**Maximize z = 5x _{1} + 9x_{2}**

subject to

-x_{1} + 5x_{2} ≤ 3

5x_{1} + 3x_{2} ≤ 27

x_{1}, x_{2} ≥ 0

Solution.

Given

-x_{1} + 5x_{2} ≤ 3
...........(i)

5x_{1} + 3x_{2} ≤ 27 ...........(ii)

Let R_{1} & R_{2} be the resources associated with
first and second constraint respectively.

The maximum value of the resources are specified in the RHS of the two
constraints, i.e., R_{1}= 3 & R_{2} = 27.

From equation (i), if we are deciding only on x_{2} and RHS
is R1, then 5x_{2} has to be less than or equal to R_{1},
i.e., x_{2} ≤ R_{1}/5.

Similarly, from equation (ii), we have

X2 ≤ R_{2}/3.

Since we are maximizing, the maximum value of x_{2} has to
be equal to the minimum of R_{1}/5 and R_{2}/3.

ƒ_{2}(R_{1}, R_{2})
= Max (9x_{2})

= 9 Max (x_{2})

= 9 Min (R_{1}/5, R_{2}/3) -------(iii)

ƒ_{1}(3, 27) = Max [5x_{1} + ƒ_{2}(3 + x_{1}, 27
– 5x_{1})] ------(iv)

0 ≤ x_{1} ≤ 27/5

From equation (iii),

ƒ_{2}(R_{1},R_{2})
= 9 Min (R_{1}/5, R_{2}/3) ----(iv)

Therefore, ƒ_{2}(3 + x_{1},
27 – 5x_{1}) = 9 Min [(3 + x_{1})/5, (27 –
5x_{1})/ 3]

From equation (iv)

ƒ_{1}(3, 27) = Max {5x_{1} + 9 Min[ (3 + x_{1})/5, (27 - 5x_{1})/3] } -----(v)

We now find the range of x_{1} for which (3 + x_{1})/5
< (27 – 5x_{1})/3.

Comparing (3 + x_{1})/5 & (27 – 5x_{1})/3,
we get

(3 + x |
= |
(27 – 5x |

3(3 + x_{1}) = 5(27 – 5x_{1})

9 + 3x_{1} = 135 – 25x_{1
}**∴** x_{1} = 4.5

From equation v), we have

ƒ_{1}(3,27) = Max [5x_{1} + 9(3 + x_{1})/5] if x_{1} ≤ 4.5

= Max [5x_{1 }+ 9(27 - 5x_{1})/3] if x_{1} >
4.5

From the above, the maximum occurs at x_{1} = 4.5

x_{2} = Min [3 + 4.5)/5 ,(27 – 5 X 4.5)/3]

= Min (7.5/5, 4.5/3) = 1.5

Hence, the optimal solution is

x_{1} = 4.5, x_{2} = 1.5

z = 5 X 4.5 + 9 X 1.5 = 22.5 + 13.5 = 36

**Maximize Z = 3x _{1} + 5x_{2}**

subject to

x_{1} ≤ 4

x_{2} ≤ 6

3x_{1} + 2x_{2} ≤ 18

x_{1}, x_{2} ≥ 0

Solution.

x_{1} ≤ 4 .........(i)

x_{2} ≤ 6 .........(ii)

3x_{1} + 2x_{2} ≤ 18 .........(iii)

Let R_{1}, R_{2} & R_{3 }be the resources
associated with first, second and third constraint respectively.

The maximum value of the resources are specified in the RHS of the two
constraints, i.e., R_{1}= 4, R_{2} = 6 & R_{3} = 18.

From equation (ii), if we are deciding only on x2 and RHS is R2 then
x_{2} has to be less than equal to R_{2}, i.e., x_{2} ≤ R_{2}.

Similarly, from equation (iii), we have

2x2 ≤ R_{3}

or x2 ≤ R_{3}/2

Since we are maximizing, the maximum value of x_{2} has to
be equal to the minimum of R_{2} & R_{3}/2.

ƒ_{2}(R_{1},R_{2},
R_{3}) = Max (5x_{2})

= 5 Max (x_{2})

= 5 Min (R_{2}, R_{3}/2) ...........(iv)

ƒ_{1}(4,6,18) = Max [3x_{1} + ƒ2_{}(4 - 2x_{1},
6, 18 - 3x_{1})]

x_{2} ≤ 2, x_{1} ≤ 6, x_{1 } ≥ 0

From equation (iv) & (v)

ƒ1_{}(R_{1},R_{2},
R_{3}) = Max [3x_{1}+ 5 Min (6, (18 - 3x_{1})/2]
...........(vi)

We now find the range of x_{1} for which

6 < (18 - 3x_{1})/2

Comparing 6 & (18 - 3x_{1})/2, we get

12 = 18 - 3x_{1
} or 3x_{1} = 6

or x_{1} = 2

From equation (vi), we have:

ƒ_{1}(4, 6, 18) = Max [3x_{1} + 5 X 6] if x ≤ 2

= Max [3x_{1 }+ 5(18 - 3x_{1})/2] if x_{1} >
2

From the above, the maximum occurs at x_{1} = 2.

x_{2 }= Min [6, (18 - 3 X 2)/2] = 6

The values for x_{1} and x_{2} are 2 and 6. The corresponding
value of the objective function is

Z = 3 X 2 + 5 X 6 = 36