The document discusses how to solve maximization and minimization word problems. It provides an example of maximizing the total area formed by cutting a rope into two pieces to form a square and rectangle. The steps are given as: (1) write the problem as two equations, (2) make one variable the subject of one equation, (3) substitute and take the derivative, (4) set the derivative equal to 0 to find the maximum/minimum. The maximum total area occurs when the side of the square is 1/2 the length of the rope. A second example finds the radius, height, and maximum volume of a cylindrical bucket given its total surface area.

1. Maxima & Minima
Problems

2. Maxima & Minima
Problems
When solving word problems we need to;

3. Maxima & Minima
Problems
When solving word problems we need to;
(1) Reduce the problem to a set of equations
(there will usually be two)

4. Maxima & Minima
Problems
When solving word problems we need to;
(1) Reduce the problem to a set of equations
(there will usually be two)
a) one will be what you are maximising/minimising

5. Maxima & Minima
Problems
When solving word problems we need to;
(1) Reduce the problem to a set of equations
(there will usually be two)
a) one will be what you are maximising/minimising
b) one will be some information given in the problem

6. Maxima & Minima
Problems
When solving word problems we need to;
(1) Reduce the problem to a set of equations
(there will usually be two)
a) one will be what you are maximising/minimising
b) one will be some information given in the problem
(2) Rewrite the equation we are trying to minimise/maximise with
only one variable

7. Maxima & Minima
Problems
When solving word problems we need to;
(1) Reduce the problem to a set of equations
(there will usually be two)
a) one will be what you are maximising/minimising
b) one will be some information given in the problem
(2) Rewrite the equation we are trying to minimise/maximise with
only one variable
(3) Use calculus to solve the problem

8. e.g. A rope 36 metres in length is cut into two pieces. The first is bent
to form a square, the other forms a rectangle with one side the
same length as the square.
Find the length of the square if the sum of the areas is to be a
maximum.

9. e.g. A rope 36 metres in length is cut into two pieces. The first is bent
to form a square, the other forms a rectangle with one side the
same length as the square.
Find the length of the square if the sum of the areas is to be a
maximum.

10. e.g. A rope 36 metres in length is cut into two pieces. The first is bent
to form a square, the other forms a rectangle with one side the
same length as the square.
Find the length of the square if the sum of the areas is to be a
maximum.
s
s

11. e.g. A rope 36 metres in length is cut into two pieces. The first is bent
to form a square, the other forms a rectangle with one side the
same length as the square.
Find the length of the square if the sum of the areas is to be a
maximum.
s r
s s

12. e.g. A rope 36 metres in length is cut into two pieces. The first is bent
to form a square, the other forms a rectangle with one side the
same length as the square.
Find the length of the square if the sum of the areas is to be a
maximum.
s r
s s
A  s 2  rs 1

13. e.g. A rope 36 metres in length is cut into two pieces. The first is bent
to form a square, the other forms a rectangle with one side the
same length as the square.
Find the length of the square if the sum of the areas is to be a
maximum.
s r
s s
A  s 2  rs 1 6 s  2r  362 

14. e.g. A rope 36 metres in length is cut into two pieces. The first is bent
to form a square, the other forms a rectangle with one side the
same length as the square.
Find the length of the square if the sum of the areas is to be a
maximum.
s r
s s
A  s 2  rs 1 6 s  2r  362 
make r the subject in 2 

15. e.g. A rope 36 metres in length is cut into two pieces. The first is bent
to form a square, the other forms a rectangle with one side the
same length as the square.
Find the length of the square if the sum of the areas is to be a
maximum.
s r
s s
A  s 2  rs 1 6 s  2r  362 
make r the subject in 2 
2r  36  6 s
r  18  3s

16. e.g. A rope 36 metres in length is cut into two pieces. The first is bent
to form a square, the other forms a rectangle with one side the
same length as the square.
Find the length of the square if the sum of the areas is to be a
maximum.
s r
s s
A  s 2  rs 1 6 s  2r  362 
make r the subject in 2 
2r  36  6 s
r  18  3s
substitute into 1

17. e.g. A rope 36 metres in length is cut into two pieces. The first is bent
to form a square, the other forms a rectangle with one side the
same length as the square.
Find the length of the square if the sum of the areas is to be a
maximum.
s r
s s
A  s 2  rs 1 6 s  2r  362 
make r the subject in 2 
2r  36  6 s
r  18  3s
substitute into 1
A  s 2  18  3s s

18. e.g. A rope 36 metres in length is cut into two pieces. The first is bent
to form a square, the other forms a rectangle with one side the
same length as the square.
Find the length of the square if the sum of the areas is to be a
maximum.
s r
s s
A  s 2  rs 1 6 s  2r  362 
make r the subject in 2 
2r  36  6 s
r  18  3s
substitute into 1
A  s 2  18  3s s
A  s 2  18s  3s 2
A  18s  2 s 2

19. A  18s  2 s 2

20. A  18s  2 s 2
dA
 18  4 s
ds

21. A  18s  2 s 2
dA
 18  4 s
ds
d2A
2
 4
ds

22. A  18s  2 s 2
dA
 18  4 s
ds
d2A
2
 4
ds
dA
Stationary points occur when 0
ds

23. A  18s  2 s 2
dA
 18  4 s
ds
d2A
2
 4
ds
dA
Stationary points occur when 0
ds
i.e.18  4 s  0
9
s
2

24. A  18s  2 s 2
dA
 18  4 s
ds
d2A
2
 4
ds
dA
Stationary points occur when 0
ds
i.e.18  4 s  0
9
s
2
9 d2A
when s  , 2  4  0
2 ds

25. A  18s  2 s 2
dA
 18  4 s
ds
d2A
2
 4
ds
dA
Stationary points occur when 0
ds
i.e.18  4 s  0
9
s
2
9 d2A
when s  , 2  4  0
2 ds
1
 when the side length of the square is 4 metres, the area is a maximum
2

26. (ii) An open cylindrical bucket is to made so as to hold the largest
possible volume of liquid.
If the total surface area of the bucket is 12 units 2, find the height,
radius and volume.

