SOLVING PROBLEMS WITH CALCULUS (LIVE)
25 MAY 2015
Section A: Summary Notes
When plotting a graph, there are 6 steps to follow:






Draw diagrams to help you make sense of the problem
Read the question carefully and “translate” what has been given into an algebraic expression
If 2 variables are given use simultaneous equations
Write down 𝑓 𝑥 𝑎𝑛𝑑 𝑓′(𝑥)
Equate the derivative to 0 to calculate the maximum or minimum
Answer the question
Calculus of Motion



Distance is given by 𝑠 𝑡
Velocity (speed) is given by 𝑠′(𝑡)
Acceleration is given by 𝑠 ′′ 𝑡
Section B: Exam practice questions
Question 1
A rectangular area is to be enclosed on three sides by a fence and an existing hedge which forms the fourth side.
a)
Find the maximum area of the rectangular area and its dimensions if 100 metres of fencing is available.
b)
The cost of fencing is R20 per metre. Suppose that the rectangular area must be 200m . What are the
dimensions such that the cost will be a minimum?
2
Question 2
The height (in metres) of a stone that is thrown into the air is given by
a)
the average velocity of the stone during the first 2 s.
b)
the velocity of the stone after 1,5s
c)
when the velocity is zero
d)
the velocity at which the stone hits the ground
e)
the acceleration of the stone.
H (t )  20t  5t 2 after t s. Determine:
Question 3
2
The area of the figure given is 20 m .
a) Find an expression for the total area in terms of
b)
Show that
l
 , r and l .
40   r 2
4r
c) Show that the perimeter P of the figure is given by: P 
20

 2r  r
r
2
d) Calculate the width of the figure when P is a minimum.
Question 4
A rectangle PQRS is drawn with P and Q on the x-axis and R and is on the parabola
If Q is the point
x ; 0 , determine:
a)
the co-ordinates of P and R in terms of x
b)
PQ and QR in terms of x
c)
the area of PQRS in terms of x
d)
the area of the largest rectangle that can be drawn.
y  2 x 2  24 .
Section C: Solutions
Question 1
a)
b)
Area:
Cost:
A( x)  xy
C x   20 x  2 x  20 y
Perimeter:
 40 x  20 y
x  x  y  100
y  2 x  100
y  100  2 x
Area:
xy  200
y
A( x)  xy
 100  2 x x
 200 
C ( x)  40 x  20

 x 
4000
 40 x 
x
 100 x  2 x 2
A / ( x)  100  4 x
100  4 x  0
x  25
Minimize cost:
 A x   100 x  2 x 2
 A(25)  1250m
2
C ( x)  40 x  4000 x 1
C / ( x)  40  4000 x  2
4000
x2
0  40 x 2  4000
0  40 
Dimensions:
x  25m
200
x
y  100  2(25)  50m
x 2  100
x  10
Dimensions:
x  10 and
y  20
Question 2
b)
a)
v ave
H (2)  H (0)

20
20(2)  5(2) 2  200  5(0) 2

2
20

2
 10m / s

c)
 
vt   H / t 
 20  10t

at 1,5s
v1,5  20  10(1,5)
 5m / 2
d)
20t  5t 2  0
20  10t  0
5t 4  t   0
10t  20
t2
t  0 or
t4
v4  H / 4
 20  104
 20m / s
e)
a  v / (t )
 H // t 
 10m / s 2
Question 3
a)
1
A  l.2r  r 2
2
1 2
 2lr  r
2
b)
A  2lr 
20  2lr 

2

r2
r2
2
40  4lr  r 2
40  r 2  4lr
40  r 2
l
4r
c)
1
2r
2
P  2l  2r  r
P  l  l  2r 
 40  r 2 
  2r  r
P  2
 4r 
40  r 2
P
 2r  r
2r
40 r 2
P

 2r  r
2r 2r
20 r
P
  r  2r
r
2
Question 4
a)
b)


B x;18  2 x 
A x; 18  2 x
2
Area  CD.AD

 2 x  18  2 x 2
2
 36 x  4 x 3
b)
A( x)  36 x  4 x 3
A / ( x)  36  12 x 2
0  36  12 x 2
x2  3
x 3
x0
M ax area  36( 3 )  4( 3 ) 3
 24 3
