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Algebra, STD2 A4 2024 HSC 26

A sheet of metal is folded to make a gutter, as shown. The cross-section of the gutter is a rectangle of width \(w\) cm and height \(h\) cm.
 

 

The area, \(A\) cm\(^{2}\), of the cross-section can be modelled by the quadratic formula

\(A=-0.5w^{2}+20w\)

A graph of this model is shown.
 

Find the width and height of the rectangle which will give the greatest possible area of the cross-section.   (3 marks)

Show Answers Only

\(w=20\ \text{cm}, h=10\ \text{cm}\)

Show Worked Solution

\(\text{Graph cuts}\ w\text{-axis at}\ \ w=0\ \ \text{and}\ \ w=40.\)

\(\text{By symmetry, maximum area occurs at}\ \ w=20.\)

\(A_{\text{max}}=-0.5 \times 20^2 + 20 \times 20 = 200\ \text{cm}^{2}\)

\(A_{\text{max}}\) \(=w \times h\)  
\(200\) \(=20 \times h\)  
\(h\) \(=10\ \text{cm}\)  

 
\(\therefore A_{\text{max}}\ \text{occurs when:}\ w=20\ \text{cm}, \ h=10\ \text{cm}\)

♦ Mean mark 43%.

Algebra, STD2 A4 2019 HSC 31

A rectangle has width `w` centimetres. The area of the rectangle, `A`, in square centimetres, is  `A = 2w^2 + 5w`.

The graph  `A = 2w^2 + 5w`  is shown.
 


 

  1. Explain why, in this context, the model  `A = 2w^2 + 5w`  only makes sense for the bold section of the graph.  (1 mark)

  2. The area of the rectangle is 18 cm². Calculate the perimeter of the rectangle.  (2 marks)

Show Answers Only
  1. `text(The width of a rectangle cannot be negative.)`
  2. `22\ text(cm)`
Show Worked Solution

a.   `text(The width of a rectangle cannot be negative.)`

♦ Mean mark 44%.

 

b.   `text(When)\ A = 18, w = 2`

♦ Mean mark 42%.

`text(Let)\ h =\ text(height of rectangle)`

`18` `= 2 xx h`
`h` `= 9\ text(cm)`

 

`:.\ text(Perimeter)` `= 2 xx (2 + 9)`
  `= 22\ text(cm)`

Algebra, STD2 A4 2012 HSC 30b

A golf ball is hit from point `A` to point `B`, which is on the ground as shown. Point `A` is 30 metres above the ground and the horizontal distance from point `A` to point `B` is  300 m.
 

The path of the golf ball is modelled using the equation 

`h = 30 + 0.2d-0.001d^2` 

where 

`h` is the height of the golf ball above the ground in metres, and 

`d` is the horizontal distance of the golf ball from point `A` in metres.

The graph of this equation is drawn below.

  

  1. What is the maximum height the ball reaches above the ground?    (1 mark)

  2. There are two occasions when the golf ball is at a height of 35 metres.

     

    What horizontal distance does the ball travel in the period between these two occasions?   (1 mark)

  3. What is the height of the ball above the ground when it still has to travel a horizontal distance of 50 metres to hit the ground at point `B`?   (1 mark)

  4. Only part of the graph applies to this model.

     

    Find all values of `d` that are not suitable to use with this model, and explain why these values are not suitable.   (2 marks)

Show Answers Only
  1. `40 text(m)`
  2. `140 text(m)`
  3. `text(17.5 m)`
  4. `d < 0\ text(and)\ d>300`
Show Worked Solution

i.   `text(Max height) = 40 text(m)`

COMMENT: With a mean mark of 92% in (i), a classic example of low hanging fruit in later questions.

 

ii.   `text(From graph)`

`h = 35\ text(when)\ x = 30\ text(and)\ x = 170`

`:.\ text(Horizontal distance)` `= 170-30`
  `= 140\ text(m)`

 

iii.   `text(Ball hits ground at)\ x = 300`

MARKER’S COMMENT: Responses for (iii) in the range  `17<=\ h\ <=18`  were deemed acceptable estimates read off the graph.

`=>text(Need to find)\ y\ text(when)\ x = 250`

`text(From graph,)\ y = 17.5 text(m)\ text(when)\ x = 250`

`:.\ text(Height of ball is 17.5 m at a horizontal)`

`text(distance of 50m before)\ B.`

 

iv.   `text(Values of)\ d\ text(not suitable).`

♦♦♦ Mean mark (iv) 12%
MARKER’S COMMENT: Many students did not refer to the domain `d>300` as unsuitable to the model.

`text(If)\ d < 0 text(, it assumes the ball is hit away)`

`text(from point)\ B text(. This is not the case in our)`

`text(example.)`

`text(If)\ d > 300 text(,)\ h\ text(becomes negative which is)`

`text(not possible given the ball cannot go)`

`text(below ground level.)`