Question

players. Trevor runs a factory that makes Blu-ray players. Each R80 takes 6 ounces of plastic and 2 ounces of metal. Each G150 requires 2 ounces of plastic and 6 ounces of metal. The factory has 156 ounces of plastic, 276 ounces of metal available, with a maximum of 16 R80 that can be built each week. If each R80 generates $13 in profit, and each G150 generates $14, how many of each of the Blu-ray players should Trevor have the factory make each week to make the most profit? requires 2 ounceso R80: G150: Best profit: Get help: Video

Answer 1

Solution

[NOTE:

Answers are given below. Detailed Working and Back-up Theory follow at the end.]

R80: 12 units, Answer 1

G150: 42 units, Answer 2

Best profit: $744 Answer 3

Back-up Theory

The solution is obtained by graphical method. The optimum solution occurs at one of the corner points of the convex polygon representing the feasible region.

Now to work out the solution,

Formulation

Let x = number of R80 and y = number of G150.

Then, the LPP is:

Maximize Z = 13x + 14y [profit function]

Subject to,

6x + 2y ≤ 156 [Plastic constraint]

2x + 6y ≤ 156 [Metal constraint]

x ≤ 16 [Demand constraint]

x, y ≥ 0 [Non-negativity constraint]

Solution by graphical method

Treating each of the above inequalities as equations, each representing a straight line, the feasible region is found to be the convex polygon, OPQRSO, where O = (0, 0), P = (0, 46), Q = (12,42),

R = (16, 30), S = (16, 0).

The value of Z at each of these corner points is:

Z_O = 0, Z_P = 644, Z_Q = 744, Z_R = 628, Z_S = 208.

Since Z_Q = 744 is maximum, the optimum solution is: x = 12, y = 42, Z = 744. ANSWER

DONE