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v1 Algebra, STD2 A4 2017 HSC 28e

Sage brings 60 cartons of unpasteurised milk to the market each week. Each carton currently sells for $4 and at this price, all 60 cartons are sold each weekend.

Sage considers increasing the price to see if the total income can be increased.

It is assumed that for each $1 increase in price, 6 fewer cartons will be sold.

A graph showing the relationship between the increase in price per carton and the income is shown below.

 


 

  1. What price per carton should be charged to maximise the income?   (1 mark)

  2. What is the number of cartons sold when the income is maximised?   (1 mark)

  3. The cost of running the market stall is $40 plus $1.50 per carton sold.

    Calculate Sage's profit when the income earned from a day selling at the market is maximised.   (2 marks)

Show Answers Only

a.   `$7`

b.   `42`

c.   `$191`

Show Worked Solution

a.   `text(Graph is highest when increase = $3)`

`:.\ text(Carton price)\ = 4 + 3= $7`
 

b.   `text(Cartons sold)\ =60-(3 xx 6)=42`
  

c.   `text{Cost}\ = 42 xx 1.50 + 40 = $103`

`:.\ text(Profit when income is maximised)`

`= (42 xx 7)-103`

`= $191`

Algebra, STD2 A4 2020 HSC 19

A fence is to be built around the outside of a rectangular paddock. An internal fence is also to be built.

The side lengths of the paddock are `x` metres and `y` metres, as shown in the diagram.
 

 
A total of 900 metres of fencing is to be used. Therefore  `3x + 2y = 900`.
 
The area, `A`, in square metres, of the rectangular paddock is given by  `A =450x - 1.5x^2`.

The graph of this equation is shown.
  

  1. If the area of the paddock is `30 \ 000\ text(m)^2`, what is the largest possible value of `x`?   (1 mark)

  2. Find the values of `x` and `y` so that the area of the paddock is as large as possible.   (2 marks)

  3. Using your value from part (b), find the largest possible area of the paddock.   (1 mark)

Show Answers Only
  1. `200 \ text(m)`
  2. `x = 150 \ text(m and) \ y = 225 \ text(m)`
  3. `33 \ 750 \ text(m)^2`
Show Worked Solution

a.     `text(From the graph, an area of)\ 30\ 000\ text(m)^2`

♦ Mean mark part (a) 39%.

  `text(can have an)\ x text(-value of)\ \ x=100 or 200\ text(m.)`

`:. x_text(max) = 200 text(m)`
 

b.    `A_text(max) \ text(occurs when) \ \ x = 150`

♦♦ Mean mark part (b) 34%.

`text(Substitute)\ \ x=150\ \ text(into)\ \ 3x + 2y = 900:`

`3 xx 150 + 2y` `= 900`
`2y` `= 450`
`y` `= 225`

 
`therefore \ text(Maximum area when) \ \ x = 150 \ text(m  and) \ \ y = 225 \ text(m)`

♦ Mean mark part (c) 40%.
c.    `A_(max)` `= xy`
    `= 150 xx 225`
    `= 33 \ 750 \ text(m)^2`

Algebra, STD2 A4 2017 HSC 28e

A movie theatre has 200 seats. Each ticket currently costs $8.

The theatre owners are currently selling all 200 tickets for each session. They decide to increase the price of tickets to see if they can increase the income earned from each movie session.

It is assumed that for each one dollar increase in ticket price, there will be 10 fewer tickets sold.

A graph showing the relationship between an increase in ticket price and the income is shown below.
 


 

  1. What ticket price should be charged to maximise the income from a movie session?  (1 mark)

  2. What is the number of tickets sold when the income is maximised?  (1 mark)

  3. The cost to the theatre owners of running each session is $500 plus $2 per ticket sold.

     

    Calculate the profit earned by the theatre owners when the income earned from a session is maximised.  (2 marks)

Show Answers Only
  1. `$14`
  2. `140`
  3. `$1180`
Show Worked Solution

i.   `text(Graph is highest when increase = $6)`

♦ Mean mark 50%.

`:.\ text(Ticket price)\ = 8 + 6= $14`
 

ii.   `text(Solution 1)`

♦ Mean mark 45%.

`text(Tickets sold)\ =200-(4 xx 10)=140`
 

`text(Solution 2)`

`text(Tickets)\ = text(max income)/text(ticket price) = 1960/14= 140`
 

iii.  `text{Cost}\ = 140 xx $2 + $500= $780`

`:.\ text(Profit when income is maximised)`

`= 1960-780`

`= $1180`