Download Worksheet 2.2

Maths Quest Maths B Year 12 for Queensland
WorkSHEET 2.2
Chapter 2 Applications of differentiation WorkSHEET 2.2
1
Applications of differentiation
Name: _________________________
1
Write an expression for each derivative
indicated:
1
dP
;
(a) P  h 4  h
2
dh
(b)
T  2we 3w
;
dT
dw
1 4
h  h
2
1
dP
1 
 2h 3  h 2
dh
2
(a)
P
(b)
T  2we3 w
Let
u  2w
v  e 3w
du
dv
2
 3e  3w
dw
dw
dT
dv
du
u
v
dw
dw
dw
3w
 2 w  3e  e  3 w  2
 2e  3 w 1  3w
2
The monthly profit, P ($), of a manufacturing
P   4w3  300w  1000
company is related to the number of workers, w
dP
 12w2  300  0 for maximum profit .
where P  4w3  300w  1000.
dw
w2  25
How many workers should be employed for
maximum profit and what is the profit?
w5
P  45  3005  1000  2000
3
Therefore 5 workers should be employed for
maximum profit $2000.
3
The cost C ($ / h) of running a ferry at a
constant speed of v (km / h) is
5625 4v 2
C

.
v
75
At what speed will the cost be a minimum?
C
5625 4v 2

v
75
dC
8v
 5625v  2 
 0 for minimum cost
dv
75
5625 8v

v2
75
5625
 75
v3 
8
v  37.5 km / h
Maths Quest Maths B Year 12 for Queensland
4
Chapter 2 Applications of differentiation WorkSHEET 2.2
The cost of production, C ($), of producing n
1
pairs of scissors is given by C  n 2  300 .
8
If each pair of scissors is sold for $15, how
many pairs of scissors should be produced for
maximum profit? What is the profit?
1
P  15n  ( n 2  300)
8
dP
1
 15  n  0
dn
4
1
15  n
4
n  60
1

P  15  60    602  300   $150
8

60 pairs of scissors should be produced for a
profit of $150.
5
A farmer plans to use a river as one boundary
of a rectangular paddock. If the farmer has
960 metres of fencing to be used to fence the
other 3 sides, what dimensions should the
paddock be to ensure maximum area?
Let x  width of paddock, length of paddock is
960  2x.
Area  A(x)  x( 960  2 x)  960 x  2 x 2
A x   960  4 x  0 for maximum area
4 x  960
x  240 metres  width
length  960  2  240  480 metres
6
What positive number when cubed and when
added to its own reciprocals cubed leads to a
minimum value?
Let n be the numbers and T be the total.
1
T n   n 3  3
n
3
T n   n  n 3
T / n   3n 2  3n  4  0 for a minimum value
3
3n 2  4
n
6
n 1
n 1
n  1 is the positive number
7
The sum of two numbers is 80. What are the
two numbers if the sum of their squares is a
minimum?
Let x be one number.
The other number = 80 x.
2
T  x   x 2  80  x 
 x 2  6400  160 x  x 2
 2 x 2  160 x  6400
T
/
 x   4 x  160  0 for a minimum
4 x  160
x  40
Therefore both numbers are 40.
2
Maths Quest Maths B Year 12 for Queensland
8
Chapter 2 Applications of differentiation WorkSHEET 2.2
A closed box is to be constructed that has the
following conditions:
(i) a square base
(ii) a volume of 27 000 cm3
(iii) minimum surface area
Let x be length of base and h be the height.
Volume = x2h  27 000
27 000
So h 
x2
Surface area = S  2 x 2  4 xh
What are the dimensions of the box?
Substitute for h to get
27 000
S  2x 2  4x 
x2
S  2 x 2  108 000 x 1
dS
 4 x  108 000 x  2  0
dx
108 000
4x 
x2
4 x 3  108 000
x 3  27 000
x  30
Dimensions of box are 30 cm  30 cm 30 cm.
3
Maths Quest Maths B Year 12 for Queensland
9
Find the minimum distance the line
2 x  3 y  6  0 is from the origin.
Chapter 2 Applications of differentiation WorkSHEET 2.2
4
Let x, y  be the point on the line that is closest
to the origin. The distance will be
Now 2 x  3 y  6  0
2x  3y  6
x
x2  y 2
1
3 y  6
2
Let distance = L =
x2  y 2

1
3 y  6 2  y 2
4

9 y 2  36 y  36  4 y 2
4
1
13 y 2  36 y  36
2
dL
0
This is a minimum when
dy



1

dL 1 1
   26 y  36  13 y 2  36 y  36 2
dy 2 2
26 y  36
0
4 13 y 2  36 y  36
26 y  36
18
13
1  18
12

x  3  6  
2  13
13

y
distance =
x2  y 2 
144 324 2 117


169 169
13