contestada

FIBA manufactures two types of trucks: 1 and 2. Each truck must go through thepainting shop and assembly shop. If the painting shop were completely devoted to painting Type 1trucks, then 800 per day could be painted; if the painting shop were completely devoted to paintingType 2 trucks, then 700 per day could be painted. If the assembly shop were completely devotedto assembling truck 1 engines, then 1,500 per day could be assembled; if the assembly shop werecompletely devoted to assembling truck 2 engines, then 1,200 per day could be assembled. EachType 1 truck contributes$300to profit; each Type 2 truck contributes$500. Formulate the aboveas a linear optimization problem.

Respuesta :

MrRoyal

Formulating a linear optimization problem involves minimizing or maximizing an objective function.

The linear optimization problem is:

Max [tex]z= 300x + 500y[/tex]

Subject to:

[tex]\frac{x}{800} + \frac{y}{700} \le 1[/tex]

[tex]\frac{x}{1500} + \frac{y}{1200} \le 1[/tex]

[tex]x,y \ge 0[/tex]

Let:

[tex]x \to[/tex] Type 1

[tex]y \to[/tex] Type 2

Painting shop

We have:

[tex]x= 800[/tex]

[tex]y = 700[/tex]

The total number of type 1 and type 2 trucks that can be painted is not given. This means that; we represent this constraint using proportions.

So, we have:

[tex]\frac{x}{800} + \frac{y}{700} \le 1[/tex]

Assembly shop

We have:

[tex]x =1500[/tex]

[tex]y = 1200[/tex]

Using proportions, we have:

[tex]\frac{x}{1500} + \frac{y}{1200} \le 1[/tex]

Profit contributed

We have:

[tex]x = 300[/tex]

[tex]y = 500[/tex]

So, the objective function is:

[tex]z= 300x + 500y[/tex]

Hence, the linear optimization problem is:

Max [tex]z= 300x + 500y[/tex]

Subject to:

[tex]\frac{x}{800} + \frac{y}{700} \le 1[/tex]

[tex]\frac{x}{1500} + \frac{y}{1200} \le 1[/tex]

[tex]x,y \ge 0[/tex]

Read more about linear optimization problems at:

https://brainly.com/question/16103572