`

Given `b > 0`

`b = 2a^2`

♦♦ Mean mark (e) 33%.
MARKER’S COMMENT: Common errors: Students found `\int_{-b}^{\frac{8 a^4+b^2}{b}}\left(y_n\right) d x` and failed to subtract `g(x)` or had incorrect terminals.

Calculus, MET1 2023 VCAA 9

The shapes of two walking tracks are shown below.

Track 1 is described by the function $f(x)=a-x(x-2)^2$.

Track 2 is defined by the function $g(x)=12x-bx^2$.

The unit of length is kilometres.

Given that $f(0)=12$ and $g(1)=9$, verify that $a=12$ and $b=-3$. (1 mark)
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Verify that $f(x)$ and $g(x)$ both have a turning point at $P$.
Give the co-ordinates of $P$. (2 marks)
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A theme park is planned whose boundaries will form the triangle $\Delta OAB$ where $O$ is the origin, $A$ is at $(k, 0)$ and $B$ is at $(k, g(k))$, as shown below, where $k \in (0, 4)$.
Find the maximum possible area of the theme park, in km². (3 marks)

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Show Answers Only

a. $\text{See worked solution}$

b. $P(2, 12)\ \text{Also see worked solution}$

c. $\dfrac{128}{9}$

Show Worked Solution

a.	$f(0)$	$=a-0(0-2)^2=12$
	$\therefore\ a$	$=12$

$g(1)$	$=12\times 1+b\times 1^2$
$12+b$	$=9$
$\therefore b$	$=-3$

b.	$f(x)$	$=12-x(x-2)^2$
		$=-x^3+4x^2-4x+12$
	$f^{′}(x)$	$=-(3x^2-8x+4)$

$\text{For turning point }f^{′}(x)=0:$

$ -(3x^2-8x+4)$	$=0$
$-(3x-2)(x-2)$	$=0$

$\therefore\ x=\dfrac{2}{3}\ \text{or}\ 2$

$f(2)$	$=12-2(2-2)^2=12$
$f\Big{(}\dfrac{2}{3}\Big{)}$	$=12-\dfrac{2}{3}\Big(\dfrac{2}{3}-2\Big)^2<12$

$\therefore\ \text{Given graph of }f(x)\ \text{relevant turning point is }(2, 12).$

$g(x)=12x-3x^2\ \Rightarrow\ g^{′}(x)=12-6x$

$\text{For turning point}\ \ g^{′}(x)=0:$

$12-6x=0\ \ \Rightarrow \ x=2$

$g(x)$	$=12x-3x^2$
$g(2)$	$=12\times 2-3\times 2^2=12$

$\therefore\ g(x)\ \text{also has a turning point at }(2, 12).$

♦ Mean mark (b) 48%.

c. $\text{Area of triangle }\Delta OAB$

$A(k)$	$=\dfrac{1}{2}\times k\times g(k)$
	$=\dfrac{k}{2}\times (12k-3k^2)$
	$=6k^2-\dfrac{3k^3}{2}$

$\text{Max area when }A^{′}(k)=0:$

$A^{′}(k)=12k-\dfrac{9k^2}{2}=\dfrac{1}{2}(24k-9k^2)$

$\dfrac{1}{2}k(24-9k)=0$

$\therefore\ k=\dfrac{24}{9}=\dfrac{8}{3}$

$\text{Maximum area occurs when}\ k=\dfrac{8}{3}:$

$A(k)$	$=\dfrac{1}{2}k\times (12k-3k^2)$
$A\bigg(\dfrac{8}{3}\bigg)$	$=\dfrac{1}{2}\times\dfrac{8}{3}\times\bigg (12\bigg(\dfrac{8}{3}\bigg)-3\bigg(\dfrac{8}{3}\bigg)^2\bigg)$
	$=\dfrac{4}{3}\bigg(32-\dfrac{64}{3}\bigg)$
	$=\dfrac{4}{3}\bigg(\dfrac{96}{3}-\dfrac{64}{3}\bigg)$
	$=\dfrac{4}{3}\times\dfrac{32}{3}$
	$=\dfrac{128}{9}$

♦♦♦ Mean mark (c) 27%.
MARKER’S COMMENT: Many students made arithmetic errors substituting the fractional values of $k$ into $A(k)$ to find max area.

Calculus, MET1-NHT 2018 VCAA 9

Let diagram below shows a trapezium with vertices at `(0, 0), (0, 2), (3, 2)` and `(b, 0)`, where `b` is a real number and `0 < b < 2`.

On the same axes as the trapezium, part of the graph of a cubic polynomial function is drawn. It has the rule `y = ax(x - b)^2`, where `a` is a non-zero real number and `0 ≤ x ≤ b`.

At the local maximum of the graph, `y = b`.

Find `a` in terms of `b`. (3 marks)

The area between the graph of the function and the `x`-axis is removed from the trapezium, as shown in the diagram.

Show that the expression for the area of the shaded region is `b + 3 - (9b^2)/(16)` square units. (3 marks)
Find the value of `b` for which the area of the shaded region is a maximum and find this maximum area. (2 marks)

Show Answers Only

`(27)/(4b^2)`
`text(Proof (See Worked Solution))`
`(31)/(9)`

Show Worked Solution

a. `y`	`= ax(x – b)^2`
`y′`	`= a(x – b)^2 + 2ax(x – b)`
	`= a(x – b)(x – b + 2x)`
	`= a(x – b)(3x – b)`

`text(S) text(olve) \ \ y′ = 0 \ \ text(for) \ x:`

`x = b \ \ text(or) \ \ (b)/(3)`

`y_text(max) = b \ \ text{(given) and occurs at} \ \ x = (b)/(3)`

`b`	`= a((b)/(3))((b)/(3) – b)^2`
`b`	`= a((b)/(3))((4b^2)/(9))`
`b`	`= a((4b^3)/(27))`

