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Calculus, MET2 2022 VCAA 19 MC

A box is formed from a rectangular sheet of cardboard, which has a width of `a` units and a length of `b` units, by first cutting out squares of side length `x` units from each corner and then folding upwards to form a container with an open top.

The maximum volume of the box occurs when `x` is equal to

  1. `\frac{a-b+\sqrt{a^2-a b+b^2}}{6}`
  2. `\frac{a+b+\sqrt{a^2-a b+b^2}}{6}`
  3. `\frac{a-b-\sqrt{a^2-a b+b^2}}{6}`
  4. `\frac{a+b-\sqrt{a^2-a b+b^2}}{6}`
  5. `\frac{a+b-\sqrt{a^2-2 a b+b^2}}{6}`
Show Answers Only

`D`

Show Worked Solution

`V(x)=x(b-2 x)(a-2 x)=abx-2(a+b)x^2+4x^3`

Maximum occurs when `V\^{\prime} (x)=0`

`V\^{\prime} (x)=ab-4(a+b)x+12x^2=0`

Solving for `x` using CAS.

`x=\frac{a+b \+ \sqrt{a^2-a b+b^2}}{6}`  or  `x=\frac{a+b \- \sqrt{a^2-a b+b^2}}{6}`

Testing shape of the cubic using `b = 2` and `a = 1`

Maximum occurs at the smaller value of `x`

`:.\  x=\frac{a+b \- \sqrt{a^2-a b+b^2}}{6}`

`=>D`


♦♦ Mean mark 34%.
MARKER’S COMMENT: 34% of students incorrectly chose option B. 

Calculus, MET2 2021 VCAA 1

A rectangular sheet of cardboard has a width of `h` centimetres. Its length is twice its width.

Squares of side length `x` centimetres, where  `x > 0`  are cut from each of the corners, as shown in the diagram below.
 

The sides  of this sheet of cardboard are then folded up to make a rectangular box with an open top, as shown in the diagram below.

Assume that the thickness of the cardboard is negligible and that  `V_text{box} > 0`.
 

A box is to be made from a sheet of cardboard with  `h` = 25 cm.

  1. Show that the volume, `V_text{box} > 0`, in cubic centimetres, is given by  `V_text{box} (x) = 2x (25-2x)(25-x)`.   (1 mark)

  2. State the domain of  `V_text{box}`.   (1 mark)

  3. Find the derivative of  `V_text{box}`  with respect to `x`.   (1 mark)

  4. Calculate the maximum possible volume of the box and for which value of `x` this occurs.   (3 marks)

  5. Waste minimisation is a goal when making cardboard boxes.
  6. Percentage wasted is based on the area of the sheet of cardboard that is cut out before the box is made.
  7. Find the percentage of the sheet of cardboard that is wasted when `x = 5`.   (2 marks)

Now consider a box made from a rectangular sheet of cardboard where  `h>0` and the box's length is still twice its width.

    1. Let  `V_text{box}` be the function that gives the volume of the box.
    2. State the domain `V_text{box}` in terms of `h`.   (1 mark)

    3. Find the maximum volume for any such rectangular box, `V_text{box}`, in terms of  `h`.   (3 marks)

  2. Now consider making a box from a square sheet of cardboard with side lengths of `h` centimetres.
  3. Show that the maximum volume of the box occurs when  `x = h/6`.   (2 marks)

Show Answers Only
  1. `text{Show Worked Solutions}`
  2. `(0, 12.5)`
  3. `text{Show Worked Solutions}`
  4. `{15625 sqrt3}/{9}  \ text{or} \ {15625}/{3 sqrt3}`
  5. `8%`
  6. i.  `text{Domain} \ (V_text{box}) = (0, h/2)`
  7. (sqrt3 h^3)/9`
  8. `text(See Worked Solutions)`
Show Worked Solution
a.   `V` `= x (h-2x)(2h-2x)`
    `= x (25-2x)(50-2x)`
    `= 2x (25-2x)(25-x)`

 

b.   `text{Sketch} \ y = 2x (25-2x)(25-x) \ text{by CAS:} `

♦ Mean mark (b) 42%.

`text{Domain is} \ (0, 12.5)`
 

c.    `(dV)/dx = 12x^2-300x + 1250`
 

d.    `text{Solve} \ V^{′}(x) = 0 \ text{for} \ x :`

`x = {-25 (sqrt3-3)}/6 \ or \ x = {25(sqrt3 + 3)}/6`

`:. \ x =  {-25 (sqrt3-3)}/6 \ , \ \ x ∈ (0, 12.5)`

`:. \ V_text{max} = {15\ 625 sqrt3}/9  \ text{or} \  \ {15\ 625}/{3 sqrt3}`
 

e.   `text{Area cut out} = 4 xx 5^2 = 100\ text(cm)^2`

`text{% Wasted}` `= {100}/{25 xx 50} xx 100`
  `= 8text(%)`

♦ Mean mark (f.i.) 33%.

 
f.i.  `text{Domain} \ (V_text{box}) = (0, h/2), \ \ text{(using part b)}`
 

f.ii.  `V = x (h-2x)(2h-2x)`

♦ Mean mark (f.ii.) 45%.