27. (ii) An open cylindrical bucket is to made so as to hold the largest
possible volume of liquid.
If the total surface area of the bucket is 12 units 2, find the height,
radius and volume.
r
h

28. (ii) An open cylindrical bucket is to made so as to hold the largest
possible volume of liquid.
If the total surface area of the bucket is 12 units 2, find the height,
radius and volume.
r V  r 2 h1
h

29. (ii) An open cylindrical bucket is to made so as to hold the largest
possible volume of liquid.
If the total surface area of the bucket is 12 units 2, find the height,
radius and volume.
r V  r 2 h1
12  r 2  2rh 2 
h

30. (ii) An open cylindrical bucket is to made so as to hold the largest
possible volume of liquid.
If the total surface area of the bucket is 12 units 2, find the height,
radius and volume.
r V  r 2 h1
12  r 2  2rh 2 
make h the subject in 2 
h

31. (ii) An open cylindrical bucket is to made so as to hold the largest
possible volume of liquid.
If the total surface area of the bucket is 12 units 2, find the height,
radius and volume.
r V  r 2 h1
12  r 2  2rh 2 
make h the subject in 2 
h
2rh  12  r 2
12  r 2
h
2r

32. (ii) An open cylindrical bucket is to made so as to hold the largest
possible volume of liquid.
If the total surface area of the bucket is 12 units 2, find the height,
radius and volume.
r V  r 2 h1
12  r 2  2rh 2 
make h the subject in 2 
h
2rh  12  r 2
12  r 2
h
2r
substitute into 1

33. (ii) An open cylindrical bucket is to made so as to hold the largest
possible volume of liquid.
If the total surface area of the bucket is 12 units 2, find the height,
radius and volume.
r V  r 2 h1
12  r 2  2rh 2 
make h the subject in 2 
h
2rh  12  r 2
12  r 2
h
2r
substitute into 1
 12  r 2 
V  r 2  
 2r 

34. (ii) An open cylindrical bucket is to made so as to hold the largest
possible volume of liquid.
If the total surface area of the bucket is 12 units 2, find the height,
radius and volume.
r V  r 2 h1
12  r 2  2rh 2 
make h the subject in 2 
h
2rh  12  r 2
12  r 2
h
2r
substitute into 1
 12  r 2 
V  r 2  
 2r 
1
V  6r  r 3
2

35. 1
V  6r  r 3
2

36. 1
V  6r  r 3
2
dV 3
 6  r 2
dr 2

37. 1
V  6r  r 3
2
dV 3
 6  r 2
dr 2
d 2V
2
 3r
dr

38. 1
V  6r  r 3 Stationary points occur when
dV
0
2 dr
dV 3
 6  r 2
dr 2
d 2V
2
 3r
dr

39. 1
V  6r  r 3 Stationary points occur when
dV
0
2 dr
dV 3 3 2
 6  r 2 i.e. 6  r  0
dr 2 2
d 2V 3 2
2
 3r r  6
dr 2
r2  4
r  2

40. 1
V  6r  r 3 Stationary points occur when
dV
0
2 dr
dV 3 3 2
 6  r 2 i.e. 6  r  0
dr 2 2
d 2V 3 2
2
 3r r  6
dr 2
r2  4
d 2V r  2
when r  2, 2  6  0
dr

41. 1
V  6r  r 3 Stationary points occur when
dV
0
2 dr
dV 3 3 2
 6  r 2 i.e. 6  r  0
dr 2 2
d 2V 3 2
2
 3r r  6
dr 2
r2  4
d 2V r  2
when r  2, 2  6  0
dr
 when r  2, V is a maximum

42. 1
V  6r  r 3 Stationary points occur when
dV
0
2 dr
dV 3 3 2
 6  r 2 i.e. 6  r  0
dr 2 2
d 2V 3 2
2
 3r r  6
dr 2
r2  4
d 2V r  2
when r  2, 2  6  0
dr
 when r  2, V is a maximum
12  2 2
when r  2, h 
22 
2

43. 1
V  6r  r 3 Stationary points occur when
dV
0
2 dr
dV 3 3 2
 6  r 2 i.e. 6  r  0
dr 2 2
d 2V 3 2
2
 3r r  6
dr 2
r2  4
d 2V r  2
when r  2, 2  6  0
dr
 when r  2, V is a maximum
12  2 2
when r  2, h  V   2  2 
2
22 
 8
2

44. 1
V  6r  r 3 Stationary points occur when
dV
0
2 dr
dV 3 3 2
 6  r 2 i.e. 6  r  0
dr 2 2
d 2V 3 2
2
 3r r  6
dr 2
r2  4
d 2V r  2
when r  2, 2  6  0
dr
 when r  2, V is a maximum
12  2 2
when r  2, h  V   2  2 
2
22 
 8
2
 maximum volume is 8 units 3 when the radius and height are both 2 units

45. Exercise 10H; evens (not 30)
Exercise 10I; evens