Chapter 3.9, Problem 2P

Explanation of Solution

Given:

Let $x_1$ be the number of finished tables, $x_b$ be the number of unfinished tables.

Let $y_1$ be the number of finished chairs, $y_2$ be the number of unfinished chairs.

Objective function:

$\begin{array}{l}\text{Profit=Revenue-Cost}\\ \text{=}\left[\begin{array}{l}\text{(revenue of finished table)}-\text{(cost/ft of finished table)}\\ \text{+(revenue of unfinished table)}-\text{(cost/ft of unfinished table)}\\ \text{+(revenue of finished chair)}-\text{(cost/ft of finished chair)}\\ \text{+(revenue of unfinished chair)}-\text{(cost/ft of unfinished chair)}\end{array}\right]\end{array}$

  $\begin{array}{l}=(140-40)x_1+(70-40)x_2+(110-30)y_1+(60-30)y_2\\ =100x_1+30x_2+80y_1+30y_2\end{array}$

$\text{Maximize }{\text{z=100x}}_{\text{1}}{\text{+30x}}_{\text{2}}{\text{+80y}}_{\text{1}}{\text{+30y}}_{\text{2}}$

Considering the constraints,

Constraint 1 and 2: At most, 4000 woods are available for purchase.

Constraint 3: At most, 6000 hours of labor are available to manufacture

Expressing the constraint 1 in terms of variables:

$\begin{array}{l}\left[\begin{array}{l}\text{(40 ft of wood required to manufacture table)+}\\ \text{(30 ft of wood required to manufacture table)}\end{array}\right]\le 40000\\ 40(x_1+x_2)+30(y_1+y_2)\le 40000\end{array}$

Expressing