`:. \ a = (27)/(4b^2)`

b. `text(Let)\ \ A_1 = \ text(area under the graph)`

`A_1`	`= int_0^b ax (x – b)^2 dx`
	`= a int_0^b x(x^2 – 2bx + b^2)\ dx`
	`= (27)/(4b^2) int_0^b x^3 – 2bx^2 + b^2 x \ dx`
	`= (27)/(4b^2) [(x^4)/(4) – (2bx^3)/(3) + (b^2 x^2)/(2)]_0^b`
	`= (27)/(4b^2) ((b^4)/(4) – (2b^4)/(3) + (b^4)/(2))`
	`= (27)/(4b^2) ((b^4)/(12))`
	`= (9b^2)/(16)`

`text(Let) \ \ A_2 = \ text(area of triangle)`

`A_2 = (1)/(2) xx 2 xx (3 – b) = 3 – b`

`:. \ text(Shaded region)`	`= 6 – (9b^2)/(16) – (3 – b)`
	`= b + 3 – (9b^2)/(16)`

c. `A = b + 3 – (9b^2)/(16)`

`(dA)/(db) = 1 – (9b)/(8)`

`(dA^2)/(db^2) = – (9)/(8) <0`

`=>\ text{SP (max) when} \ \ (dA)/(db) = 0`

`(9b)/(8) = 1 \ => \ \ b = (8)/(9)`

`:. A_(max)`	`= (8)/(9) + 3 – {9((8)/(9))^2}/{16}`
	`= (35)/(9) – (64)/(9 xx 16)`
	`= (31)/(9)`

Calculus, MET2-NHT 2019 VCAA 4

A mining company has found deposits of gold between two points, `A` and `B`, that are located on a straight fence line that separates Ms Pot's property and Mr Neg's property. The distance between `A` and `B` is 4 units.

The mining company believes that the gold could be found on both Ms Pot's property and Mr Neg's property.

The mining company initially models he boundary of its proposed mining area using the fence line and the graph of

`f : [0, 4] → R, \ f(x) = x(x - 2)(x - 4)`

where `x` is the number of units from point `A` in the direction of point `B` and `y` is the number of units perpendicular to the fence line, with the positive direction towards Ms Pot's property. The mining company will only mine from the boundary curve to the fence line, as indicated by the shaded area below.

Determine the total number of square units that will be mined according to this model. (2 marks)

The mining company offers to pay Mr Neg $100 000 per square unit of his land mined and Ms Pot $120 000 per square unit of her land mined.

Determine the total amount of money that the mining company offers to pay. (1 mark)

The mining company reviews its model to use the fence line and the graph of

`p : [0, 4] → R, \ p(x) = x(x - 4 + (4)/(1 + a)) (x - 4)`

where `a > 0`.

Find the value of `a` for which `p(x) = f(x)` for all `x`. (1 mark)
Solve `p′(x) = 0` for `x` in terms of `a`. (2 marks)

Mr Neg does not want his property to be mined further than 4 units measured perpendicular from the fence line.

Find the smallest value of `a`, correct to three decimal places, for this condition to be met. (2 marks)
Find the value of `a` for which the total area of land mined is a minimum. (3 marks)
The mining company offers to pay Ms Pot $120 000 per square unit of her land mined and Mr Neg $100 000 per square unit of his land mined.

Determine the value of `a` that will minimize the total cost of the land purchase for the mining company. Give your answer correct to three decimal places. (2 marks)

Show Answers Only

`8`
`$880\ 000`
`1`
`x = ((8a + 4) ± 4 sqrt(a^2 + a + 1))/(3a + 3)`
`0.716`
`1`
`0.886`

Show Worked Solution

a. `A`	`= int_0^2 x(x – 2)(x – 4)\ dx – int_2^4 x(x – 2)(x – 4)\ dx`
	`= 4 – (-4)`
	`= 8`

b. `text(Total payment)`	`= 4 xx 100 000 + 4 xx 120 000`
	`= $880\ 000`

c. `text(If) \ \ p(x) = f(x)`

`x – 2`	`= x – 4 + (4)/(1 + a)`
`(4)/(1 + a)`	`=2`
`:. a`	`=1`

d. `p′(x) = (3(a + 1) x^2 – 8(2a + 1) x + 16a)/(a + 1) \ \ \ text{(by CAS)}`

`text(S) text(olve) \ \ p′(x) = 0 \ \ text(for) \ \ x :`

`x = ((8a + 4) ± 4 sqrt(a^2 + a + 1))/(3a +3) \ , \ a > 0`

e. `text(If no mining further than 4 units,) \ \ p(x) ≥ – 4`

`text(Max distance from Mr Neg’s fence occurs when) \ \ 2< x <4 .`

`text((i.e. the higher) \ x text(-value when) \ \ p′(x) = 0)`

`text(At) \ \ x = ((8a + 4) + 4 sqrt(a^2 + a + 1))/(3a + 3) \ ,`

`text(S) text(olve for) \ \ a \ \ text(such that) \ \ p(x) = -4`

`=> a = 0.716 \ \ text((to 3 d.p.))`

f. `p(x) \ text(intersects) \ x text(-axis at) \ \ x = 4 – (4)/(1 + a) = (4a)/(1 + a)`

`A`	`= int_0^{(4a)/(1 +a)} p(x)\ dx – int_{(4a)/(1 +a)}^4 p(x)\ dx`
	`= (64(1 + 2a + 2a^3 + a^4))/(3(1 + a)^4)`

`text(S) text(olve) \ \ A′(a) = 0 \ \ text(for) \ \ a :`

`a = 1`

g. `C(a) = 120\ 000 int_0^{(4a)/(1 +a)} p(x)\ dx – 100\ 000 int_{(4a)/(1 +a)}^4 p(x)\ dx`

`text(S) text(olve) \ \ C′(a) = 0 \ \ text(for) \ \ a :`

`a = 0.886 \ \ text((to 3 d.p.))`

Calculus, MET2 2019 VCAA 5

Let `f: R -> R, \ f(x) = 1 - x^3`. The tangent to the graph of `f` at `x = a`, where `0 < a < 1`, intersects the graph of `f` again at `P` and intersects the horizontal axis at `Q`. The shaded regions shown in the diagram below are bounded by the graph of `f`, its tangent at `x = a` and the horizontal axis.

