An agricultural company has 80 tons of Type I fertilizer and 120 tons of Type II fertilizer. The company mixes both types of fertilizer into 3 raw material blends. Blend X requires 3 parts of Type I and 1 part of Type II. Blend Y requires 1 part of Type I and 4 parts of Type II. Blend Z requires half a part of Type I and 2 parts of Type II. Blend X is sold at $2000 per ton, Blend Y at $1500 per ton, and Blend Z is sold at $2500 per ton. Use the following decision variables:

x: number of tons to mix of Blend X y: number of tons to mix of Blend Y z: number of tons to mix of Blend Z

to formulate the company's mathematical optimization problem.

A. max 2000x + 1500y + 2500z subject to: 3x + y + 0.5z <= 80; x + 4y + 2z <= 120; x, y, z >= 0; x, y, z are integers

B. max 2000x + 1500y + 2500z subject to: 3x + y + 0.5z <= 120; x + 4y + 2z <= 80; x, y, z >= 0; x, y, z are integers

C. max 2000x + 1500y + 2500z subject to: 3x + y + 0.5z <= 80; x + 4y + 2z <= 120; x, y, z >= 0

D. max 2000x + 1500y + 2500z subject to: 3x + y + 0.5z <= 120; x + 4y + 2z <= 80; x, y, z >= 0

Answer rating: 100% (QA)

A max 200... View the full answer