Question

A craftsman produces two products: floor lamps and table lamps. Production of one floor lamp
requires 70 minutes of his labor and materials that cost P210. Production of one table lamp
requires 50 minutes of labor and the materials cost P 190. The craftsman wishes to work no
more than 40 hours each week, and his financial resources allows him to pay no more than
P 8000 for materials each week. He can sell as many lamps as he can make and his profit is
P 395 per floor lamp and P 345 per table lamp.
1. Identify the variables.
2. Set up the objective function.
3. Give the constraints in mathematical expression.
4. Graph the constraints and identify the solution.
5. How many units of each product should be produced weekly in order to maximize the
craftsman’s profit?

Answer 1

Answer:

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Answer 2

Given:

Amount of time required to make one floor lamp = 70 minutes

Cost required to purchase materials for floor lamp = P 210

Amount of time required to make one table lamp = 50 minutes

Cost required to purchase materials for table lamp = P 190

Amount of time craftsman can work for a week = 40 hours

Cost required to purchase materials for both lamps = P 8000

Profit gained per floor lamp = P 395

Profit gained per table lamp = P 345

To find:

1. Variables.

2. Objective function.

3. Constraints in mathematical expression.

4. Graph the constraints.

5. Identify solution.

6. Units of each product to be produced weekly for maximum profit.

Solution:

1. Those variables which are changeable and are impactful are called decision variables.

Here, the variables are the no. of floor lamps and table lamps that can be produced weekly. The craftsman can produce $x$ no. of floor lamps and $y$ no. of table lamps weekly.

2. Objective function is the linear function of the objective. Here, the objective is to maximise profit. Profit gained per floor lamp is P 395 and profit gained per table lamp is P 345.

$P=395x+345y$ is the objective function.

3. Constraints mean any kind of limitation.

Here, the constraints are the amount of time and the purchase cost of materials for the craftsman to produce the lamps. The craftsman cannot work more than 40 hours per week, i.e time should 40 hours or lesser than 40. Here, total time (40) is given in hours whereas individual time is given in minutes. Hence, we convert hours into minutes.

∴ $70x+50y\leq 2400$                                                  $(1h=60min\\40h=40*60=2400 min)$

Similarly, he cannot purchase for materials more than P 8000, i.e., cost should be 8000 or less than that.

∴ $210x+190y\leq 8000$

Hence, the mathematical formulation of this problem is:

Maximize, $P=395x+345y$

Subject to: $70x+50y\leq 2400$

$210x+190y\leq 8000$

$x\geq 0,y\geq 0$

4. The constraints are graphed as shown in the figure below and we obtain four corner points: A(0,42), B(20,20), C(34,0) and O(0,0). Testing these corner points on $P=395x+345y$ to check which gives maximum profit.

For A(0,42)

$x=0,y=42$

$P=395(0)+345(42)=14490$

For B(20,20)

$x=20,y=20$

$P=395(20)+345(20)=7900 + 6900=14800$

For C(34,0)

$x=34,y=0$

$P=395(34)+345(0)=13430$

For O(0,0)

$x=0,y=0$

$P=395(0)+345(0)=0$

5. Point B (20,20) produces the highest profit and hence, we can conclude that on producing 20 floor lamps and 20 table lamps weekly, the craftsman will yield a maximum profit of P 14800.

The variables are no. of floor lamps $(x)$ and no. of table lamps $(y)$ that can be produced weekly.

The objective function is $P=395x+345y$.

The constraints in mathematical expression are $70x+50y\leq 2400$, $210x+190y\leq 8000$, $x\geq 0,y\geq 0$

On producing 20 floor lamps and 20 table lamps weekly, the craftsman will yield a maximum profit of P 14800.