Question
A skip is a container used to carry garbage away from a construction site. For safety reasons, the garbage must not extend beyond the top of the skip. The maximum volume of garbage to be removed is therefore equal to the volume of the skip.
A particular design of skip can be modelled as a prism with a trapezoidal cross-section. For the skip to be transported, it must have a rectangular base of length 10 m and width 3 m. The length of the sloping edge is fixed at 4 m, and it makes an angle of \( \theta \) with the horizontal. The following diagram shows such a skip.
(a) Find the volume of this skip:
(i) if the length of the top edge of the skip is 11 m.
(ii) if the height of the skip is 3.2 m.
(iii) if \( \theta = 60^\circ \).
(b) Show that the volume, \( V \, \text{m}^3 \), of the skip is given by:
\[ V = 24 \sin(\theta)(5 + \cos(\theta)) \]
(c) Explain, in context, why \( \theta \neq 0 \).
(d) (i) Sketch the graph of \( V = 24 \sin(\theta)(5 + \cos(\theta)) \), \( 0 < \theta < 90^\circ \).
(ii) Find the maximum volume of the skip and the value of \( \theta \) for which this maximum volume occurs.
▶️Answer/Explanation
Detailed solution
(a) (i) Volume of the skip when the top edge is 11 m:
The skip is modelled as a prism with a trapezoidal cross-section. The volume \( V \) is given by:
\[ V = \text{Area of trapezoid} \times \text{Length of the skip} \]
The trapezoid has two parallel sides: the base (10 m) and the top edge (11 m). The height of the trapezoid can be found using the Pythagorean theorem, as the sloping edge is 4 m and the horizontal difference between the top and base is \( 11 – 10 = 1 \) m.
\[ \text{Height} = \sqrt{4^2 – 1^2} = \sqrt{16 – 1} = \sqrt{15} \]
The area of the trapezoid is:
\[ \text{Area} = \frac{1}{2} \times (10 + 11) \times \sqrt{15} = \frac{21}{2} \times \sqrt{15} \]
Multiply by the length of the skip (3 m) to get the volume:
\[ V = \frac{21}{2} \times \sqrt{15} \times 3 = \frac{63}{2} \times \sqrt{15} \approx 122 \, \text{m}^3 \]
So, the volume is 122 m³.
(a) (ii) Volume of the skip when the height is 3.2 m:
Given the height of the skip is 3.2 m, we can find the horizontal projection of the sloping edge using the Pythagorean theorem:
\[ \text{Horizontal projection} = \sqrt{4^2 – 3.2^2} = \sqrt{16 – 10.24} = \sqrt{5.76} = 2.4 \, \text{m} \]
The top edge of the skip is then:
\[ \text{Top edge} = 10 + 2.4 = 12.4 \, \text{m} \]
The area of the trapezoid is:
\[ \text{Area} = \frac{1}{2} \times (10 + 12.4) \times 3.2 = \frac{22.4}{2} \times 3.2 = 35.84 \, \text{m}^2 \]
Multiply by the length of the skip (3 m) to get the volume:
\[ V = 35.84 \times 3 = 107.52 \, \text{m}^3 \approx 108 \, \text{m}^3 \]
So, the volume is 108 m³.
(a) (iii) Volume of the skip when \( \theta = 60^\circ \):
When \( \theta = 60^\circ \), the height of the skip is:
\[ \text{Height} = 4 \sin(60^\circ) = 4 \times \frac{\sqrt{3}}{2} = 2\sqrt{3} \, \text{m} \]
The horizontal projection of the sloping edge is:
\[ \text{Horizontal projection} = 4 \cos(60^\circ) = 4 \times 0.5 = 2 \, \text{m} \]
The top edge of the skip is then:
\[ \text{Top edge} = 10 + 2 = 12 \, \text{m} \]
The area of the trapezoid is:
\[ \text{Area} = \frac{1}{2} \times (10 + 12) \times 2\sqrt{3} = \frac{22}{2} \times 2\sqrt{3} = 22\sqrt{3} \, \text{m}^2 \]
Multiply by the length of the skip (3 m) to get the volume:
\[ V = 22\sqrt{3} \times 3 = 66\sqrt{3} \approx 114 \, \text{m}^3 \]
So, the volume is 114 m³.
