Truckco manufactures two types of trucks: 1 and 2. Each truck must go through the painting shop and assembly shop. If the painting shop were completely devoted to painting Type 1 trucks, then 800 per day could be painted; if the painting shop were completely devoted to painting Type 2 trucks, then 700 per day could be painted. If the assembly shop were completely devoted to assembling truck 1 engines, then 1,500 per day could be assembled; if the assembly shop were completely devoted to assembling truck 2 engines, then 1,200 per day could be assembled. Each Type 1 truck contributes $300 to profit; each Type 2 truck contributes $500. formulate an LP that will maximize Truckco’s profit.

Answer:

Step-by-step explanation:

Let y represent type 2 truck and x represent type 1 truck. The painting can be represented by the inequality:

[tex]\frac{x}{800}+\frac{y}{700}\leq 1[/tex]                  (1)

The assembling can be represented by the inequality:

[tex]\frac{x}{1500}+\frac{y}{1200} \leq 1[/tex]                (2)

Also, x, y ≥ 0

The objective function is given as:

max z = 300x + 500y

Plotting 1 and 2 using online geogebra

From the graph, the optimal value is at (0,700). Hence:

max z = 300(0) + 500(700) = $350000

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