Question

A furniture company has 400 board ft of teak wood and can sustain up to 450 hours of labor each week. Each chair produced requires 5 ft of wood and 10 hours of labor, and each table requires 20 ft of wood and 15 hours of labor. If a chair yields a profit of $45 and a table yields a profit of $80, what are the numbers of chairs and tables that should be produced each week in order to maximize the company's profit?

Answer 1

Step-by-step explanation:

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In physics, a force is any influence that, when unopposed, will change the motion of an object. A force can cause an object with mass to change its velocity, i.e., to accelerate. Force can also be described intuitively as a push or a pull. A force has both magnitude and direction, making it a vector quantity. Wikipedia

Answer 2

Given:

Length of wooden board to produce chairs and tables = 400 ft

No. of labour hours required each week = 450

Length of board required to produce a chair = 5 ft

No. of labour hours required to produce a chair = 10

Length of board required to produce a table = 20 ft

No. of labour hours required to produce a table = 15

Profit earned from a chair = $45

Profit earned from a table = $80

To find:

No. of chairs and tables required to produce each week to maximise the profit.

Solution:

The no. of chairs and tables required to produce each week are our decision variables. Let $x$ be the no. of chairs and $y$ be the no. of tables produced in order to maximise the profit.

Here, the constraints are the maximum amount of wooden board and labour hours required for the construction of a chair and table. The company has a wooden board of length 400 ft out of which chairs and tables have to be produced each week.

∴ $5x+20y\leq 400$

Similarly, the company can sustain up to 450 hours of labour each week for chairs and tables.

∴ $10x+15y\leq 450$

The profit earned from a chair is $45 and the profit earned from a table is $80. The total profit earned from $x$ no. of chairs and $y$ no. of tables is

$P=45x+80y$, which is the objective function.

Non-negativity constraints: $x\geq 0,y\geq 0$

The mathematical formulation of this linear programming problem is:

Maximize: $P=45x+80y$

subject to: $5x+20y\leq 400$

$10x+15y\leq 450$

$x\geq 0,y\geq 0$

To calculate the optimal solution for this problem, we use the graphical method. In this method, we identify the feasible region and the corner points of this region. The graph is shown in the figure below. The corner points are A(0,20), B(24,14), C(45,0) and O(0,0). Testing these corner points on $P=45x+80y$ to determine which gives the maximum profit.

For A(0,20)

$P=45(0)+80(20)=1600$

For B(24,14)

$P=45(24)+80(14)=1080+1120=2200$

For C(45,0)

$P=45(45)+80(0)=2025$

For O(0,0)

$P=45(0)+80(0)=0$

Point B(24,14) gives the highest profit. Hence, we can say that for a maximum profit of $2200 the company has to produce 24 chairs and 14 tables each week.

The company has to produce 24 chairs and 14 tables each week in order to maximize the company's profit.