`text{Solve} \ V^{′}(x) = 0 \ text{for} \ x:`

`x = {-h (sqrt3-3)}/{6} \ \ text{or} \ \ x = {h(sqrt{3} + 3)}/{6}`

`V_text{max} \ text{occurs when} \ \ x = {-h(sqrt3-3)}/{6} \ \ text{(see part a)}`

`V_text{max} = (sqrt3 h^3)/9`

 

g.    `V = x(h-2x)(h-2x)`

♦ Mean mark part (g) 40%.

`text{Solve} \ V^{′}(x) = 0 \ text{for} \ x:`

`x = h/2 \ \ text{or} \ \ x =  h/6`

`V (h/6) = {2h^3}/{27} \ , \ V(h/2) = 0`

`:. V_text{max} \ text{occurs when} \ \ x = {h}/{6}`

Calculus, MET2 2019 VCAA 6 MC

A rectangular sheet of cardboard has a length of 80 cm and a width of 50 cm. Squares, of side length `x` centimetres, are cut from each of the corners, as shown in the diagram below.
 

 

A rectangular box with an open top is then constructed, as shown in the diagram below.
 

 

The volume of the box is a maximum when `x` is equal to

  1. `10`
  2. `20`
  3. `25`
  4. `100/3`
  5. `200/3`
Show Answers Only

`A`

Show Worked Solution
`text(Base of box)` `= (80-2x) xx (50-2x)`
  `= 4000-260x + 4x^2`
`text(Volume)` `= 4x^3-260x^2 + 4000x`

 
`(dV)/(dx) = 12x^2-520x + 4000`

`(dV)/(dx) = 0\ \ text(when)\ \ x = 10`

`=>   A`

Calculus, MET2 2010 VCAA 3

An ancient civilisation buried its kings and queens in tombs in the shape of a square-based pyramid, `WABCD.`

The kings and queens were each buried in a pyramid with  `WA = WB = WC = WD = 10\ text(m).`

Each of the isosceles triangle faces is congruent to each of the other triangular faces.

The base angle of each of these triangles is `x`, where  `pi/4 < x < pi/2.`

Pyramid `WABCD` and a face of the pyramid, `WAB`, are shown here.
 

VCAA 2010 3a

`Z` is the midpoint of `AB.`

  1. i. Find `AB` in terms of `x`.   (1 mark)

  2. ii. Find `WZ` in terms of `x`.   (1 mark)

  3. Show that the total surface area (including the base), `S\ text(m)^2`, of the pyramid, `WABCD`, is given by 
  4.      `S = 400(cos^2 (x) + cos (x) sin (x))`.   (2 marks)

  5. Find `WY`, the height of the pyramid `WABCD`, in terms of `x`.    (2 marks)

  6. The volume of any pyramid is given by the formula  `text(Volume) = 1/3 xx text(area of base) xx text(vertical height)`.
  7. Show that the volume, `T\ text(m³)`, of the pyramid `WABCD`  is  `4000/3 sqrt(cos^4 x-2 cos^6 x)`.   (1 mark)

Queen Hepzabah’s pyramid was designed so that it had the maximum possible volume.

  1. Find  `(dT)/(dx)`  and hence find the exact volume of Queen Hepzabah’s pyramid and the corresponding value of `x`.   (4 marks)

Queen Hepzabah’s daughter, Queen Jepzibah, was also buried in a pyramid. It also had

`WA = WB = WC = WD = 10\ text(m.)`

The volume of Jepzibah’s pyramid is exactly one half of the volume of Queen Hepzabah’s pyramid. The volume of Queen Jepzibah’s pyramid is also given by the formula for `T` obtained in part d.

  1. Find the possible values of `x`, for Jepzibah’s pyramid, correct to two decimal places.   (2 marks)

Show Answers Only

a.i. `20 cos (x)`

a.ii.`10 sin (x)`

b.   `text(Proof)\ \ text{(See Worked Solutions)}`

c.   `10 sqrt (sin^2(x)-cos^2 (x))`

d.   `text(Proof)\ \ text{(See Worked Solutions)}`

e.   `x = cos^-1 (sqrt 3/3) -> T_max = (4000 sqrt 3\ m^3)/27`

f.   `x dot = 0.81 or x dot = 1.23`

Show Worked Solution
a.i.   `cos x` `= (1/2 AB)/10`
  `:. AB` `= 20 cos(x)`

 

  ii.   `sin (x)` `= (wz)/10`
  `:. wz` `= 10 sin (x)`

 

b.   `text(Area base)` `= (20 cos (x))^2`
    `= 400 cos^2(x)`
  `4 xx text(Area)_Delta` `= 4 xx (1/2 xx 20 cos (x) xx 10 sin (x))`
    `= 400 cos (x) sin (x)`
♦ Mean mark 47%.