Find the equation of the tangent to the graph of `f` at `x = a`, in terms of `a`. (1 mark)
Find the `x`-coordinate of `Q`, in terms of `a`. (1 mark)
Find the `x`-coordinate of `P`, in terms of `a`. (2 marks)

Let `A` be the function that determines the total area of the shaded regions.

Find the rule of `A`, in terms of `a`. (3 marks)
Find the value of `a` for which `A` is a minimum. (2 marks)

Consider the regions bounded by the graph of `f^(-1)`, the tangent to the graph of `f^(-1)` at `x = b`, where `0 < b < 1`, and the vertical axis.

Find the value of `b` for which the total area of these regions is a minimum. (2 marks)
Find the value of the acute angle between the tangent to the graph of `f` and the tangent to the graph of `f^(-1)` at `x = 1`. (1 mark)

Show Answers Only

`y = 2a^3 – 3a^2x + 1`
`(2a^3 + 1)/(3a^2)`
`-2a`
`(80a^6 + 8a^3 – 9a^2 + 2)/(12a^2)`
`10^(-1/3)`
`9/10`
`tan^(-1)(1/3)`

Show Worked Solution

a. `f(x) = 1 – x^3,\ \ f^{\prime}(x) = -3x^2`

`m_text(tang) = -3a^2\ \ text(through)\ \ (a, 1 – a^3)`

`y = 2a^3 – 3a^2x + 1`

b. `text(Solve for)\ \x:`

`2a^3 – 3a^2x + 1 = 0`

`x = (2a^3 + 1)/(3a^2)`

c. `text(Solve for)\ \ x:`

`2a^3 – 3a^2x + 1 = 1 – x^3`

`x=-2a`

`:. x text(-coordinate of)\ P = -2a`

d. `P(-2a, 8a^3 + 1),\ \ Q((2a^3 + 1)/(3a^2), 0)`

`A`	`= text(Area of triangle)-int_(-2a)^1 f(x)\ dx`
	`= 1/2((2a^3 + 1)/(3a^2) + 2a)(8a^3 + 1)-int_(-2a)^1 1-x^3\ dx`
	`= (80a^6 + 8a^3 – 9a^2 + 2)/(12a^2)\ \ text{(by CAS)}`

e. `text(Solve for)\ a: \ (dA)/(da) = 0\ \ text{(by CAS)}`

`a = 10^(-1/3)`

f. `text(Consider)\ f(x):\ \ f(10^(-1/3)) = 9/10`

`A_text(min)\ text(for)\ \ f(x)\ \ text(occurs at)\ \ (10^(-1/3), 9/10)`

`=> A_text(min)\ text(for)\ \ f^(-1)(x)\ \ text(occurs when)\ \ x=9/10`

`:. b = 9/10`

g. `f^{\prime} (1) = -3`

`text(Gradient of)\ \ f^(-1)(x)\ \ text(at)\ \ x = 1\ \ text(is vertical line.)`

`tan theta`	`= 1/3`
`theta`	`= tan^(-1)(1/3)`
	`=18.4°\ \ text{(to 1 d.p.)}`

Calculus, MET1 2019 VCAA 7

The graph of the relation `y = sqrt (1 - x^2)` is shown on the axes below. `P` is a point on the graph of this relation, `A` is the point `(-1, 0)` and `B` is the point `(x, 0)`.

Find an expression for the length `PB` in terms of `x` only. (1 mark)
Find the maximum area of the triangle `ABP`. (3 marks)

Show Answers Only

`PB = sqrt(1 – x^2)`
`(3 sqrt 3)/8`

Show Worked Solution

a. `PB = sqrt(1 – x^2)`

b.	`A`	`= 1/2 ⋅ AB ⋅ PB`
		`= 1/2 (x + 1) ⋅ (1 – x^2)^(1/2)`
	`(dA)/(dx)`	`= 1/2[(x + 1) ⋅ 1/2 ⋅ -2x ⋅ 1/sqrt(1 – x^2) + sqrt(1 – x^2)]`
		`= 1/2((-x^2 – x + 1 – x^2)/sqrt(1 – x^2))`
		`= (-2x^2 – x + 1)/(2 sqrt(1 – x^2))`

`text(Find max when)\ \ (dA)/(dx) = 0`

`2x^2 + x – 1 = 0`

`(2x – 1)(x + 1) = 0`

`x = 1/2 qquad (x =\ text{–1 is a min)}`

`:. A_max`	`= 1/2 (3/2)(1 – 1/4)^(1/2)`
	`= 3/4 ⋅ sqrt(3/4)`
	`= (3 sqrt 3)/8`

Calculus, MET1 SM-Bank 17

The diagram shows a point `T` on the unit circle `x^2+y^2=1` at an angle `theta` from the positive `x`-axis, where `0<theta<pi/2`.

The tangent to the circle at `T` is perpendicular to `OT`, and intersects the `x`-axis at `P`, and the line `y=1` intersects the `y`-axis at `B`.