(b) Deriving the volume formula:
The volume \( V \) of the skip is given by:
\[ V = \text{Area of trapezoid} \times \text{Length of the skip} \]
The area of the trapezoid is:
\[ \text{Area} = \frac{1}{2} \times (\text{Base} + \text{Top edge}) \times \text{Height} \]
Here, the base is 10 m, the top edge is \( 10 + 4 \cos(\theta) \), and the height is \( 4 \sin(\theta) \). Thus:
\[ \text{Area} = \frac{1}{2} \times (10 + 10 + 4 \cos(\theta)) \times 4 \sin(\theta) = \frac{1}{2} \times (20 + 4 \cos(\theta)) \times 4 \sin(\theta) \]
Simplify:
\[ \text{Area} = (10 + 2 \cos(\theta)) \times 4 \sin(\theta) = 40 \sin(\theta) + 8 \sin(\theta) \cos(\theta) \]
Multiply by the length of the skip (3 m):
\[ V = 3 \times (40 \sin(\theta) + 8 \sin(\theta) \cos(\theta)) = 120 \sin(\theta) + 24 \sin(\theta) \cos(\theta) \]
Factor out \( 24 \sin(\theta) \):
\[ V = 24 \sin(\theta)(5 + \cos(\theta)) \]
Thus, the volume is given by:
\[ V = 24 \sin(\theta)(5 + \cos(\theta)) \]
(c) Why \( \theta \neq 0 \):
If \( \theta = 0 \), the sloping edge would be horizontal, and the skip would have no height. This would result in zero volume, making it impossible to carry any garbage. Additionally, the contents of the skip would fall out if the angle were zero.
(d) (i) Graph of \( V = 24 \sin(\theta)(5 + \cos(\theta)) \):
The graph of \( V = 24 \sin(\theta)(5 + \cos(\theta)) \) for \( 0 < \theta < 90^\circ \) is a curve that starts at \( V = 0 \) when \( \theta = 0 \), reaches a maximum, and then decreases back to \( V = 0 \) when \( \theta = 90^\circ \).
(d) (ii) Maximum volume and corresponding \( \theta \):
To find the maximum volume, we differentiate \( V \) with respect to \( \theta \) and set the derivative equal to zero:
\[ V = 24 \sin(\theta)(5 + \cos(\theta)) \]
Differentiate using the product rule:
\[ \frac{dV}{d\theta} = 24 [\cos(\theta)(5 + \cos(\theta)) + \sin(\theta)(-\sin(\theta))] = 24 [5 \cos(\theta) + \cos^2(\theta) – \sin^2(\theta)] \]
Set \( \frac{dV}{d\theta} = 0 \):
\[ 5 \cos(\theta) + \cos^2(\theta) – \sin^2(\theta) = 0 \]
Using \( \sin^2(\theta) = 1 – \cos^2(\theta) \), we get:
\[ 5 \cos(\theta) + \cos^2(\theta) – (1 – \cos^2(\theta)) = 0 \implies 5 \cos(\theta) + 2 \cos^2(\theta) – 1 = 0 \]
Solve the quadratic equation for \( \cos(\theta) \):
\[ 2 \cos^2(\theta) + 5 \cos(\theta) – 1 = 0 \]
Using the quadratic formula:
\[ \cos(\theta) = \frac{-5 \pm \sqrt{25 + 8}}{4} = \frac{-5 \pm \sqrt{33}}{4} \]
Only the positive root is valid since \( \cos(\theta) \) must be between 0 and 1:
\[ \cos(\theta) = \frac{-5 + \sqrt{33}}{4} \approx 0.186 \]
Thus, \( \theta \approx \cos^{-1}(0.186) \approx 79.3^\circ \).
Substitute \( \theta = 79.3^\circ \) into the volume formula:
\[ V = 24 \sin(79.3^\circ)(5 + \cos(79.3^\circ)) \approx 122 \, \text{m}^3 \]
So, the maximum volume is 122 m³, and it occurs at \( \theta \approx 79.3^\circ \).
……………………………Markscheme……………………………….
Ans:
(a) (i) \( V = 122 \, \text{m}^3 \)
(a) (ii) \( V = 108 \, \text{m}^3 \)
(a) (iii) \( V = 114 \, \text{m}^3 \)
(b) \( V = 24 \sin(\theta)(5 + \cos(\theta)) \)
(c) If \( \theta = 0 \), the skip would have zero volume, and the contents would fall out.
(d) (i) Graph of \( V = 24 \sin(\theta)(5 + \cos(\theta)) \)
(d) (ii) Maximum volume = 122 m³ at \( \theta = 79.3^\circ \)