This document discusses how to solve maxima and minima word problems. It explains that problems should be reduced to two equations, one for the quantity being maximized/minimized and one for given information. The equation for maximizing/minimizing should be rewritten with one variable. Calculus is then used to solve the problem by finding the derivative and setting it equal to zero to find critical points. An example problem is included where the maximum area of two shapes made from a rope is found.

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
s s
r

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
s s
r
 12
rssA 

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
s s
r
 12
rssA   23626  rs

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
s s
r
 12
rssA   23626  rs
 2insubjectthemake r

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
s s
r
 12
rssA   23626  rs
 2insubjectthemake r
sr
sr
318
6362



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
s s
r
 12
rssA   23626  rs
 2insubjectthemake r
sr
sr
318
6362


 1intosubstitute

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
s s
r
 12
rssA   23626  rs
 2insubjectthemake r
sr
sr
318
6362


 1intosubstitute
 sssA 3182


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
s s
r
 12
rssA   23626  rs
 2insubjectthemake r
sr
sr
318
6362


 1intosubstitute
 sssA 3182

2
22
218
318
ssA
sssA



19. 2
218 ssA 

20. 2
218 ssA 
s
ds
dA
418

21. 2
218 ssA 
s
ds
dA
418
42
2

ds
Ad

22. 2
218 ssA 
s
ds
dA
418
42
2

ds
Ad
0occur whenpointsStationary 
ds
dA

23. 2
218 ssA 
s
ds
dA
418
42
2

ds
Ad
0occur whenpointsStationary 
ds
dA
2
9
0418i.e.


s
s

24. 2
218 ssA 
s
ds
dA
418
42
2

ds
Ad
0occur whenpointsStationary 
ds
dA
2
9
0418i.e.


s
s
04,
2
9
when 2
2

ds
Ad
s

25. 2
218 ssA 
s
ds
dA
418
42
2

ds
Ad
0occur whenpointsStationary 
ds
dA
2
9
0418i.e.


s
s
04,
2
9
when 2
2

ds
Ad
s
maximumaisareathemetres,
2
1
4issquaretheoflengthsidewhen the

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 , find the height,
radius and volume.
2
units12

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 , find the height,
radius and volume.
2
units12
h
r

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 , find the height,
radius and volume.
 12
hrV 
2
units12
h
r

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 , find the height,
radius and volume.
 12
hrV 
 2212 2
rhr  
2
units12
h
r

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 , find the height,
radius and volume.
 12
hrV 
 2212 2
rhr  
 2insubjectthemake h
2
units12
h
r

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 , find the height,
radius and volume.
 12
hrV 
 2212 2
rhr  
 2insubjectthemake h
r
r
h
rrh
2
12
122
2
2


 
2
units12
h
r

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 , find the height,
radius and volume.
 12
hrV 
 2212 2
rhr  
 2insubjectthemake h
r
r
h
rrh
2
12
122
2
2


 
 1intosubstitute
2
units12
h
r

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 , find the height,
radius and volume.
 12
hrV 
 2212 2
rhr  
 2insubjectthemake h
r
r
h
rrh
2
12
122
2
2


 
 1intosubstitute





 

r
r
rV
2
12 2
2

2
units12
h
r

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 , find the height,
radius and volume.
 12
hrV 
 2212 2
rhr  
 2insubjectthemake h
r
r
h
rrh
2
12
122
2
2


 
 1intosubstitute





 

r
r
rV
2
12 2
2

2
units12
h
r
3
2
1
6 rrV  

35. 3
2
1
6 rrV  

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

37. 2
2
3
6 r
dr
dV
 
r
dr
Vd
32
2

3
2
1
6 rrV  

38. 2
2
3
6 r
dr
dV
 
r
dr
Vd
32
2

0occur whenpointsStationary 
dr
dV3
2
1
6 rrV  

39. 2
2
3
6 r
dr
dV
 
r
dr
Vd
32
2

0occur whenpointsStationary 
dr
dV
2
4
6
2
3
0
2
3
6i.e.
2
2
2




r
r
r
r


3
2
1
6 rrV  

40. 2
2
3
6 r
dr
dV
 
r
dr
Vd
32
2

0occur whenpointsStationary 
dr
dV
2
4
6
2
3
0
2
3
6i.e.
2
2
2




r
r
r
r


06,2when 2
2
 
dr
Vd
r
3
2
1
6 rrV  

41. 2
2
3
6 r
dr
dV
 
r
dr
Vd
32
2

0occur whenpointsStationary 
dr
dV
2
4
6
2
3
0
2
3
6i.e.
2
2
2




r
r
r
r


06,2when 2
2
 
dr
Vd
r
maximumais,2when Vr 
3
2
1
6 rrV  

42. 2
2
3
6 r
dr
dV
 
r
dr
Vd
32
2

0occur whenpointsStationary 
dr
dV
2
4
6
2
3
0
2
3
6i.e.
2
2
2




r
r
r
r


06,2when 2
2
 
dr
Vd
r
maximumais,2when Vr 
3
2
1
6 rrV  
 
2
22
212
,2when
2


 hr

43. 2
2
3
6 r
dr
dV
 
r
dr
Vd
32
2

0occur whenpointsStationary 
dr
dV
2
4
6
2
3
0
2
3
6i.e.
2
2
2




r
r
r
r


06,2when 2
2
 
dr
Vd
r
maximumais,2when Vr 
3
2
1
6 rrV  
 
2
22
212
,2when
2


 hr    


8
22
2

V

44. 2
2
3
6 r
dr
dV
 
r
dr
Vd
32
2

0occur whenpointsStationary 
dr
dV
2
4
6
2
3
0
2
3
6i.e.
2
2
2




r
r
r
r


06,2when 2
2
 
dr
Vd
r
maximumais,2when Vr 
3
2
1
6 rrV  
 
2
22
212
,2when
2


 hr    


8
22
2

V
units2bothareheightandradiuswhen theunits8isvolumemaximum 3


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