Show that the equation of the line `PT` is `xcostheta+ysin theta=1`. (2 marks)
Find the length of `BQ` in terms of `theta`. (1 mark)
Show that the area, `A`, of the trapezium `OPQB` is given by
`A=(2-sintheta)/(2costheta)` (2 marks)
Find the angle `theta` that gives the minimum area of the trapezium. (3 marks)

Show Answers Only

`text{Proof (See Worked Solutions)}`
`(1-sin theta)/cos theta`
`text{Proof (See Worked Solutions)}`
`theta=pi/6\ \ text(radians)`

Show Worked Solution

`text(Find)\ T:`

♦♦ Mean mark 20%

`text(S)text(ince)\ \ cos theta=x/1\ \ \ text(and)\ \ \ sin theta=y/1`

`:. T\ (cos theta, sin theta)`

`text(Gradient of)\ OT=sin theta/cos theta`

`:.\ text(Gradient)\ PT=-cos theta/sin theta\ \ text{(} _|_ text{lines)}`

`text(Equation of)\ PT\ text(where)`

`m=-cos theta/sin theta,\ \ text(and through)\ \ (cos theta, sin theta)`

`text(Using)\ \ y-y_1`	`=m(x-x_1)`
`y-sin theta`	`=-cos theta/sin theta(x-cos theta)`
`y sin theta-sin^2 theta`	`=-x cos theta+cos^2 theta`
`x cos theta+y sin theta`	`=sin^2 theta+cos^2 theta`
`x cos theta+y sin theta`	`=1\ \ \ \ \ text(… as required)`

ii. `text(Find)\ Q:`

`Q\ => text(intersection of)\ xcos theta+y sin theta=1\ \ text(and)\ \ y=1`

`x cos theta+sin theta`	`=1`
`x cos theta`	`=1-sin theta`
`x`	`=(1-sin theta)/cos theta`

`:.\ text(Length of)\ BQ\ text(is)\ \ (1-sin theta)/cos theta\ text(units)`

iii. `text(Show Area)\ \ OPQB=(2-sin theta)/(2cos theta)`

`A=1/2h(a+b)\ \ text(where)\ \ h=OB=1\ \ a=OP\ \ text(and)`

`b=BQ=(1-sin theta)/cos theta`

`text(Find length)\ OP:`

`P => xcos theta+ysin theta=1 \ text(cuts)\ \ x text(-axis)`

♦♦ Mean mark 24%

`xcos theta`	`=1`
`x`	`=1/cos theta`
`=>\ text(Length)\ OP`	`=1/cos theta`

`text(Area)\ OPQB`	`=1/2xx1(1/cos theta+(1-sin theta)/cos theta)`
	`=1/2((2-sin theta)/cos theta)`
	`=(2-sin theta)/(2cos theta)\ \ text(u²)\ \ \ text(… as required)`

iv. `text(Find)\ theta\ text(such that Area)\ OPQB\ text(is a MIN)`

`A`	`=(2-sin theta)/(2cos theta)`
`(dA)/(d theta)`	`=(2cos theta(-cos theta)-(2- sin theta)(-2 sin theta))/(4cos^2 theta)`
	`=(4 sin theta-2sin^2 theta-2 cos^2 theta)/(4 cos^2 theta)`
	`=(4sin theta-2(sin^2 theta+cos^2 theta))/(4 cos^2 theta)`
	`=(4sin theta-2)/(4cos^2 theta)`
	`=(2sin theta-1)/(2 cos^2 theta)`

Mean mark 19%
IMPORTANT: Look for any opportunity to use the identity `sin^2 theta“+cos^2 theta=1` as it is an examiner’s favourite and can often be the key to simplifying difficult trig equations.

`text(MAX or MIN when)\ (dA)/(d theta)=0`

`=>2sin theta-1`	`=0`
`sin theta`	`=1/2`
`theta`	`=pi/6\ \ \ \ \0<theta<pi/2`

`text(Test for MAX/MIN)`

IMPORTANT: Is the 1st or 2nd derivative test easier here? Students must evaluate and choose, knowing that examiners have often made one significantly easier than the other.

`text(If)\ theta=pi/12\ \ (dA)/(d theta)<0`

`text(If)\ theta=pi/3\ \ (dA)/(d theta)>0\ \ =>text(MIN)`

`:.\text(Area)\ OPQB\ text(is a MIN when)\ theta=pi/6`.

Calculus, MET1 SM-Bank 35

The diagram shows two parallel brick walls `KJ` and `MN` joined by a fence from `J` to `M`. The wall `KJ` is `s` metres long and `/_KJM=alpha`. The fence `JM` is `l` metres long.

A new fence is to be built from `K` to a point `P` somewhere on `MN`. The new fence `KP` will cross the original fence `JM` at `O`.

Let `OJ=x` metres, where `0<x<l`.

Show that the total area, `A` square metres, enclosed by `DeltaOKJ` and `DeltaOMP` is given by
`A=s(x-l+l^2/(2x))sin alpha`. (3 marks)
Find the value of `x` that makes `A` as small as possible. Justify the fact that this value of `x` gives the minimum value for `A`. (3 marks)
Hence, find the length of `MP` when `A` is as small as possible. (1 mark)

Show Answers Only

`text{Proof (See Worked Solutions)}`
`l/sqrt2`
`(sqrt2-1)s\ \ text(metres)`

Show Worked Solution

`A=text(Area)\ Delta OJK+text(Area)\ Delta OMP`

♦♦♦ Although mean mark data for question parts is not available before 2009, students found this question extremely challenging.

`text(Using sine rule)`

`text(Area)\ Delta OJK=1/2\ x s sin alpha`

`text(Area)\ DeltaOMP =>text(Need to find)\ \ MP`

`/_OKJ`	`=/_MPO\ \ text{(alternate angles,}\ MP\ text(\|\|)\ KJtext{)}`
`/_PMO`	`=/_OJK=alpha\ \ text{(alternate angles,}\ MP\ text(\|\|)\ KJtext{)}`

`:.\ DeltaOJK\ text(|||)\ Delta OMP\ \ text{(equiangular)}`

`=>x/s`	`=(l-x)/(MP)\ `	` text{(corresponding sides of similar triangles)}`
`MP`	`=(l-x)/x *s`

`text(Area)\ Delta\ OMP`	`=1/2 (l-x)* MP * sin alpha`
	`=1/2 (l-x)((l-x))/x s sin alpha`

`:. A`	`=1/2 xs sin alpha+1/2 (l-x)((l-x))/x* s sin alpha`
	`=s sin alpha(1/2 x+1/2 (l-x)*((l-x))/x)`
	`=s sin alpha(1/2 x+(l-x)^2/(2x))`
	`=s sin alpha(1/2 x+l^2/(2x)-l+1/2 x)`
	`=s(x-l+l^2/(2x))sin alpha\ \ \ \ text(… as required)`

ii. `text(Find)\ x\ text(such that)\ A\ text(is a minimum)`

MARKER’S COMMENT: Students who could not complete part (i) are reminded that they can still proceed to part (ii) and attempt to differentiate the result given.
Note that `l` and `alpha` are constants when differentiating.

