Chapter 3.9, Problem 6P

Explanation of Solution

Given:

Let $x_{ij}$ be the amount of drugs with chemical of type “i” that has the drug type “j” for $i=1,2$ and $j=1,2$

Let $P_1$ be the process 1 and $P_2$ be the process 2.

Objective function:

$\begin{array}{l}\text{Profit=}\left[\begin{array}{l}\text{(cost of drug 1)}-\text{(amount of drug 1 produced by blending 2 chemicals)}\\ \text{+(cost of drug 2)}-\text{(amount of drug 2 produced by blending 2 chemicals)}\end{array}\right]\\ =6(x_{11}+x_{21})+4(x_{12}+x_{22})\end{array}$

$\text{Maximize }\text{z=}6(x_{11}+x_{21})+4(x_{12}+x_{22})$

Considering the constraints,

Constraint 1: Drug 1 must contain at least 65% of chemical 1.

Constraint 1: Drug 2 must contain at least 55% of chemical 1

Constraint 3: At most 100 oz of raw material available.

Constraint 4: At most 120 hours of skilled labor available.

Constraint 5: Chemical 1 used should be less than amount of chemical 1 produced.

Constraint 5: Chemical 2 used should be less than amount of chemical 2 produced.

Expressing the constraint 1 in terms of variables:

$\begin{array}{l}\text{Chemical 1 in drug 1}\ge 65\%\\ x_{11}\ge 0.65(x_{11}+x_{21})\end{array}$

Expressing the constraint 2 in terms of variables:

$\begin{array}{l}\text{Chemical 1 in drug 2}\ge 55\%\\ x_{12}\ge 0.55(x_{12}+x_{22})\end{array}$

Expressing the constraint 3 in terms of variables:

$\begin{array}{l}\left[\begin{array}{l}\text{requirement of raw material in process 1}\\ \text{+requirement of raw material in process 2}\end{array}\right]\le 100\\ 3P_1+2P_2\le 100\end{array}$

Expressing the constraint 4 in terms of variables:

$\begin{array}{l}\left[\begin{array}{l}\text{requirement of skilled labor in process 1}\\ \text{+requirement of skilled labor in process 2}\end{array}\right]\le 120\\ \end{array}$