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Calculus, MET1 2008 VCAA 9

A plastic brick is made in the shape of a right triangular prism. The triangular end is an equilateral triangle with side length `x` cm and the length of the brick is `y` cm.
 

met1-2008-vcaa-q91

The volume of the brick is 1000 cm³.

  1. Find an expression for `y` in terms of `x`.   (2 marks)

  2. Show that the total surface area, `A` cm², of the brick is given by
  3.    `A = (4000sqrt3)/x + (sqrt3 x^2)/2`   (2 marks)

  4. Find the value of `x` for which the brick has minimum total surface area. (You do not have to find this minimum.)   (3 marks)

Show Answers Only
  1. `y = (4000 sqrt3)/(3 x^2)`
  2. `text(See Worked Solutions)`
  3. `10 root(3)(4)`
Show Worked Solution

a.   `text(Cross-section is an equilateral triangle and)`

`text(angles of)\ Delta\ text(are)\ 60^@.`

`sin 60^@` `= h/x`
`:. h` `= sqrt3/2 x`

 

`text(Volume)` `= 1000`
`(1/2 x)(sqrt3/2 x)y` `= 1000`
`sqrt3/4 x^2y` `= 1000`
`:. y` `= 4000/(sqrt3 x^2)`
  `=(4000 sqrt3)/(3 x^2)`

 

♦ Mean mark 45%.
b.     `A` `= 2 xx (sqrt3/4 x^2) + 3 xx (xy)`
    `= sqrt3/2 x^2 + 3x(4000/(sqrt3 x^2))`
    `= (12\ 000)/(sqrt3 x) xx sqrt3/sqrt3 + sqrt3/2 x^2`
    `= (4000sqrt3)/x + (x^2 sqrt3)/2`

 

♦ Mean mark 42%.

c.   `A = 4000 sqrt3 x^(−1) + sqrt3/2 x^2,quadx > 0`

`text(Stationary Point when)\ \ (dA)/(dx)=0,`

`-4000sqrt3 x^(−2) + sqrt3x` `=0`
`(4000sqrt3)/(x^2)` `= sqrt3x`
`x^3` `= 4000`
`:. x` `= root(3)(4000)`
  `= 10 root(3)(4)`

Calculus, MET1 2010 VCAA 11

A cylinder fits exactly in a right circular cone so that the base of the cone and one end of the cylinder are in the same plane as shown in the diagram below. The height of the cone is 5 cm and the radius of the cone is 2 cm.

The radius of the cylinder is `r` cm and the height of the cylinder is `h` cm.

 

vcaa-2010-meth-11a

For the cylinder inscribed in the cone as shown above

  1. find `h` in terms of `r`.   (2 marks)

The total surface area, `S` cm², of a cylinder of height `h` cm and radius `r` cm is given by the formula

`S = 2 pi r h + 2 pi r^2`.

  1. find `S` in terms of `r`.   (1 mark)

  2. find the value of `r` for which `S` is a maximum.   (2 marks)

Show Answers Only
  1. `(10 -5r)/2`
  2. `10 pi r-3 pi r^2,\ \ 0 < r < 2`
  3. `5/3\ text(cm)`
Show Worked Solution

a.   `text(Sketch cross-section:)`

vcaa-2010-meth-11ai

`text(Using ratios of similar triangles,)`

♦ Mean mark 43%.

`Delta ABC\ text(|||)\ Delta PQC`

`5/h` `= 2/(2-r)`
`2h` `= 10-5r`
`:. h` `= (10-5r)/2`

 

♦ Mean mark 47%.
b.    `S` `= 2 pi r ((10-5 r)/2) + 2 pi r^2`
    `= 10 pi r-5 pi r^2 + 2 pi r^2`
    `= 10 pi r-3 pi r^2,\ \ 0 < r < 2`

 

c.   `text(SP’s occur when)\ \ (dS)/(dr) = 0,`

♦ Mean mark 46%.
`10 pi-6 pi r` `= 0`
`3r` `= 5`
`:. r` `= 5/3\ text(cm)`

Calculus, MET2 2014 VCAA 2

On 1 January 2010, Tasmania Jones was walking through an ice-covered region of Greenland when he found a large ice cylinder that was made a thousand years ago by the Vikings.

A statue was inside the ice cylinder. The statue was 1 m tall and its base was at the centre of the base of the cylinder.

VCAA 2014 2a

The cylinder had a height of `h` metres and a diameter of `d` metres. Tasmania Jones found that the volume of the cylinder was 216 m³. At that time, 1 January 2010, the cylinder had not changed in a thousand years. It was exactly as it was when the Vikings made it.

  1. Write an expression for `h` in terms of `d`.   (2 marks)

  2. Show that the surface area of the cylinder excluding the base, `S` square metres, is given by the rule  `S = (pi d^2)/4 + 864/d`.   (1 mark)

Tasmania found that the Vikings made the cylinder so that `S` is a minimum.

  1. Find the value of `d` for which `S` is a minimum and find this minimum value of `S`.   (2 marks)

  2. Find the value of `h` when `S` is a minimum.   (1 mark)

Show Answers Only
  1. `h = 864/(pi d^2)`
  2. `text(Proof)\ \ text{(See Worked Solutions)}`
  3. `12/(pi^(1/3))\ text(m);quad108pi^(1/3)\ text(square metres)`
  4. `6/(root(3)(pi))m`
Show Worked Solution
a.    `V` `= pir^2h`
  `216` `= pi(d/2)^2h`
  `:. h` `= (864)/(pi d^2)`

 

b.    `S` `= A_text(top) + A_text(curved surface)`
    `=pi xx (d/2)^2 + pi xx d xx h`

 
`text(Substitute)\ \ h = 864/(pid^2):`

`:. S` `= (pid^2)/4 + pi(d)(864/(pid^2))`
  `= (pid^2)/4 + 864/d`

 

c.  `(dS)/(dd) = (pi d)/2 -864/d^2`

`text(Stationary point when)\ \ (dS)/(dd)=0,`

`text(Solve:)\ \ (pi d)/2 -864/d^2=0\ \ text(for)\ d`

`d=12/(pi^(1/3))\ text(m)`

`:. S_text(min)=S(12/pi^(1/3)) = 108pi^(1/3)\ text(m²)`

 

d.   `text(Substitute)\ \ d=12/(pi^(1/3))\ \ text{into part (a):}`

♦ Mean mark (d) 42%.
MARKER’S COMMENT: An exact answer required here!

`h= 864/(pi(12/(pi^(1/3)))^2)`

   `= 6/(root(3)(pi))\ \ text(m)`