`A`	`=s(x-l+l^2/(2x))sin alpha`
`(dA)/(dx)`	`=s(1-l^2/(2x^2))sin alpha`

`text(MAX/MIN when)\ (dA)/(dx)=0`

`s(1-l^2/(2x^2))sin alpha`	`=0`
`l^2/(2x^2)`	`=1`
`2x^2`	`=l^2`
`x^2`	`=l^2/2`
`x`	`=l/sqrt2,\ \ \ x>0`

`(d^2A)/(dx^2)=s((l^2)/(2x^3))sin alpha`

`text(S)text(ince)\ \ 0<alpha<90°\ \ =>\ sin alpha>0,\ \ l>0\ \ text(and)\ \ x>0`

`(d^2A)/(dx^2)>0\ \ \ =>text(MIN at)\ \ x=l/sqrt2`

iii. `text(S)text(ince)\ \ MP=((l-x))/x s\ \ text(and MIN when)\ \ x=l/sqrt2`

`MP`	`=((l-l/sqrt2)/(l/sqrt2))s xx sqrt2/sqrt2`
	`=((sqrt2 l-l))/l s`
	`=(sqrt2-1)s\ \ text(metres)`

`:.\ MP=(sqrt2-1)s\ \ text(metres when)\ A\ text(is a MIN.)`

Calculus, MET2 2017 VCAA 15 MC

A rectangle `ABCD` has vertices `A(0, 0)`, `B(u, 0)`, `C(u, v)` and `D(0, v)`, where `(u, v)` lies on the graph of `y = -x^3 + 8`, as shown below.

The maximum area of the rectangle is

`root3(2)`
`6 root3(2)`
`16`
`8`
`3root3(2)`

Show Answers Only

`B`

Show Worked Solution

`A`	`=u(-u^3+8)`
	`=8u-u^4`
`(dA)/(du)`	`=8-4u^3`

`(dA)/(du)=0\ \ text(when):`

`4u^3`	`=8`
`u^3`	`=2`
`u`	`=root3(2)`

`:.A_text(max)=8root3(2)-(root3(2))^4 = 6 root3(2)`

`=> B`

Calculus, MET2 2016 VCAA 14 MC

A rectangle is formed by using part of the coordinate axes and a point `(u, v)`, where `u > 0` on the parabola `y = 4 - x^2.`

Which one of the following is the maximum area of the rectangle?

`4`
`(2 sqrt 3)/3`
`(8 sqrt 3 - 4)/3`
`8/3`
`(16 sqrt 3)/9`

Show Answers Only

`E`

Show Worked Solution

`text(Area)`	`=u xx (4-u^2)`
	`=4u-u^3`
`(dA)/(du)`	`=4-3u^2`

`text(Maximum area occurs when)\ \ (dA)/(du)=0,`

♦♦ Mean mark 37%.

`4-3u^2`	`=0`
`u^2`	`=4/3`
`u`	`= (2 sqrt 3)/3`

`:. A_text(max)`	`=(2 sqrt 3)/3 xx (4-(2 sqrt 3)^2/9)`
	`=(2 sqrt 3)/3 xx 8/3`
	` = (16 sqrt 3)/9`

`=> E`

Calculus, MET1 2006 VCAA 9

A rectangle `XYZW` has two vertices, `X` and `W`, on the `x`-axis and the other two vertices, `Y` and `Z`, on the graph of `y = 9 - 3x^2`, as shown in the diagram below. The coordinates of `Z` are `(a, b)` where `a` and `b` are positive real numbers.

Find the area, `A`, of rectangle `XYZW` in terms of `a`. (1 mark)
Find the maximum value of `A` and the value of `a` for which this occurs. (3 marks)

Show Answers Only

`A = 18a – 6a^3`
`A_max = 12\ text(u² when)\ \ a = 1`

Show Worked Solution

♦ Mean mark 37%.

a.	`A`	`=\ text(length × height)`
		`=2a (9 – 3a^2)`
		`= 18a – 6a^3`

b. `text(Stationary point when)\ \ (dA)/(da) = 0`

♦♦ Mean mark 29%.

`18 – 18 a^2`	`=0`
`a^2`	`=1`
`a`	`= 1,\ \ \ \ \ a > 0`

`text(Find)\ \ A\ \ text(when)\ \ \ a = 1:`

`A = 18 (1) – 6 (1)^3 = 12`

`:. A_max = 12\ text(u² when)\ \ a = 1`

Calculus, MET1 2013 VCAA 10

Let `f: [0, oo) -> R,\ \ f(x) = 2e^(-x/5).`

A right-angled triangle `OQP` has vertex `O` at the origin, vertex `Q` on the `x`-axis and vertex `P` on the graph of `f`, as shown. The coordinates of `P` are `(x, f(x)).`

Find the area, `A`, of the triangle `OPQ` in terms of `x`. (1 mark)
Find the maximum area of triangle `OQP` and the value of `x` for which the maximum occurs. (3 marks)
Let `S` be the point on the graph of `f` on the `y`-axis and let `T` be the point on the graph of `f` with the `y`-coordinate `1/2`.

