Question

The WorldLight Company produces two light fixtures (products 1 and 2) that require both metal frame parts and electrical components. Management wants to determine how many units of each product to produce so as to maximize profit. For each unit of product 1, 1 unitof frame parts and 2 units of electrical components are required. For each unit of product 2, 3 units of frame parts and 2 units of electrical components are required. The company has 200 units of frame parts and 300 units of electrical components. Each unit of product 1 gives a profit of $2, and each unit of product 2, up to 70 units, gives a profit of $4. Any excess over 60 units of product 2 brings no profit, so such an excess has been ruled out. Formulate a linear programming model for this problem. Use the graphical method to solve this model. What is the resulting total profit?

Answer 1

Answer:

Explanation:

a) x1 = number of unit product 1 to produce , and

x2 number of unit product 2 to produce

A linear program that will maximize world light profit is the following

maximize $x_1+2x_2$ subject to $x_1+3x_2\leq 200$

$2x_1+2x_2\leq 300\\\\x_2\leq 60\\\\x_1\geq 0\\\\x_2\geq 0$

Unit 1 is used both in products in 1 : 3 ratio which can be a maximum of 200 unit 2 is used in 2 : 2 ratio which can be maximum of 300

So, this can be written as the inequations

Profit functio is p = 0ne dollar on product A and two dollar on product B

= x + 2y

Now , we find a feasible area whose extremeties will give the maximum profit for, the  graph is ( see attached file )

So on the graph, we can get the other extremeties of the shaded regional so which will not give maximum profit ,

Thus , the maximum possible profit is

p = ($1 * 125) + ($2 * 25)

= $175