Chapter 3.9, Problem 12P

Explanation of Solution

Formulation of LP:

Let,

${\text{x}}_{\text{1}}$ (100s) be the number of litters sold of chemical B

${\text{x}}_{\text{2}}$ (100s) be the number of litters sold of chemical C

${\text{x}}_{\text{3}}$ (100s) be the number of litters sold of chemical D

${\text{x}}_{\text{4}}$ (100s) be the number of litters chemical A purchased.

The profit per 100 litter on chemical B is $12.

for ${\text{x}}_{\text{1}}$ (100s) litters the profit would be 12${\text{x}}_{\text{1}}$ (100s)

Also, the profit on ${\text{x}}_{\text{2}}$ and ${\text{x}}_{\text{3}}$ litters are 16${\text{x}}_{\text{2}}$ and 26${\text{x}}_{\text{3}}$ respectively on chemical C and D.

The total selling price $=12{\text{x}}_{\text{1}}+16{\text{x}}_{\text{2}}+26{\text{x}}_{\text{3}}$

The cost per 100 litter on chemical A is $6.

If the user purchases ${\text{x}}_{\text{4}}$ (100s)litters of chemical A, then the cost would be ${\text{6x}}_{\text{4}}$.

For processing ${\text{x}}_{\text{4}}$ (100s)litters of A and ${\text{x}}_2$ litters of C, the processing cost is $3{\text{x}}_{\text{4}}+1{\text{x}}_2$

Subtracting the purchasing cost and processing cost from selling price, the actual profit is found.

Therefore, the objective function is,

Maximize, $Z=12x_1+16x_2+26x_3-(6x_4+3x_4+1x_2)$

Constraint 1:

For, 100 litters of A, 3 hours of labour is needed $\Rightarrow$

For ${\text{x}}_{\text{4}}$ litters of A, ${\text{3x}}_{\text{4}}$ hours of labour is needed.

Similarly, for 100 litters of C, 1 hour of labour is needed $\Rightarrow$

For, ${\text{x}}_2$ litters of C, ${\text{x}}_2$ hours of labour needed.

Therefore, total labour time required is ${\text{x}}_{\text{2}}+3{\text{x}}_4$

Since, only 200 labour hours are available, the constraint is $\le$ type