Find the area of the region bounded by the graph of `f` and the line segment `ST`. (3 marks)

Show Answers Only

`x e^(-x/5)`
`5/e\ text(u²)`
`25/4 log_e (4) – 15/2\ text(u²)`

Show Worked Solution

a.	`text(Area)`	`= 1/2 xx b xx h`
		`= 1/2x(2e^(-x/5))`
	`:. A`	`= xe^(-x/5)`

b. `text(Stationary point when)\ \ (dA)/(dx) = 0,`

♦ Mean mark 35%.

`x(-1/5 e^(-x/5)) + e^(-x/5)`	`= 0`
`e^(-x/5)[1 – x/5]`	`= 0`
`:. x`	`= 5\ \ text{(as}\ \ e^(-x/5) >0,\ \ text(for all)\ x text{)}`

`text(When)\ \ x = 5,\ \ A`	`= xe^(-x/5)`
	`= 5e^-1`

`:. A_max = 5/e\ text(u², when)\ \ x = 5`

c. `text(Find)\ \ S:\ \ F(0) = 2.`

♦♦ Mean mark 32%.

`:. S(0, 2)`

`text(Find)\ \ T:\ \ \ `	`2e^(-x/5)`	`= 1/2`
	`e^(-x/5)`	`= 1/4`
	`-x/5`	`= log_e (1/4)`
	`x`	`= 5 log_e (4)`

`:. T(5log_e(4), 1/2)`

`:.\ text(Area)`	`= text(Area)\ SOAT – int_0^(5 log_e(4)) (2e^(-x/5)) dx`
	`=1/2h(a+b) + 10 [e^(-x/5)]_0^(5 log_e (4))`
	`= 5/2 log_e (4) (2 + 1/2) + 10 [e^(-log_e (4)) – e^0]`
	`= 25/4 log_e (4) +10 (1/4 – 1)`
	`= 25/4 log_e (4) – 15/2\ text(u²)`

Calculus, MET1 2014 VCAA 10

A line intersects the coordinate axes at the points `U` and `V` with coordinates `(u, 0)` and `(0, v)`, respectively, where `u` and `v` are positive real numbers and `5/2 <= u <= 6`.

When `u = 6`, the line is a tangent to the graph of `y = ax^2 + bx` at the point `Q` with coordinates `(2, 4)`, as shown.

If `a` and `b` are non-zero real numbers, find the values of `a` and `b`. (3 marks)
The rectangle `OPQR` has a vertex at `Q` on the line. The coordinates of `Q` are `(2, 4)`, as shown.
1. Find an expression for `v` in terms of `u`. (1 mark)
2. Find the minimum total shaded area and the value of `u` for which the area is a minimum. (2 marks)
3. Find the maximum total shaded area and the value of `u` for which the area is a maximum. (1 mark)

Show Answers Only

`a = −3/2, b = 5`
1. `v=(4u)/(u-2)`
2. `8\ text{u² (when}\ u=4)`
3. `17\ text{u²}\ \ (text{when}\ u=5/2)`

Show Worked Solution

a. `text(S)text{ince (2, 4) lies on parabola:}`

♦ Mean mark 48%.
MARKER’S COMMENT: The most common error incorrectly stated that (6,0) lay on the parabola.

`4`	`= a(2)^2 + b(2)`
`4`	`= 4a + 2b`
`2`	`= 2a + b\ …\ (1)`

`y`	`=ax^2+bx`
`dy/dx`	`=2ax+b`

`text(At)\ \ x=2,`

`m_(QU)`	`=m_(text(tang))`
`(4 – 0)/(2 – 6)`	`= 2a(2) + b`
`−1`	`= 4a + b\ …\ (2)`

`text(Subtract)\ \ (2) – (1)`

`−3 = 2a`

`:. a = −3/2, b = 5`

b.i.

`text(Solution 1)`

`text(Using similar triangles:)`

`Delta VOU\ text(|||)\ Delta QPU`

♦♦ Mean mark 32%.

`v/4`	`= u/(u -2)`
`:. v`	`= (4u)/(u -2)`

`text(Solution 2)`

`m_(VQ)`	`=m_(UQ)`
`(v-4)/(0-2)`	`=(0-4)/(u-2)`
`v`	`=8/(u-2) + 4`
	`=(8+4(u-2))/(u-2)`
	`=(4u)/(u-2)`

♦♦♦ Part (b)(ii) mean mark 18%.

b.ii.	`text(Area)`	`= 1/2uv – 2 xx 4`
		`= 1/2u((4u)/(u -2)) – 8`
		`= (2u^2)/(u -2) – 8`

`text(SP occurs when)\ \ (dA)/(du)=0,`

`(4u(u – 2) – 2u^2)/((u – 2)^2)`	`= 0`
`2u^2 – 8u`	`= 0`
`2u(u – 4)`	`= 0`

`u = 0quadtext(or)quadu = 4`

`:. u = 4, \ \ u ∈ [5/2,6]`

`text(When)\ u=4,\ \ A`	`=(2 xx 4^2)/(4-2) – 8`
	`= 8\ text(u²)`

`text(Test areas at end points,)`

`text(When)\ u=5/2,\ A=17`

`text(When)\ u=6,\ A=10`

`:. A_(text(min))=8\ text{u² (when}\ u=4)`

b.iii. `text{As only one (local minimum) stationary point}`

♦♦♦ Mean mark 8%.

`text(exists over)\ 5/2 <= u <= 6, text(the maximum area)`

`text(must occur at an endpoint.)`

`:. A_(text(max)) = 17\ text{u²}\ \ (text{when}\ u=5/2)`

Calculus, MET1 2015 VCAA 10

The diagram below shows a point, `T`, on a circle. The circle has radius 2 and centre at the point `C` with coordinates `(2, 0)`. The angle `ECT` is `theta`, where `0 < theta <= pi/2`.

