Applications Unit Test Revision
Apps Assessment Standard 1.4
(Exercise 6Q)
1. A rectangular block has a square base of side x cm. Its surface area is 150 𝑐𝑚2.
The volume of the block is given by the formula
1
2
(𝒙) = 54𝑥 − 𝑥3
Find the value of x which maximises the volume of the block.
(5)
2. A wind shelter has a back, top and two square sides. The total amount of canvas used in
the shelter is 96 m2
The volume of the shelter is V cm3, given by the formula
V(x) = x(48 – x2)
Find the value of x which gives a maximum volume.
(5)
3. The volume of a wooden cuboid with a square base can be represented by the formula:
2 3
𝑥
3
Find the value of x which maximises the volume of the box.
𝑉(𝑥) = 162𝑥 −
(Exercise 9L, Q3, plus finding limits)
4. The diagram opposite shows the graph of y = 4x - x2
Calculate the shaded area.
(#2.1,4)
(5)
5. The diagram opposite shows the graph of
y = .x2 – 5x.
Calculate the shaded area.
(#2.1,4)
6. The diagram shows part of the graph
of y = 6x + 2x2
Calculate the shaded area.
(#2.1,4)
(Exercise 9P)
7. The line with equation y = -x + 8 and the curve with equation y = -x2 + 4x + 8 are shown
below:
(5)
The line and the curve meet at points x = 0 and x = 5. Calculate the shaded area.
8. The line with equation y = x + 3 and the curve with equation y = x2 – 2x + 3 are shown
below.
The line and the curve meet at the where x = 0 and x = 3. Calculate the shaded area. (5)
Answers
Applications Assessment Standard 1.4
1. one term correct 54 or … −
3
3
2
𝑥2
𝑉`(𝑥) = 54 − 2 𝑥 2 = 0 stated explicitly
x = -6 or +6
use 2nd derivative or nature table
maximum at x = 6
1 begin to differentiate
2 complete derivative and equate to 0
3 solve for x
4 justify nature of stationary points
5 interpretation
2. one term correct 48 or … − 3𝑥 2
1 begin to differentiate
𝑉`(𝑥) = 4 − 3𝑥 2 = 0 stated explicitly 2 complete derivative and equate to 0
x = 4 or −4
3 solve for x
use 2nd derivative or nature table
4 justify nature of stationary points
maximum at x = 4
5 interpretation
3. one term correct 4 or … − 3𝑥 2
𝑉`(𝑥) = 162 − 2𝑥 2 = 0 stated explicitly
x = 9 or −9
use 2nd derivative or nature table
maximum at x = 9
1 begin to differentiate
2 complete derivative and equate to 0
3 solve for x
4 justify nature of stationary points
5 interpretation
4. x = 0 and 4
#2.1 know to and find limits
4

1 know to integrate and interpret limits
.... dx
0
𝑥3 4
]0
3
43
[(2(42 ) − ( 3 ))
32
2
2 integrate
[2𝑥 2 −
Area =
5.
3
03
− (2(02 ) − ( 3 ))]
3 substitute limits
4 process limits
units
x = 0 and 5.
#2.1 know to and find limits
5
∫0 … 𝑑𝑥
1 know to integrate and interpret limits
[
2 integrate
𝑥3
3
53
−
[( 3 −
5𝑥 2 5
]0
2
5(5)2
Area = -
03
)−(3 −
2
125
6
units2
5(0)2
2
)]
3 substitute limits
4 process limits
6.
x = 0 and -3
#2.1 know to and find limits
0
∫−3 … 𝑑𝑥
1 know to integrate and interpret limits
𝑥3
2 integrate
[3𝑥 2 + 2 3 ]0-3
03
−33
[(3(02 ) + 2 ( 3 )) − (3(−32 ) + 2 (
3
5
∫0 … 𝑑𝑥
1 know to integrate and interpret limits
5
2 use ‘upper – lower’
∫0 (−𝑥 2 + 4𝑥 + 8) − (−𝑥 + 8) 𝑑𝑥
[
5𝑥 2
2
[((
−
𝑥3 5
]0
3
5(52 )
2
Area =
3 integrate
5(03 )
53
) − ( 3 )) − ((
2
03
) − ( 3 ))]
125
6
∫0 … 𝑑𝑥
1 know to integrate and interpret limits
3
2 use ‘upper – lower’
∫0 (𝑥 + 3) − (𝑥 2 − 2𝑥 + 3) 𝑑𝑥
[
4 substitute limits
5 process limits
units2
3
8.
3 substitute limits
4 process limits
Area = 9 units2
7.
))]
3𝑥 2
2
[((
−
𝑥3 3
]0
3
3(32 )
2
3 integrate
33
3(03 )
) − ( 3 )) − ((
9
Area = 2 units2
2
03
) − ( 3 ))]
4 substitute limits
5 process limits