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Maxima & Minima
    Problems
Maxima & Minima
         Problems
When solving word problems we need to;
Maxima & Minima
          Problems
When solving word problems we need to;
(1) Reduce the problem to a set of equations
    (there will usually be two)
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
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
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
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
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.
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.
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
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
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
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        6s  2r  362
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       6s  2r  362
                     make r the subject in 2
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       6s  2r  362
                     make r the subject in 2
                          2r  36  6s
                           r  18  3s
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       6s  2r  362
                     make r the subject in 2
                          2r  36  6s
                          r  18  3s
                        substitute into 1
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       6s  2r  362
                     make r the subject in 2
                          2r  36  6s
                          r  18  3s
                        substitute into 1
                        A  s 2  18  3s s
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       6s  2r  362
                     make r the subject in 2
                          2r  36  6s
                          r  18  3s
                        substitute into 1
                        A  s 2  18  3s s
                        A  s 2  18s  3s 2
                         A  18s  2s 2
A  18s  2s 2
A  18s  2s 2
dA
    18  4s
ds
A  18s  2s 2
 dA
      18  4s
 ds
d2A
   2
      4
ds
A  18s  2s 2
 dA
      18  4s
 ds
d2A
   2
      4
ds
                                          dA
             Stationary points occur when    0
                                          ds
A  18s  2s 2
 dA
      18  4s
 ds
d2A
   2
      4
ds
                                            dA
             Stationary points occur when      0
                                            ds
                           i.e. 18  4s  0
                                      9
                                   s
                                      2
A  18s  2s 2
 dA
      18  4s
 ds
d2A
   2
      4
ds
                                            dA
             Stationary points occur when      0
                                            ds
                           i.e. 18  4s  0
                                      9
                                   s
                                      2
          9 d2A
  when s  , 2  4  0
          2 ds
A  18s  2s 2
 dA
      18  4s
 ds
d2A
   2
      4
ds
                                            dA
             Stationary points occur when      0
                                            ds
                           i.e. 18  4s  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
(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.
(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
(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
(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
(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
(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
(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
(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 
(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
1
V  6r  r 3
         2
1
 V  6r  r 3
           2
dV        3
    6  r 2
dr        2
1
  V  6r  r 3
            2
dV         3
      6  r 2
 dr        2
d 2V
   2
      3r
dr
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
1
  V  6r  r 3   Stationary points occur when
                                                 dV
                                                    0
            2                                    dr
dV         3                          3
      6  r 2             i.e. 6  r 2  0
 dr        2                          2
d 2V                                  3 2
   2
      3r                             r  6
dr                                    2
                                         r2  4
                                          r  2
1
  V  6r  r 3       Stationary points occur when
                                                     dV
                                                        0
            2                                        dr
dV         3                              3
      6  r 2                 i.e. 6  r 2  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
1
  V  6r  r 3          Stationary points occur when
                                                        dV
                                                           0
            2                                           dr
dV         3                                 3
      6  r 2                    i.e. 6  r 2  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
1
  V  6r  r 3          Stationary points occur when
                                                        dV
                                                           0
            2                                           dr
dV         3                                 3
      6  r 2                    i.e. 6  r 2  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
1
  V  6r  r 3          Stationary points occur when
                                                        dV
                                                           0
            2                                           dr
dV         3                                 3
      6  r 2                    i.e. 6  r 2  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
1
   V  6r  r 3          Stationary points occur when
                                                         dV
                                                            0
             2                                           dr
 dV         3                                 3
       6  r 2                    i.e. 6  r 2  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
Exercise 10H; evens (not 30)

    Exercise 10I; evens

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11X1 T12 06 maxima & minima (2010)

  • 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 6s  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 6s  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 6s  2r  362 make r the subject in 2 2r  36  6s 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 6s  2r  362 make r the subject in 2 2r  36  6s 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 6s  2r  362 make r the subject in 2 2r  36  6s 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 6s  2r  362 make r the subject in 2 2r  36  6s r  18  3s substitute into 1 A  s 2  18  3s s A  s 2  18s  3s 2 A  18s  2s 2
  • 19. A  18s  2s 2
  • 20. A  18s  2s 2 dA  18  4s ds
  • 21. A  18s  2s 2 dA  18  4s ds d2A 2  4 ds
  • 22. A  18s  2s 2 dA  18  4s ds d2A 2  4 ds dA Stationary points occur when 0 ds
  • 23. A  18s  2s 2 dA  18  4s ds d2A 2  4 ds dA Stationary points occur when 0 ds i.e. 18  4s  0 9 s 2
  • 24. A  18s  2s 2 dA  18  4s ds d2A 2  4 ds dA Stationary points occur when 0 ds i.e. 18  4s  0 9 s 2 9 d2A when s  , 2  4  0 2 ds
  • 25. A  18s  2s 2 dA  18  4s ds d2A 2  4 ds dA Stationary points occur when 0 ds i.e. 18  4s  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  6  r 2 i.e. 6  r 2  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  6  r 2 i.e. 6  r 2  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  6  r 2 i.e. 6  r 2  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  6  r 2 i.e. 6  r 2  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  6  r 2 i.e. 6  r 2  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  6  r 2 i.e. 6  r 2  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