The diagram also shows the tangent to the circle at `T`. This tangent is perpendicular to `CT` and intersects the `x`-axis at point `X` and the `y`-axis at point `Y`.

Find the coordinates of `T` in terms of `theta`. (1 mark)
Find the gradient of the tangent to the circle at `T` in terms of `theta`. (1 mark)
The equation of the tangent to the circle at `T` can be expressed as
`qquad cos(theta)x + sin(theta)y = 2 + 2cos(theta)`
i. Point `B`, with coordinates `(2, b)`, is on the line segment `XY`.
Find `b` in terms of `theta`. (1 mark)
ii. Point `D`, with coordinates `(4, d)`, is on the line segment `XY`.
Find `d` in terms of `theta`. (1 mark)
Consider the trapezium `CEDB` with parallel sides of length `b` and `d`.
Find the value of `theta` for which the area of the trapezium `CEDB` is a minimum. Also find the minimum value of the area. (3 marks)

Show Answers Only

`T(2 + 2costheta, 2 sintheta)`
`(−1)/(tan(theta))`
i. `2/(sintheta)`
ii. `(2 – 2 costheta)/(sintheta)`
`theta = pi/3`
`A_text(min) = 2sqrt3\ text(u²)`

Show Worked Solution

a.	`cos theta`	`= (CM)/(CT)`
		`=(CM)/2`
	`CM`	`= 2costheta`

♦♦♦ Part (a) mean mark 20%.
MARKER’S COMMENT: Many students did not include the “+2” in the `x`-coordinate.

`sintheta`	`= (TM)/2`
`TM`	`= 2sintheta`

`:. T\ text(has coordinates)\ \ (2 + 2costheta, 2 sintheta)`

♦♦♦ Part (b) mean mark 16%.

b.	`m_(CT)`	`=(TM)/(CM)`
		`=(2 sin theta)/(2 cos theta)`
		`=tan theta`

	`:.m_(XY)`	`=- 1/tan theta,\ \ \ (CT ⊥ XY)`

c.i. `text(Substitute)\ \ (2,b)\ \ text(into equation:)`

`2costheta + bsintheta`	`= 2 + 2costheta`
`:. b`	`= 2/(sintheta)`

c.ii. `text(Substitute)\ \ (4,d)\ \ text(into equation:)`

♦ Part (c)(ii) mean mark 47%.

`4costheta + dsintheta`	`= 2 + 2costheta`
`d sin theta`	`=2 – 2cos theta`
`:.d`	`= (2 – 2 costheta)/(sintheta)`

♦♦♦ Part (d) mean mark 19%.

d.	`text(A)_text(trap)`	`= 1/2 xx 2 xx (b + d)`
		`= 2/(sintheta) + (2 – 2costheta)/(sintheta)`
		`= (4 – 2costheta)/(sintheta)`

`text(Stationary point when)\ \ (dA)/(d theta)=0`

`(2sin^2theta – costheta(4 – 2costheta))/(sin^2theta)`	`= 0`
`2sin^2theta – 4costheta + 2cos^2theta`	`= 0`
`2[sin^2theta + cos^2theta] – 4costheta`	`= 0`
`2 – 4costheta`	`= 0`
`costheta`	`= 1/2`
`theta`	`= pi/3,\ \ \ \ theta ∈ (0, pi/2)`

`A(pi/3)`	`= (4 – 2(1/2))/(sqrt3/2)`
	`=3 xx 2/sqrt3`
	`= 2sqrt3`

`:. A_text(min) = 2sqrt3\ text(u²)`

Calculus, MET2 2013 VCAA 4

Part of the graph of a function `g: R -> R, \ g (x) = (16 - x^2)/4` is shown below.

Points `B` and `C` are the positive `x`-intercept and `y`-intercept of the graph `g`, respectively, as shown in the diagram above. The tangent to the graph of `g` at the point `A` is parallel to the line segment `BC.`
i. Find the equation of the tangent to the graph of `g` at the point `A.` (2 marks)
ii. The shaded region shown in the diagram above is bounded by the graph of `g`, the tangent at the point `A`, and the `x`-axis and `y`-axis.
Evaluate the area of this shaded region. (3 marks)
Let `Q` be a point on the graph of `y = g(x)`.
Find the positive value of the `x`-coordinate of `Q`, for which the distance `OQ` is a minimum and find the minimum distance. (3 marks)

The tangent to the graph of `g` at a point `P` has a negative gradient and intersects the `y`-axis at point `D(0, k)`, where `5 <= k <= 8.`

Find the gradient of the tangent in terms of `k.` (2 marks)
i. Find the rule `A(k)` for the function of `k` that gives the area of the shaded region. (2 marks)
ii. Find the maximum area of the shaded region and the value of `k` for which this occurs. (2 marks)
iii. Find the minimum area of the shaded region and the value of `k` for which this occurs. (2 marks)

Show Answers Only

1. `A: y = 5 – x`
2. `text(See Worked Solutions)`
`2sqrt3\ text(units when)\ x = 2sqrt2`
`gprime(2sqrt(k – 4)) = −sqrt(k – 4)`
1. `A(k) = (k^2)/(2sqrt(k – 4)) – 32/3, k ∈ [5,8]`
2. `A_text(max) = 16/3quadtext(when)quadk = 8`
3. `A_text(min) = (64sqrt3)/9 – 32/3quadtext(when)quadk = 16/3`

Show Worked Solution

a.i. `B(4,0), C(0,4), A(a,(16 – a^2)/4)`

`m_(BC) = m_text(tan) = (4 – 0)/(0 – 4) = −1`

`g(x)`	`= (16 – x^2)/4`
`g′(x)`	`=- x/2`

`text(S)text(ince)\ \ g′(a) = −1,`

`=>a = 2`

`text(T)text(angent passes through)\ \ (2,3),`

`y-3`	`=-(x-2)`
`y`	`=-x + 5`

`text(Alternatively, using technology:)`

`[text(CAS: tangent Line)]\ (g(x),x,2)`

a.ii. `text(Solution 1)`

♦ Mean mark 37%.
MARKER’S COMMENT: A common incorrect answer:
`int_0^5((-x+5)-g(x))dx=35/12`. Know why it’s wrong!

