Answer to Question #283428 in Operations Research for Lucky

Question

The furniture company inexpensive tables and chairs. The production process of each is similar in the painting department each table take 4 hours of carpentry and 2 hours in the painting department. Each chair requires 3 hours of carpentry and 1 hour painting department. During the current product period 240 hours of carpentry time are available and 100 hrs in the painting is available. Each table sold yields of profit of $7 and chair produced sold for $5 profit. Find the best combination of table and chairs to manufacture in order to reach the maximum number of profit? 

Answer 1

The company wants to determine the best combination of tables and chairs to produce to reach the maximum profit.

The objective is to Maximize profit.

The constraints are

1. The hours of carpentry time used cannot exceed 240 hours per week.

2. The hours of painting and varnishing time used cannot exceed 100 hours per week.

The decision variables representing the actual decisions we will make are

$x=$ number of tables to be produced per week

$y =$ number of chairs to be produced per week

We create the LP objective function in terms of $x$ and $y$

Maximize profit $=\$7x+\$5y$

Develop mathematical relationships for the two constraints

For carpentry

$$4x+3y\leq240$$

For painting and varnishing

$$2x+y\leq100$$

Both of these constraints restrict production capacity and affect total profit.

The values for $x$ and $y$ must be nonnegative

$$x\geq0, y\geq 0$$

Maximize profit $z=7x+5y$

subject to

$$4x+3y\leq240$$ $$2x+y\leq100$$ $$x\geq0, y\geq 0$$

Graphical solution

Point $A(0, 80)$

$$z=7(0)+5(80)=400$$

Point $B(30, 40)$

$$z=7(30)+5(40)=410$$

Point $C(50, 0)$

$$z=7(50)+5(0)=350$$

Point $O(0, 0)$

$$z=7(0)+5(0)=0$$

We should manufacture 30 tables and 40 chairs to reach the maximum profit with value of $\$410.$