 

`:. S` `= 400 cos^2 (x) + 400 cos (x) sin (x)`
  `= 400 (cos^2 (x) + cos (x) sin (x))\ \ text(… as required.)`

 

c.   `text(Using)\ \ Delta WYZ,`

 vcaa-graphs-fur2-2010-ci

`text(Using Pythagoras,)`

`WY` `= sqrt (10^2 sin^2 (x)-10^2 cos^2 (x))`
  `= 10 sqrt (sin^2(x)-cos^2 (x))`
♦♦♦ Mean mark part (d) 22%.

 

d.   `T` `= 1/3 xx text(base) xx text(height)`
    `= 1/3 xx (400 cos^2 (x)) xx (10 sqrt(sin^2 (x)-cos^2 (x)))`
    `= 4000/3 sqrt (cos^4 (x) (sin^2 (x)-cos^2 (x))`

 

`text(Using)\ \ sin^2 (x) = 1-cos^2 (x),`

`T` `= 4000/3 sqrt (cos^4 (x) (1-cos^2 (x)-cos^2 (x))`
  `= 4000/3 sqrt (cos^4 (x)-2 cos^6 (x))`

 

e.   `(dT)/(dx) = (8000 cos (x) sin (x) (3 cos^2 (x)-1))/(3 sqrt(1-2 cos^2 (x)))`

♦ Mean mark part (e) 45%.

 

`text(Stationary point when,)`

`(dT)/(dx) = 0\ \ text(for)\ \ x in (pi/4, pi/2)`

`:. x = cos^-1 (sqrt 3/3)`

`:.T_max` `=T(cos^-1 (sqrt 3/3))`
  `= (4000 sqrt 3)/27\ \ text(m³)`

♦♦♦ Mean mark part (f) 16%.

 

f.   `text(Solve)\ \ T(x) = (2000 sqrt 3)/27\ \ text(for)\ \ x in (pi/4, pi/2)`

`:. x = 0.81  or  x = 1.23\ \ text{(2 d.p.)}`

Calculus, MET2 2014 VCAA 15 MC

Zoe has a rectangular piece of cardboard that is 8 cm long and 6 cm wide. Zoe cuts squares of side length `x` centimetres from each of the corners of the cardboard, as shown in the diagram below.

VCAA 2014 15mc

Zoe turns up the sides to form an open box.

VCAA 2014 15mci

The value of `x` for which the volume of the box is a maximum is closest to

  1. `0.8`
  2. `1.1`
  3. `1.6`
  4. `2.0`
  5. `3.6`
Show Answers Only

`B`

Show Worked Solution
`V` `=\ text(Base × Height)`
  `=x(8 – 2x)(6 – 2x),qquadx ∈ (0,3)`

 

`text(Solve:)\ \ V′=0,`

♦ Mean mark 44%.

`[text(CAS:)\ ftext(Max) (x(8 – 2x)(6 – 2x),x,0,3)]`

`V_text(max) -> x` `= 1.13`

`=>   B`

Calculus, MET2 2012 VCAA 1

A solid block in the shape of a rectangular prism has a base of width `x` cm. The length of the base is two-and-a-half times the width of the base.

 VCAA 2012 1a

The block has a total surface area of 6480 sq cm.

  1. Show that if the height of the block is `h` cm,  `h = (6480-5x^2)/(7x).`   (2 marks)

  2. The volume, `V` cm³, of the block is given by  `V(x) = (5x(6480-5x^2))/14.`
  3. Given that  `V(x) > 0`  and  `x > 0`, find the possible values of `x`.   (2 marks)

  4. Find  `(dV)/(dx)`, expressing your answer in the form  `(dV)/(dx) = ax^2 + b`, where `a` and `b` are real numbers.   (3 marks)

  5. Find the exact values of `x` and `h` if the block is to have maximum volume.   (2 marks)

Show Answers Only
  1. `text(Proof)\ \ text{(See Worked Solutions)}`
  2. `x ∈ (0,36)`
  3. `(dV)/(dx) = −75/14 x^2 + (16\ 200)/7`
  4. `(120sqrt3)/7`
Show Worked Solution

a.   `text(Total surface area) = 6480`

`6480` `= 2 xx (5/2 x xx h) + 2xx(5/2x^2) + 2 xx (xh)`
`6480` `= 5xh + 5x^2 + 2xh`
`7xh` `= 6480-5x^2`
`:. h` `= (6480-5x^2)/(7x)\ …\ text(as required)`

 

b.   `V (x) > 0quadtext(for)quadx > 0,`

`=>(6480-5x^2)` `>0`
`5x^2` `<6480`
`x` `<36`

`:. x ∈ (0,36)`

 

c.    `V(x)` `=(5x(6480-5x^2))/14`
    `=(-25x^3)/14 + (16\ 200x)/7`
  `(dV)/(dx)` `=- 75/14 x^2 + (16\ 200)/7`

 

d.   `text(S.P. when)\ (dV)/(dx) = 0quadtext(for)quad x > 0`

`text(Solve:)\ \-75/14 x^2 + (16\ 200)/7=0\ \ text(for)\ x,`

`=> x = 12sqrt3`

`text(Substitute)\ \ x = 12sqrt3quadtext(for)quadh,`

`:. h` `= (6480-5(12sqrt3)^2)/(7(12sqrt3))`
  `= (120sqrt3)/7`