`D(5,0), E(0,5)`

`text(Area)`	`= DeltaEOD – int_0^4 g(x)\ dx`
	`= 1/2 xx 5 xx 5 – 32/3`
	`= 11/6\ text(u²)`

`text(Solution 2)`

`text(Area)`	`= int_0^5 (-x+5)\ dx – int_0^4 g(x)\ dx`
	`= 11/6\ text(u²)`

b. `Q(x, (16 – x^2)/4), O(0,0)`

♦♦♦ Mean mark 20%.

`z`	`= OQ`
	`= sqrt(x^2 + ((16 – x^2)/4)^2), x > 0`

`text(Max or min when)\ \ (dz)/(dx)=0,`

`text(Solve:)\ (dz)/dx = 0quadtext(for)quadx > 0`

`=> x = 2sqrt2`

`=>z(2sqrt2) = 2sqrt3`

`:. text(Min distance of)\ \ 2sqrt3\ \ text(units when)\ \ x = 2sqrt2`

c. `text(Let)\ \ P(p, (16-p^2)/4)`

♦♦♦ Mean mark 8%.

`m_text(tan)\ text(at)\ P = – p/2`

`m_(PD) = ((16-p^2)/4 – k)/(p-0)`

`text(Equating gradients,)`

`text(Solve:)\ \ ((16-p^2)/4 – k)/p=-p/2\ \ text(for)\ p,`

`=> p=2sqrt(k – 4)`

`:.\ text(Gradient of tangent:)`

`gprime(2sqrt(k – 4)) = −sqrt(k – 4)`

d.i. `text(Equation of tangent:)`

♦♦♦ Mean mark 8%.

`y = −sqrt(k – 4)x + k`

`text(CAS: tangent Line)\ (g(x),x,2sqrt(k – 4))`

`xtext(-int of tangent:)\ x = k/(sqrt(k – 4))`

`:. A(k)`	`= 1/2 xx (k/(sqrt(k – 4))) xx k – int_0^4 g(x)\ dx`
	`= (k^2)/(2sqrt(k – 4)) – 32/3, \ \ k ∈ [5,8]`

d.ii. `text(Solve:)\ \ A(k)=0quadtext(for)quadk ∈ [5,8]`

♦♦♦ Mean mark 5%.

`=> k=16/3`

`text(Sketch the graph of)\ \ A(k)\ \ text(for)\ \ 5<=k<=8`

`:. A_text(max) = 16/3quadtext(when)quadk = 8`

d.iii. `A_text(min)\ text(occurs at the turning point)\ (k=16/3).`

♦♦♦ Mean mark 3%.

`A_text(min)`	`=A(16/3)`
	`=(64sqrt3)/9 – 32/3`

Calculus, MET2 2014 VCAA 21 MC

The trapezium `ABCD` is shown below. The sides `AB, BC` and `DA` are of equal length, `p.` The size of the acute angle `BCD` is `x` radians

The area of the trapezium is a maximum when the value of `x` is

`pi/12`
`pi/6`
`pi/4`
`pi/3`
`(5 pi)/12`

Show Answers Only

`D`

Show Worked Solution

`sin x`	`= h/p`
`:. h`	`= p sinx`

`cosx`	`= w/p`
`:.w`	`= pcos x`

♦♦ Mean mark 28%.

`A_text(trap)`	`= 1/2h xx [p + (p + 2w)]`
	`= (p sin x)/2 [2p + 2pcosx]`
	`= p^2sin x (1 + cosx), \ \ x ∈ (0,pi/2)`

`text(Find when)\ \ (dA)/(dp)=0,\ \ x ∈ (0,pi/2)`

`:. x = pi/3`

`=> D`

\(A\)	\(=\dfrac{h}{2}(a+b)\)
	\(=\dfrac{4x}{2}\sqrt{100-x^2}\quad\text{(Using Pythagoras)}\)
	\(=2x\sqrt{100-x^2}\)
\(\dfrac{dA}{dx}\)	\(=\dfrac{4(50-x^2)}{\sqrt{100-x^2}}\quad\text{(Using product rule)}\)

\(\dfrac{4(50-x^2)}{\sqrt{100-x^2}}\)	\(=0\quad(x \neq 10)\)
\(50-x^2\)	\(=0\)
(\(\sqrt{50}-x)(\sqrt{50}+x)\)	\(=0\)
\(\therefore\ x\)	\(=\sqrt{50}=5\sqrt{2}\quad (0<x<10)\)

\(x\)	\(7\)	\(\sqrt{50}\)	\(7.2\)
\(A^{\prime}\)	0.56	0	-1.06
\(\text{Gradient}\)	+	0	–

a.	\(f(0)\)	\(=a-0(0-2)^2=12\)
	\(\therefore\ a\)	\(=12\)

\(g(1)\)	\(=12\times 1+b\times 1^2\)
\(12+b\)	\(=9\)
\(\therefore b\)	\(=-3\)

b.	\(f(x)\)	\(=12-x(x-2)^2\)
		\(=-x^3+4x^2-4x+12\)
	\(f^{′}(x)\)	\(=-(3x^2-8x+4)\)

\( -(3x^2-8x+4)\)	\(=0\)
\(-(3x-2)(x-2)\)	\(=0\)

\(f(2)\)	\(=12-2(2-2)^2=12\)
\(f\Big{(}\dfrac{2}{3}\Big{)}\)	\(=12-\dfrac{2}{3}\Big(\dfrac{2}{3}-2\Big)^2<12\)

\(g(x)\)	\(=12x-3x^2\)
\(g(2)\)	\(=12\times 2-3\times 2^2=12\)